11. Parametric Hypothesis Testing (cont.) and Testing Goodness of Fit

The next content material is offered below a ingenious Commons license your support will support MIT OpenCourseWare proceed to offer high first-class academic assets without spending a dime to make a donation or to view further substances from hundreds and hundreds of MIT guides discuss with MIT opencourseware at ocw.Mit.Edu so we’ve got been speakme about this chi-square scan and the title ok square comes from the fact that we build a test statistic that has asymptotic distribution given by the chi-squared distribution let’s just supply it yet another shot adequate and this exams who has truly ever encountered the chi-rectangular experiment external of a stats rapid room all right so some individuals have write it’s a quite long-established test that you just could encounter and it used to be virtually to experiment if given some knowledge with a fixed probability mass function proper so discrete distribution you desired to test if it was once given the PMF used to be equal to a set value P naught or if it was different from P naught and the way the chi-rectangular arose right here was once by means of watching at waltz scan and very nearly in case you write a wealth is the person who has the chi-rectangular because the limiting distribution and for those who invert the covariance matrix the asymptotic covariance matrix so you compute the Fisher understanding which on this special case does now not exist for the Milton multinomial distribution but we determined the trick on how to do this we eliminate the part that forbid it to be invertible then we found this chi-rectangular distribution so in a way we now have this test statistic which you might have realized as a black box laundry list however going via the math which would were rather disagreeable I renowned but really informed you why you must do that unique normalization so in view that some of you requested a bit of more you know practical examples of how these matters work let me show you a couple the first one is the you want to answer the question well you already know when must I be born to be positive right and some men and women think in zodiac and so Fortune journal surely accumulated the signs of 256 residences of the fortune 500 correct those were taken randomly and so they had been amassed there and you can find this this the depend of quantity of CEOs which have a distinct zodiac sign and if this was once fully uniformly distributed you will have to truly get a quantity that is round 256 divided by means of 12 which in this case is 21 factor 33 okay you can see that there is numbers that you understand are traditionally in the vicinity but see appear at this guy that is correct that’s 29 so whose piece is here all proper and all right so supply me your know-how and we’ll meet again in 10 years and so sincerely you might need to experiment if certainly the fact that it is uniformly dispensed is a legitimate assumption now this is obviously a random variable I select a random CEO and that i measure what is zodiac sign is and i need to be aware of if this so it can be a mostly over I do not know 12 zodiac indicators and that i need to be aware of if it can be uniform or no longer correct uniform sounds like it should be the fame quo if you are affordable and and maybe there is really something that moves away so you understand we might do that in view of this data is there evidence that one is distinctive this is one other illustration where you would need to apply the chi-square test all right in order I said the benchmark distribution was once the uniform distribution for the zodiac solar and that is commonly the one I gave you 1 over ok 1 over okay in view that good that is form of the zero you recognize the principal point for all distributions proper that’s the point that’s proper at the core of what we name the simplex but that you may have a different benchmark that style of is sensible proper so for example that is an exact data set have been 200 and fifty and seventy-5 jurors have been recognized with identified racial staff have been collected and you sincerely would wish to know if juries on this nation are virtually representative of the genuine population and so here of course the population is not uniformly dispensed in keeping with racial team and the best way you actually do it is you truly go on Wikipedia for example and you appear on the demographic the U.S. And you in finding that the share of white is seventy two% like 7% Hispanic is 12 and other is good about 9% so that’s a whole of 1 and that is what we truely measured for some jurors so for this man that you could truly run the chi-square scan correct you have the estimated share which comes from this primary line you have the tested proportion p0 that includes the 2nd line and also you would wish to investigate if these matters truly correspond to one another k so i am not gonna do it for you but I variety of invite you to do it and experiment and notice how this compares to the quantiles of the proper chi-rectangular distribution and spot what you can conclude from these to fix all right so this used to be the multinomial k so this is essentially what we did we computed the MLE they usually’re the correct constraint and that was our test statistic that converges to the chi-square distribution so should you’ve obvious it before that’s all that was given to you now we all know why the normalization here is P naught J and not P no longer J squared or square root of P now J and even 1 right I imply it’s not clear that this must be the right normalization but we know that is what comes from taking the correct normalization which comes from the Fisher know-how all proper adequate so the item I desired to move on to so we now have essentially included okay squared test are there any questions about ok quiz there’s ok squared scan and for these of you who were not right here on Thursday i am particularly simply I don’t pretend I just did it correct that’s something we did final Thursday all proper the are there any query that arose when you had been studying your notes things that you simply failed to comprehend sure so so it is it is now not so it is no longer the Fisher and the Fisher know-how does now not exist on this case and so there is no proper name for this is the pseudo-inverse of the artificial revenge matrix and that’s what it’s and there is I don’t know if I acknowledged it last time however there’s this complete discipline that uses you understand for persons who are relatively aspire to differential geometry but are caught in it can be that is division and there may be this factor known as information geometry which is basically learning the manifolds related to the Fisher knowledge matrix and the the metric that is related to Fisher knowledge proper and so these of course may also be decrease dimensional manifolds no longer handiest distorts the geometry however forces the whole thing to survive the shrink dimension which is what happens whilst you Fisher knowledge does not exist and so you realize there may be a bunch of things that you can be trained what this manifold appears like and many others however no there is no designated terminology right here about about going right here to be fair inside the scope of this type this is the one case the place you Milton you multinomial case is the one case where you most of the time see a lack of Fisher know-how matrix and that is just given that we’ve this further constraints that the one of the crucial parameter should be one and when you have an additional constraint that appears like it’s actually removed one degree of freedom this may occur inevitably and so maybe what you are able to do is restore match Reaper beginner eyes so if I honestly repair American formed in function of P 1 to P ok minus one after which one minus the sum this do not have occur because i have handiest a okay-dimensional space so there may be tricks round this to make it exist if you want it to exist another question all right so let’s transfer on to students T checks we recounted that last time correct so almost you’ve ordinarily achieved it more even in homework than you’ve gotten achieved it in in lectures but simply swiftly that is well-nigh the experiment that’s the test when now we have an exact knowledge that comes from a traditional distribution there’s no vital limit theorem that exists this is rather to account for the fact that for smaller pattern sizes it might be the case it can be no longer precisely real that after I seem at xn bar minus mu / so if I look at xn bar minus mu divided by Sigma times square root of n then this thing should have n zero 1 distribution roughly correct by the vital restrict theorem so that is for n huge but when any small then it’s still authentic when the information is n nu Sigma squared then it’s proper that square root of n so right here it can be approximately and that is invariably proper however I don’t know Sigma in practice right maybe nu it comes from my perhaps nu comes from my new knot right perhaps some thing from the the scan statistic where mu honestly is right here however for this guy i’m gonna have inevitably to search out an estimator and now in this case for small n that is no longer authentic and what the T statistic is doing is very nearly telling you what the distribution of this guy is all correct so the real what you will have to say is that now this guy has a T distribution with n minus 1 degrees of freedom that is clearly the laundry record stats that you would learn it says simply look at one more desk that is what it is however we truly outlined was eight what a T distribution was once and a T distribution is essentially some thing that has the equal distribution as some n zero 1 divided by using the square root of a chi-squared with d levels of freedom divided by way of d and that’s a T distribution with levels of freedom correct and those two need to be unbiased k and so what I have to check is that this man over there’s of this way k so let’s seem on the numerator good square root of n xn bar minus mu what’s the distribution of this factor is it an N 0 1 & 0 Sigma squared proper so i am not going to put it right here so if i want this guy to be + 0 1 I have got to divide with the aid of Sigma that’s what we’ve over there so that’s my n zero 1 that’s gonna play the function of this man right here so if I wish to go a bit additional I have got to simply say k now I need to have square root of N and that i have to find whatever right here that looks like my case square root of Chi square divided by ya no that’s just the in distribution so I do not know that’s let’s just write it like that if you need I imply yeah that is now not quite appropriate do you have most often you write only one distribution on the proper hand aspect of this little factor so no longer just this tricky perform of distributions I mean this is more like to provide an explanation for ok and so most often you consider you should say is that T is equal to this X divided by means of rectangular root of Z divided by D the place X has common distribution Z has chi-squared distribution with the levels of freedom sir okay so what do i want right here good I have to have something which appears like my Sigma hat correct so in some way inevitably i will have got to have Sigma hat now of course I need to divide this by using my Sigma so that my Sigma goes away and so now this thing here sorry I should move on to the right okay and so this thing right here so Sigma hat is is square root of SN and now i’m practically there so this factor is truely equal to square root of N proper however this factor here is surely not a so this factor right here follows a distribution which is in reality a chi-square rectangular root of a chi-squared distribution divided by way of and yeah that’s the square root k square distribution with n minus 1 degrees of freedom divided via n considering Sigma hat is equal to 1 over n sum from I equal 1 to N X I minus X bar squared and we simply stated that this section right here used to be a chi-squared disbursed we did not simply say it we stated it a few lectures again that this thing with the chi-square distribution and the truth that the presence of this xn bar X X bar right here was truly taking away one measure of freedom from this sum ok so this man here has the equal distribution because the Chi rectangular n minus 1 divided with the aid of n so now I need to truly still arrange this thing a bit of bit to have a T distribution I should no longer see n here however I must see n minus 1 correct the D right here is the identical as this D right here and so let me make the correction so that this clearly occurs well if I actually write this to be equal to so if I write square root of n minus 1 as on the slide times xn bar minus mu divided by way of good let me write it as rectangular root of SN which is my Sigma hat then what this factor is truly equal to good it’s it can be it follows a n 0 1 divided via the rectangular root of my chi-square distribution with n minus 1 degrees of freedom and here the fact that I multiplied by using rectangular root of n minus 1 and i’ve the rectangular root of n here is almost the same as dividing right here by means of n minus 1 proper and that’s my TN distribution my t-distribution with n minus 1 levels of freedom simply by definition of what this factor is k all correct yes this guy i am sorry that is Sigma rectangular thank you that’s simply that’s the estimator of the variance not the estimator of the ordinary deviation when I wish to divide I divide through average deviation thank you some other query or remark the estimator for the variance oh sure you are right so there is a Sigma rectangular here that what you are asking yes yeah certainly and that’s the place it get cancels right here it will get cancelled right here k so that is rather a Sigma squared occasions ok square okay so the fact that it can be Sigma rectangular is solely in view that i can pull out Sigma squared and just make those guys in zero one correct all right so that’s my T distribution now that I genuinely have a pivotal distribution what I do is that I type the statistic here I known as it TN tilde ok and what is this thing i do know that this has a pivotal distribution so for instance i do know that the chance that T and tilde in absolute worth exceeds some number that i am gonna name Q alpha over 2 for the TN minus 1 is the same as alpha proper so that is sincerely recall the T distribution has the equal shape as the Gaussian distribution what i am discovering is for this T distribution some quantity Q alpha over 2 of TN minus 1 and minus Q alpha over 2 of TN minus 1 so those are distinctive from the Gaussian one such that the field beneath the curve right here is alpha over 2 on each and every part in order that the likelihood that my absolute price exceeds this number is equal to alpha and that is what i’ll use to reject the scan correct so now my scan turn out to be for a 0 say mu is the same as some new zero versus H 1 mu isn’t equal to MU zero is well the rejection vicinity is going to be equal to the set on which square root of n minus 1 instances xn bar minus mu 0 this time divided by using rectangular root of SN exceeds in absolute worth exceeds Q oh sorry there may be already here exceeds Q alpha over 2 of TN minus 1 k so I reject when this thing raises the same as the Gowen case besides that rather than reading my quintiles from the Gaussian desk I breed them from the pupil desk it’s just the same thing so they’re just gonna be a little bit bit farther proper so this guy here is just gonna be a bit greater than the one for the Gaussian one for the reason that it can be gonna require me a little more proof in my information to be capable to reject because I ought to account for the fluctuations of Sigma hat ok so of direction pupils scan is used in all places men and women use handiest Chi assessments correct that is essentially for those who seem at any knowledge factor any output despite the fact that you had 500 observations when you seem at the statistical application output is gonna say t-experiment and the purpose why you see t-test is seeing that so that by hook or by crook it’s felt like it’s now not asymptotic you do not need to genuinely are you aware to be in particular cautious and anyway if n is equal to five 1000’s is the two curves are above each other it’s basically the same thing so it doesn’t rather trade something so why not using the t-scan so it can be now not asymptotic it does not require vital material to kick in and so in distinctive it may be run when you’ve got 15 observations of course the situation of the scholar tested that it depends on the assumption that the pattern is Gaussian and that is some thing we really have to keep in intellect if in case you have a small pattern dimension there isn’t any magic occurring it’s now not like pupil t-experiment allows you to eliminate this asymptotic normality it form of assumes that it’s developed-in it assumes that your data has a Gaussian distribution so when you’ve got 15 observations what are you gonna do you want to scan if the mean is equal to zero or not equal to zero however you have most effective 15 observations you have to by some means expect that your information is Gaussian but if the info is given to you this isn’t math you truly need to verify that it is Gaussian and so we’re gonna have to to find checks that given some data tells us whether it’s Gaussian or not if i have 15 observations eight of them are equal to plus one and 7 of them are equal to minus one and it is pretty unlikely that you are gonna be capable to conclude that your knowledge has a Gaussian distribution nonetheless in case you see some kind of unfold round some value you kind a histogram might be and it sort of looks find it irresistible’s a Gaussian one you would need to say it’s Gaussian right and so how can we make this more quantitative well the unhappy answer to this question is that there will be some tasks that make it quantitative however here in case you feel about it for one second what will be your null speculation you’re not I posses us for the reason that it’s one point it’ll be that it is gal seein and then it’s the alternative is gonna be that it’s now not Gaussian so what it method is that for the primary time to your information life you are gonna wish to conclude that’s the h0 is the proper one you’re definitely not gonna want to see that it’s now not Gaussian given that then the whole thing you realize is sort of falling apart and so it can be variety of a bizarre thing the place you are variety of gonna be seeking checks that don’t have any power truly you’re gonna want test that and that is what that’s the character the quantity of choices the number of approaches that you could be not Gaussian is so huge that all tests are kind of certain to have very low power and so that is why men and women are lovely glad with the suggestion that matters is Gaussian for the reason that it’s very hard to discover a scan that is gonna reject this hypothesis all proper and so we’re even gonna in finding some tests which are visual where you are gonna be able to claim good sort of appears Gaussian to me if it is all allows for it makes it possible for you to to deal with the borderline circumstances lovely effectively we’ll see genuinely a special instance all proper so this thought of trying out whether knowledge comes from a special distribution is called goodness-of-fit is this distribution a good match for my information that is the goodness of match scan we’ve got simply noticeable a goodness of fit experiment what was once it yeah the the the chi-rectangular test right the chi-rectangular test we were given a candidate PMF and we were testing if this used to be a good fit for our information that was a goodness-of-match test so of direction multinomial is one example however rather what we have now behind our mind is I want to test if my data’s calcium that is clearly the ordinary factor and just like you at all times see t-scan as the commonplace output from statistical program whether you ask for it or now not there can be a scan for normality whether you ask it or now not from any statistical software output all right so once more this a notable situation fellows you might have given it there may be a random variable X and you’re given iid copies of X proper X 1 to xn they come from the identical distribution and you’re gonna ask the next query that as X have a ordinary usual distribution correct so for T distribution that is definitely the form of questions you may want to ask does X have a uniform distribution on 0 1 right that is specific from the distribution 1 over okay one among our kids the steady new inspiration of uniform uniformity all correct and for illustration you might want to test that so there may be really a exceptional endeavor which is when you appear on the p-values proper so we have now defined what the p values were and the p worth is a number between zero & 1 right and also you would definitely ask your self what’s the distribution of the p-price beneath the null right so it’s a the p worth is a random random number it is the likelihood right so the p worth let’s seem at a the following experiment a zero mu is the same as zero versus H 1 mu is say no longer equal to zero and i do know that the p worth is so i’m going to kind what i’m gonna seem at xn bar minus mu divided by way of x square root of n divided by shall we say that we all know Sigma for one second all proper then the p price is the likelihood that this is say higher than rectangular root of n little xn bar minus mu minus 0 definitely on this case correct / Sigma the place this guy is the observed proper ok so now you might say good how is that a random variable correct there is just a quantity it is only a probability of something but then i will view this as a operate of this guy here once I plug plug it back it when I plug it back to be a random variable so what I mean by way of this is that if I look at this value right here this if i bathroom loo if I say that Phi is the CDF of n 0 1 right so the p value is the probability that it exceeds this so that is the likelihood that i am on either here or right here k no it is no longer correct this can be a large X and this is a small X this is just where you plug to your information do i’ve right the p worth of the brotherly you may have more proof towards your L to your null than what you already have ok so now i will write in an interpretive cumulative distribution function correct so that is what this is Phi of this man which is minus this factor right here good it can be in actual fact two instances this man Phi of minus square root of n xn bar divided by way of Sigma proper that’s the worth of my tv that’s my p price if you happen to provide me information i’m gonna compute the common and plug it in there and it is gonna spit out the p price all people concurs so now i will view this if I start now watching again I say well the place does this knowledge come from well it could be a random variable proper it it got here from the attention of this factor so i can try to i will be able to think of this worth or now this is a random variable seeing that I simply plugged in a random variable in right here all correct so now I view my p price as a random variable proper so i hold on switching from small x to colossal ax everyone concurs what i’m doing right here so I simply wrote it as a deterministic perform of some deterministic number and now the number it the operate stays deterministic but the number turns into random and so now i will be able to think of this as some statistic of my information and i might say well what is the distribution of this random variable now if my data is surely more commonly distributed so i’m absolutely under the null so beneath the null that signifies that xn bar times cordovan divided through Sigma has what distribution typical yeah well it was once Sigma ice i assumed I knew it proper so it is n 0 1 right I divided by way of Sigma right here adequate so now i’ve this random variable and so my random variable is now 2 Phi of minus absolute price of a Gaussian ok and i am simply within the distribution of this factor okay I might ask that any person has an concept of how you would need to sort out this factor if I requested you what is the distribution of a random variable how do you deal with this query there’s in actual fact two approaches one is to check out to seek out anything that looks just like the expectation of H of X for all H and you try to write this making use of exchange of variables as anything that looks like quintessential of H of X P of X DX and you then say well that is the density proper if which you could write this for any H then that’s the way you could possibly do it but there may be a a lot less complicated approach that does not contain altering variables and so forth you just attempt to compute the cumulative distribution perform all proper so let’s attempt to compute the likelihood that 2 Phi minus n zero 1 is lower than T and might be we can to find some thing we know adequate good that is equal to what that is the probability that a minus n 0 good let’s assume that an N 0 1 i’m sorry n 0 1 absolute worth is greater than minus Phi inverse of T over 2 proper and that’s basic that is what good it is just the identical factor that we had earlier than it is equal to so if I look again that is the chance that i’m virtually on this part or that aspect of this number and this quantity is what it is minus Phi of T over two why do i’ve a minus right here that is fine k so it is genuinely not this it is really the likelihood at my absolute price seeing that Phi inverse okay considering that Phi inverse is so i’ll seem at T between zero and so this number is ranging between 0 and 1 so it means that this quantity is ranging between 0 good the probability that some thing is less than T should be ranging between the numbers that this guy takes to so that’s between zero and a pair of right correct when you consider that this thing takes values between 0 and 2 I wish to see zero and 1 now yeah you’re correct thank you so that is constantly some number which is not up to zero so the probability that the Gaussian is not up to this quantity is normally less than chances less than zero which is one-1/2 so T handiest has to be between zero and one thanks and so now for T that is between zero and one then this guy is really becoming anything which is positive for the equal reason as earlier than and so that’s what that’s just truly two times Phi of Phi inverse of T over two ok that’s simply playing with the symmetry a bit of bit you could look on the areas below the curve and so what it approach is that those two guys cancel this is the identity and so this is equal to T so which distribution has a this has a density as a sorry which distribution has a cumulative distribution perform which is the same as T for T between 0 & 1 that’s the uniform distribution right so it means that this man follows a uniform distribution on the interval 0 1 okay and you can certainly check that for any experiment you are gonna provide you with this is gonna be the case your p-price beneath the null may have a distribution which is uniform so now if a person shows up and says here is my scan it can be top notch it simply works best I do not know i am now not gonna explain to you how I constructed it it can be a elaborate information that involves moments of order 27 and i’m like ok you understand how am I gonna scan that your test statistic absolutely is sensible good one thing i will do is to run a bunch of data draw a bunch of samples compute your experiment statistic compute the p-worth and examine if my p-worth has a uniform distribution on the interval 0 1 but for that I must have a scan that is allowed given a bunch of observations can inform me whether or not they’re sincerely distributed uniformly on the interval 0 1 and again one thing I would do is build a histogram and notice if it looks like that of a uniform however I would also you should be somewhat extra quantitative about this for 2 checks no it’s it is uniform below than all correct so due to the fact that my my scan statistic was once developed below the no and so I have to be ready to plug within the correct value in there or else gonna shift the whole lot for this targeted test that is the p-value right that’s the completing of the p-worth all proper is the likelihood that my test statistic exceeds what I’ve actually observed yeah so my my p-price is just this quantity when I just plug in the values that I become aware of here right that’s one number for each data set you are gonna provide me there’s gonna be one quantity now what i will be able to do is for generate a bunch of data units of dimension n like 200 of them after which i’m gonna have a new pattern of size 200 which is solely the sample of 200 p-values and i want to scan if those p-values have a uniform distribution okay when you consider that that is the distribution they must be having all correct k this one we have already noticeable does X have a PMF with you already know 30% 50% and 20% right that’s some thing I might try to experiment that appears like your grade factor distribution for this classification all proper i’m not precisely however that looks love it all correct so so all these matters are comprehend are known as goodness-of-match assessments the goodness of fit experiment is something that wish to understand if the information that you’ve at hand has a follows a hypothesized distribution all correct so it’s now not a parametric scan it is now not a experiment that says is my mean equal to 25 or no longer is my proportion of heads bigger than one-half or now not it’s anything that says is my distribution this specified factor all right so i’m gonna read them as goodness-of-match gof here so here there’s you don’t must have parametric modeling to try this so how do I work right so if I do not have any parametric modeling I must have whatever which is rather nonparametric right whatever that goes beyond computing the imply and the ordinary deviation something that computes some intrinsic nonparametric facet of my information and just like here we made this computation what we did is we mentioned good if I really check that the CDF of my knowledge has that of my p value is uniform then I are aware of it’s uniform so it signifies that the cumulative distribution perform has an intrinsic price about it captures the complete distribution proper the whole thing I must learn about my distribution is captured with the aid of the cumulative distribution function now i’ve an empirical manner of computing i’ve an information-driven means of computing an estimate for the cumulative distribution function which is making use of the historic statistical trick which consists in changing expectations via averages all correct in order I said the the cumulative distribution function for any distribution for any random variable is proper so f of T is the probability that X is not up to or equal to T which is equal to the expectation of the indicator that X is lower than or equal to T all right that is the definition of a chance and so here i’m just gonna substitute expectation by means of the normal right that is my common statistical trick and so my estimator FN for the distribution is going to be 1 over n sum from I equal 1 to N of those warning signs good enough and this is referred to as the empirical the CDF it can be simply the info variant of the CDF okay so I simply replaced this expectation right here by an usual ok now after I sum symptoms i’m surely counting the quantity of them that satisfy some thing so for those who look at what this guy is that is the quantity of excise that is lower than T right and so if I divide by using n it can be the percentage of observations that have their own not up to T ok that’s what the empirical distribution so that’s what’s written here the number of knowledge points which might be less than T and so this is going to be something that variety of tried to estimate one or the opposite and the legislation of huge quantity sincerely tells me that for any given T if n is gigantic enough f n of T it should be virtually f of T right on account that it can be an typical and this complete factor this complete statistical trick which consists in exchanging expectations with the aid of averages is justified by way of the regulation of tremendous number at any time when we used it that was once considering the fact that the regulation of enormous number type of warranty to us that the traditional was nearly the expectation k so law of gigantic numbers tell me that f n of T converges so that is the powerful legislation says that close to certainly absolutely f n of T goes to F of T ok and that is just for any given T everybody is there any question about this proper that averages converge to expectation that is the legislation of colossal number and just about obviously we would say in probability it is the equal that could be the weak legislation of colossal number now that is satisfactory for any given T the usual converges to the authentic correct it simply occurs that this random variable is indexed with the aid of T and i might do it for T equals 1 or 2 or 25 and just verify it again however I might want to assess it for all T’s immediately and that’s certainly a further outcomes that is known as a uniform outcomes i need this to maintain for all T at the same time and it could be the case that it really works for each T individually but now not for all T’s even as what would occur is that for T equals 1 it converges at a certain expense and for T equals 2 it converges at a little of a slower price and for T equals 3 as at a slower expense and slower rate and in order T goes to infinity the rate goes to disappear and nothing is gonna converge that would happen I would make this happen at a finite point there’s many methods the place i couldn’t make this occur correct let’s examine how that could work I would say well surely no I still must have this at infinity for some cause good enough so turns out that that is still true uniformly and that is simply a much more tricky influence than the law of massive number it is referred to as livenko Cantelli theorem and the glencoe would tell the concept tells me that for all T’s without delay FN converges to F good enough so let me simply let you know i’ll exhibit you rapidly why that is just a bit bit better than the one who we we had so take into account if soup is confusing you consider of max it can be just a max over a limiteless set and so what we know is that f n of t minus goes to F of T as n goes to infinity and shall we say close to definite oh ok and that’s the gigantic large numbers which is equivalent to pronouncing that F n of t minus f of T as n goes to infinity converges practically definitely to zero correct this is the identical thing now i need this to happen for all T’s without delay okay and so what i am gonna do oh and that is actually an identical to this and so what i’m gonna do is i am gonna make it slightly greater so right here the arrow handiest goes one way and this is where the soup for T and R of F n of T ok and that you can definitely show that this happens also close to surely ok now maybe just about most likely is a little more complicated to get a seize on does can anyone does someone have wooden does any one wish to see it and like why this declaration for the soup is precisely greater than the person who holds in my view for all T’s you want to peer that k so let’s do it so overlook about almost most likely for one second let’s just do it for in probability write the truth that f n of T converges to F F F of T for all T in chance signifies that this goes to zero as n goes to infinity for any epsilon correct for any epsilon and T we do not we’ve this that is the convergence in probability now what i want is to place a soup here proper but the probability that the soup is cut down than Epsilon probably really continuously larger than by no means go to to zero in some cases correct it could be the case that for each and every given T i will make n large enough in order that that is small in order that this likelihood turns into small but then perhaps it is an N of T proper so this right here implies that for any say probably I mustn’t put let me put the Delta right here so for any epsilon so for any T there and for any epsilon there exists n which would rely on on each epsilon and T such that the probability that F and G minus F of T exceeding Delta is not up to epsilon T there existent and any delta no that is for all Delta sorry so right so that is genuine correct so that is that is what the this restrict announcement clearly way however it might be the case that that after I take the soup over T probably that n of T is anything that looks like tea or maybe well integer part of T it would be right i don’t say whatever it can be simply an finish that depends upon T so if this n is solely T might be T over epsilon proper considering the fact that i want epsilon whatever like this good that signifies that if i would like this to preserve for all T’s at once i’m gonna have to go for the T the N that works for all T’s directly however there’s no such n that works for all T’s at once that is the one finish that works is infinity and so I can not make this happen for all of them what banker can tell D tells you it’s definitely this is not anything that holds like this that the N that is dependent upon T there’s surely one biggest n that works for all of the cheese at once and that is it adequate so just so you recognize why that is sincerely a better assertion and and that’s truly how how it works every other question yeah no good the random verbal does have finite mean and finite variance in view that the random variable is an indicator so it has the whole thing you want this is likely one of the nicest random variables this can be a burly random variable proper so here after I say a legislation of huge number that this holds the place did I write this I think I raised it yeah the one over there this is without a doubt a variety of huge numbers for Bernalillo random variables they have got the whole lot you wish to have they are bounded yes so that you imply this one for all epsilon and all T so that you repair them now that the chance that I Excel sorry that was once Delta R a transformed this epsilon to Delta at some factor oh so then the second line says that for all so this is so the way you so this i’m simply rewriting in phrases of epsilon Delta what this end goes goes to infinity manner in order that implies that for any well for any T and Delta proper so that is the identical as this guy here then right here i’m simply going again to rewriting this it says that for any epsilon there exist an N giant ample such that good n greater than this thing sincerely such that this factor is lower than Epsilon k so Glencoe Catelli tells us that now not simplest is that this thing is just right proposal point shrewd however it is usually a good proposal uniformly and all it can be pronouncing is should you honestly have been completely satisfied with just this outcome you will have to be even happier with that outcomes and this both of those outcome best inform you one factor they are just telling you that the empirical CDF is a good estimator of the CDF okay now because these F these symptoms are Bernoulli distributions i can simply do much more so let me get this guy right here okay so those guys a fan of tea this man is a Burnley distribution what is the parameter of this Bernoulli distribution what is the likelihood it takes price 1 f of T right it’s just the probability that this factor happens which is f of T so in detailed the variance of this guy is the variance of this Bernoulli so it can be F of T 1 minus F of T and i will use that in my critical restrict theorem and my title and imperative me theorem is solely gonna inform me that if I seem at the usual of random variables I put off their mean correct so you look at rectangular root of n F n of T which I might relatively write as xn bar right that’s particularly just an xn bar minus the expectation which is f of T that comes from this guy now if I divide by means of square root of the variance that is my square root P 1 minus P then this man by way of the vital limit theorem goes to sum n 0 1 okay which is the same thing as you see there except that the variance used to be placed on the other side ok do i have the same factor uniformly in T proper am i able to write something that holds uniformly in T good in case you think about it for one 2nd it’s not likely it is gonna go to good within the experience that it is not likely that the supremum of those random variables over T goes to also be a Gaussian right and the reason is that well definitely the cause is that this thing is really a stochastic method index by t a stochastic procedure is only a sequence and random variables that is within the XY say time one the one that probably the most noted is the brown and brown and motion and it is essentially a bunch of Gaussian increments right so whilst you go from T to only T somewhat after that you simply have got to just add some gasoline into the object and and here it is sincerely the identical thing that is going down and you possibly can kind of assume on the grounds that every of this man is Gaussian you could count on to peer whatever that looks like a Brownian motion on the finish however it’s no longer exactly a circular motion it can be whatever that’s called a Brownian bridge so for those who’ve seen the brown movement is anything if I make it at zero for instance so that is the worth of my Brownian motion let’s write it so this is one course one recognition of Brown let’s name it W of T as T raises so it’s let’s consider it starts at zero and it looks like anything like this correct so that’s what Brown movement appears find it irresistible’s just something that’s lovely nasty and but it it’s I imply it appears beautiful nasty it’s no longer steady etc nevertheless it’s actually very very benign in some typical approach okay and so Brown movement is just something what you will have to view this as if I sum some random variable there are Gaussian and then I look at at this farther it from farther and farther it’s gonna look like this and so here I can’t have a Brown motion in the end on the grounds that what is the worth at what is the variance of F n of T minus f of T at T is equal to 1 sorry at T is equal to infinity it can be zero right the variance goes from 0 at T is poor infinity on account that that bad infinity f of T is going to zero and at F at T as T goes to plus infinity f of T goes to 1 which means that that the variance of this guy as T goes from terrible infinity to plus infinity is Penda to be zero on each part and so my Brannon motion cannot right once I describe aberrant motion i am just like including increasingly entropy to the thing and it is going far and wide the location but right here what i want is that as i am going back it will have to return to practically zero it must be pinned right down to a unique price at the end and that’s actually known as a Brownian bridge it is a manufacturer movement that’s conditioned to come back to where it started practically now you don’t need to recognize brown bridges to appreciate what i’m gonna be telling you what the one factor I wish to communicate to you is that this man right here once I say a Brannon bridge i will be able to go to any parabola stand they can let you know all the chance homes of this stochastic method it may well tell me the chance it takes any worth at any factor in unique it might tell me the supreme amines Eero and one in every of this man it might inform me what the cumulative distribution function of this thing is can tell me what the density of this thing is can inform me the whole thing so it implies that if I need to compute possibilities on this object right here which is the maximum worth that this guy can take over a detailed interval of time which is in actual fact this random variable rights if I look at the value here it’s a random variable that fluctuates it might probably tell me the place it is with excessive likelihood can inform me the quintiles of this factor and it tell can inform me which is priceless seeing that i will be able to construct a table and use it to compute my quintiles and perform duties from it ok so that’s what without a doubt is relatively excellent it says that if I seem at this rectangular root of n FN hat now soup over T I get anything that appears like this soup of this gaussians but it’s the early soup of fuel and it’s super of a brown movement now there’s whatever you should be very cautious here I cheated slightly bit I imply I did not cheat i can do whatever i want however my notation probably a little bit confusing every person sees that this T right here will not be the identical as this T right here can anyone see that proper simply on account that initially this man’s between zero & 1 and this guy’s between these all in all of our what is this T here as a function of this T here this man is F of this guy k so really if i need it to be fully transparent and now not you understand keep the cheese of my keyboard i would write this as soup over t FN t minus F of T goes to and distribution as n goes to infinity the supremum over t again in R so this man is for T in the complete real line this guy’s 14 the entire real line but now I must write B of what F of T exactly so particularly the T right here is F of the original one and so that is a Brownian movement that is a Brownian bridge the place when T goes to infinity the Brownian bridge goes from zero to 1 and it looks like this a Brownian bridge at zero is 0 at 1 is 0 it does this but it surely doesn’t stray too a ways considering I conditioned it to come back to this factor that’s what a brown bridges okay so in distinct i can find a distribution for this man and i will use this to build a test which is known as the kolmogorov-smirnov test k so the yr is the next it can be as well if I had been if I need to chat some distribution if no longer some distribution that has a specified CDF F dot and that i plug it in beneath the null then this guy will have to have commonly the same distribution as this supremum of Brownian bridge and so if I see this to be so much larger than it should be when it is the supreme of a Brownian bridge i am simply going to reject my speculation adequate so this is the test I wish to test whether h0 F is the same as F zero and you’re going to see that most of the the goodness-of-match checks are formulated mathematically in terms of the cumulative distribution perform I could formulate them in terms of probability density function or simply write x follows and nil one however that is the way we write it we formulate them in phrases of cumulative distribution function because that’s what we have now a handle on by means of the empirical community of distribution function ok after which it is vs.H1 F is just not equal to f zero okay so now i have my empirical CDF and i’m hoping that for all T’s F n of T will have to be nearly f0 of T good enough like this I put it two on the exponent considering the fact that or else that may be i’ve F the empirical distribution operate headquartered on zero observations all proper so now I kind the following scan statistic ok so my test statistic is TN which is the supremum over T in the true line of square root of n FN of T minus F of T sorry F 0 of T so i can compute everything i know this from the information and that is the one who comes from my null hypothesis i will compute this thing and i know that if this is authentic this will have to honestly be the supremum of a Brownian bridge ordinarilly k and so the kolmogorov-smirnov test is with ease reject if this man TN in absolute price correct there’s no simply not in absolute worth this is simply already absolute valued then this guy should be what good it should be better then the Q alpha over 2 distribution that i’ve however now as an alternative than putting and nil 1 or TN this is here something notation i have first premium of Brownian bridge k identical to I did for any pivotal distribution proper that was once the identical recipe each single time I type the experiment statistic such that the asymptotic distribution didn’t rely upon anything i do know after which i’d just reject when this pivotal distribution used to be higher than something sure are you aware what it brown in bridges or k so this thing right here is consider of it as being a Gaussian right so for all T you may have a Gaussian distribution ok now a Brownian motion so if I so if I had a Brownian motion I must inform you what the so it’s essentially a brown movement is something that looks like this it’s some random variable that’s listed through T I will not say the expectation of XT to be equal to zero for all T good enough and what i need is that the increments have a specified distribution okay so what i want is that the expectation of X T minus X s follows some distribution which is n zero t minus s good enough so the increments are bigger as i am going farther in terms of variability and i also want some covariance structure between the two proper so what i would like is that the covariance between X s and X staff is virtually equal to the minimum of S&T yeah probably there must be that k so this is you open a likelihood guide that is what it can be gonna appear like so in above all you’ll find if I put zero here as T goes to int and X zero is the same as 0 it has zero variance and it’s so in distinctive it implies that XT if I appear best on the cheap one it has some ordinary distribution with variance T so this is whatever that just blows up so this guy right here it appears love it’s gonna be a brown movement since when I look on the left-hand part it has a normal distribution now there is a bunch of other things you have got to assess is the truth that you have got this covariance for illustration which i did not inform you but it surely must seem somewhat relatively like that in particular when I seem on the typical with mean 0 and and variance here then it’s clear that this man does now not have a variance that is gonna go to infinity identical to the variance of this man we all know that the variance is pressured to be back to 0 okay and so in particular now we have anything that has imply zero normally whose variance has to be zero at 0 and the variance story at t equals negative infinity and variance 1 at t equals plus infinity so variance 0 at t equals plus infinity and so i must sincerely force it to be equal to 0 at each and every end so the brown movement here tends to only go to infinity somewhere whereas this guy compelled it to come again now everything i described to you is on the dimensions 0 terrible infinity to plus infinity however on account that the whole lot depends upon f of t i will simply simply put that back right into a scale which is 0 and 1 by a chain is discreet change of variable it can be find it irresistible’s known as alternate of time for the brown motion ok yeah yeah I mean they’re undoubtedly it’s gonna be I imply it by symmetry that you could regularly infer all of the things right so yes so I do not know precisely express system for the span it can be simple that’s what i will let you know but I do not know what the specific i would on the prime of my head with the specific components is ya know good the brown one Britt that is the supreme are you correct so that is yeah so correct so this might be this form for the variance for definite on account that this is most effective marginal distribution that do not take correct the procedure is not only what is the distribution at each and every on the spot t it’s also how do these distributions have interaction with each different right in phrases of coherent but for the marginal distributions at each and every instant T proper the variance is f of T 1 minus f of T that is no longer gonna we’re no longer gonna escape that but then the covariance structure between those guys is a bit more intricate however yes you are proper for marginal that’s enough yeah yeah it would be infinity k on account that what I did I lift it yeah considering the fact that here I didn’t say the supremum of absolutely the price of a bearing bridge i just said the supremum of a brown bridge however you’re right let’s simply do this like that after which and then it can be it’s most often cleaner right so yeah in reality good I must be Q alpha so this is basically you are proper so this is consider of it as being one-sided and there may be surely no symmetry for the supremum correct so I imply the supreme is not symmetric round zero and so so you are proper I must not use alpha over two thanks any other question there will have to be alpha yeah I imply those slides had been written with I minus alpha and i have now not replaced all situations of one minus alpha with the aid of alpha I mean except this man still day good it depends upon how you want to name it but this is still the chance that Z the verrat Z exists this guy will have to be alpha ok and this can also be determined in tables and we will compute the p-worth just like we did earlier than but we must simulate it given that it is not going to rely on the cumulative distribution perform of a Gaussian find it irresistible did for the usual gastrin test it can be something that’s extra problematic and by and large you do not even try you get the statistical software to do it for you ok so simply let me let me skip a few traces that is what the table appears like for the kolmogorov-smirnov test okay so it simply tells you what’s your some quantity of observations after which you want alpha to be equal to five% say let’s assume you’ve got nine observations so if square root of an absolute value of f n of t minus f of t exceeds this factor you reject ok so what’s beautiful clear from this scan is that it appears very best and that i inform you this is the way you build it but when you’re serious about it for one 2nd it is in reality rather an stressful factor to build given that you have to take the supremum over t correct this is dependent upon computing your supremum which you realize observe possibly tremendous cumbersome I do not want you must compute this for all values T after which take the maximum of those guys it turns out that it can be virtually quite first-class that we don’t have got to absolutely do that what does the empirical distribution perform appear like good this thing keep in mind f n of T through definition was so let me go to the slide that is imperative so FN of T looks like this so what it means is that after T is between two observations then this guy is virtually retaining the equal value so if I put my observations on the true line here so let’s say i’ve one observation right here one commentary right here one statement here one of the vital relation right here and one statement right here for simplicity then this man is sincerely up to this normalization counting how many observations i’ve which are lower than T correct so when you consider that I normalized through n i do know that the smallest quantity here goes to be zero and the most important number here is going to be 1 adequate so let’s say this looks like this this is the value 1 good enough add the value for the reason that I take it not up to or equal to after I’m at X i’m without a doubt counting it so the bounce happens at X I so that’s the first observation after which I soar by how much do I soar yeah 1 over N correct k and then this value belongs to to the proper after which I do it once more i’m gonna i know i am no longer gonna work out for me however we are going to see that is truely I did lovely good ok this is what my cumulative distribution appears like now when you look at this gentle there’s this weird notation where I start striking now my indices in parenthesis X parenthesis one experiences two and many others those are called the ordered statistic it is just considering that it maybe you know when my seller is given to me I simply call the first remark the one that is on top of the table but isn’t it it doesn’t have got to be the smallest price so it maybe that this is x1 and that this is x2 then that is X 3 X four and X 5 right this possibly my observations so what I do is that I name them in this type of means that that is simply I take into account this guy X 1 which is solely particularly X 3 this is X 2 X 3 X four and X 5 these are my reordered observations in any such means that the smallest one is indexed by using one and the biggest one is indexed via n adequate so now this is truly quite nice on the grounds that what i am seeking to do is to search out the biggest deviation from this guy to the authentic cumulative distribution operate the proper cumulative distribution perform let’s assume it’s Gaussian appears like this all right anything continues for a symmetric distribution it crosses this axis at 1/2 and that is what it looks like and also you feel the the kamakura of smirnoff steps is simply telling me how a ways do I get those two curves get within the worst feasible case proper so in special here the place are they the farthest evidently at this factor and so up to rescaling this is the value i’m gonna be interested in that is how they get so far as viable from every other here anything just occurred right the farthest distance that I acquired used to be precisely at a type of dots it seems that is enough to seem at those dots and the cause is good since after this dot and until the next bounce this guy does no longer exchange however this man increases and so the the only factor the place they may be able to be the farthest aside is either to the left at one to the left of a jumper to the right of a job that’s the one place the place they are able to be a long way from each and every different and that means that only one commentary every body sees that the farthest points the points at which these two curves are the farthest from each and every other needs to be at one of the most observations and so instead than watching at a soup over all viable tees relatively all I have to do is to seem at a maximum handiest at my observations right I just have to determine at each of those aspects whether they are far now here become aware of that you just didn’t this isn’t written F n of X I the purpose is given that I genuinely understand what F n of X is f n of the eighth order commentary is just the quantity of jumps I’ve had unless this statement so here i do know that the value of F n is 1 over n right here it can be 2 over n three over N 4 over N 5 over n so i know that the values of F n at my observations and those are without a doubt the one values that f n can take are an integer divided by n and that’s why you see I minus 1 over N or I over N that is the change just before the leap and this is the change at the bounce good enough so here the important thing message is that this is now not supreme over all T’s however it’s just a maximum from 1 to n so I relatively have handiest two n values to compute this price and this worth for each and every statement that’s 2n complete I look at the maximum and that’s truely the value and it can be sincerely equal to TN these are matters it’s achieved an approximation these matters are equal that’s just the one locations the place these guys may also be highest ok yes Oh strictly less powerful so are you able to i’m now not definite what query you are asking yes yeah yeah so bear in mind right here in this scan we need to conclude to a nil in the other experiment we often want to conclude two h1 so here we simply do not need vigour in the way in which and and also you have got to also count on that doing a scan on the imply it is in general not the only factor you’re going to grow to be doing on your data after after you simply establish that it’s in general allotted then you’ve got an information set you may have style of established it is most commonly disbursed after which which you can simply run the arsenal of statistical stories and we’re gonna see regression and all sorts of prediction predictive things which are not simply simply test if the imply is equal to some thing possibly you wish to have to build a self belief interval for the imply then this isn’t the comparison rule will not be a experiment correct so you’re gonna ought to first experiment if it is average after which see if that you may certainly use the quantiles of a Gaussian distribution or a T distribution to build this self belief interval so in a technique you must do this is like you understand the flat price to enter the gaussian world after which you can do something you need to do in the gaussian world we will see surely that your query goes back to something that’s somewhat more that’s just a little main right here I said F zero is fully exact correct it’s like an N 1 5 however I failed to say is it Gauss is it regularly disbursed which is the query that each person asks right you are no longer asking is it this unique usual distribution with this detailed imply on this unique variance so how would you do it in practice well you possibly can say i’m just gonna replace the mean via the empirical mean and the variance by using the empirical variance but with the aid of doing that you are making a enormous mistake for the reason that you’re sort of depriving your test of the likelihood to reject the Gaussian hypothesis simply centered on the truth that the mean is mistaken or the variance is wrong you’ve gotten already stuck to your knowledge lovely well ok and so you’re type of like already you know tilting the sport in favor of h0 giant time so there may be surely a weight you organize for this ok so this is about pivotal statistic we have used this phrase many time and okay so that’s how i’m not gonna go into this scan it’s really this can be a recipe on how you can clearly construct the table that I showed you proper this desk that is essentially the recipe on find out how to construct it there is a different recipe to build it which is just open a book at this page very well that is slightly faster or use the program I simply wanted to exhibit you so let’s simply hold in intellect any person has a excellent memory let’s simply maintain in intellect this number that is the the edge for the comma grossman of statistic if i’ve 10 observations and i wish to do it at 5% okay it is about forty one% so that’s the quantity that it should be greater from so it seems that if you wish to scan if it is typical and now not only a precise typical this quantity goes to be unique do you feel the quantity i’m gonna read in a table that’s correct for that is gonna be bigger or smaller who says greater so the question is this is the quantity I should see if my experiment was once is X say n 0 5 proper that’s a specific distribution with a targeted F zero right so that is the number that’s i would build the comma graph smear sufficient statistic from this i would perform a experiment and investigate if my como gross man of statistic TN is bigger than this quantity or now not ok if its better i’m gonna reject no I say certainly i don’t need to scan if a zero is + zero 5 but it’s only a new Sigma squared for some mu and Sigma squared and in specific i am simply gonna plug for your hat and Sigma hat into my F zero run the identical statistic but examine it to a different number must this so the better the number the roughly likely i’m to reject the much less seemingly i’m to reject proper so if I just use that number i’d shall we embrace this is the massive quantity i would be more tempted to say it can be Gaussian when you look on the table you can get if you are making the appropriate correction on the equal number of observations 10 and the equal degree you get to 25% as opposed to forty one% that implies that you’re absolutely more likely if you happen to use the suitable tests to reject the fact that it’s traditional which is unhealthy information on the grounds that that suggests you do not have entry to the gaussian Arsenal and no person wishes to do this so absolutely it is a mistake that persons do rather a lot they use the comb over of smear experiment to experiment for normality without adjusting for the truth that they’ve plugged in the estimated imply and the estimated variance this leads to rejecting much less mainly correct I imply this is practically 1/2 of the number that we had after which there may also be blissful and more calm and said well I did the experiment and it was typical so this is without a doubt a mistake that I suppose that really at the least a quarter of the men and women do make in purpose they are saying well I wanted it to be Gaussian so i’m simply gonna make my lifestyles easier all right so that is the so-called Kolmogorov chief force experiment we are going to speak about it well no longer at present for certain there is different facts you could test that you need to use and the suggestion is to claim well we want to be aware of if the empirical distribution operate the empirical CDF is virtually the real CDF the way we did it is through forming the difference in looking on the worst feasible distance they are able to be that’s called a soup norm or L infinity norm in in functional analysis so here that is what it gave the look of proper the gap between F N and F that we measured was once simply the supremum distance over all T’s that is one way to measure distance between two services but there is an unlimited manner many ways to measure distance between capabilities one is some thing we’re much more familiar with which is d squared l 2-norm that is excellent since this has like inner internal product it has some quality homes and you might absolutely simply rather than taking the soup you would just integrate the squared distance and that is what leads to crime Ervin misses tasks and then there is an additional one that claims good maybe i do not need to integrate with out weights maybe I need to put weights that money owed for the variance and this man is known as Anderson darling for every of this test that you can determine that the syntactic distribution is going to be pivotal which means that there can be a desk at the back of some booklet that tells you what the statistic the the quintiles of rectangular root of n occasions this guy are asymptotically sincerely yeah yeah yeah so that is just to exhibit you that asymptotically is men and women oh and i will be able to point you to one distinct certain factor nevertheless it seems that this factor is genuinely pivotal for every N and that is why you may have this recipe to assemble the whole thing for the reason that it is in reality no longer proper for all possible ends additionally there may be the tip that suggests up here so no honestly this is simply anything you will have to have in mind okay so in actual fact let me let me strike what I just said this thing that you could sincerely this distribution won’t rely upon F zero for any specified n it’s simply not gonna be a Brownian bridge however a finite pattern approximation of a Brown bridge and you can simulate that simply drawing samples from it constructing a histogram and developing the quintiles for this guy oh there may be one that’s the table proper perhaps yeah let’s examine if we see it on the backside of the opposite table yeah see over 40 and we’re thirty so this is not the kolmogorov-smirnov however that’s the coma guerrilla force those numbers that you simply see here they’re within the numbers for the asymptotic thing which is some kind of Barron bridge yeah yeah so then you definitely is not going to have this excellent photograph proper this will happen at any point given that you’re gonna have this continuity’s in F and those things can happen all over the place and then oh you mean the brown bridge no no however you still need it you still want it for thus there may be there is some finite sample models of it that you should utilize that are quite more conservative so which is in first-class wave excellent news seeing that you are gonna conclude extra to h0 and there’s some you already know forget the identify it can be Keefer Wolfe of it can be the Keefer baretsky Wolfe of it can be an equality which is sincerely like happenings inequality so it can be clearly as much as unhealthy constants telling you the same influence as the Brownian bridge outcomes and those are proper at all times however for the exactly the asymptotic distribution you want continually yes sure well if you know what they’re you need to use comb-over of smirnoff but when you do not know what they are you’re gonna plug in as soon as you are gonna estimate the mean and the variance from the information you’re gonna you should utilize the one the place see you next time which is known as comb-over of Lili drive don’t must consider about it too much we’ll speak about it don’t Thursday yeah any other question so we were out of time so I consider we should stop right here and we’ll see ya well we’ll resume on Thursday

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