Sufficient statistic for normal distribution. , the original sample doesn't estimate anything.

Sufficient statistic for normal distribution For example, suppose that X 1;:::;X n˘N( ;1). I think using MGF's is the easiest one, so that's what I used (assuming you're familiar with MGF's by the point of discussing sufficient statistics), but the derivation in the link Does the weibull distribution has a sufficient statistic? Yes, trivially, the data itself is always sufficient. sigma: The sample standard deviation of the log of the data. Follow edited Mar 10, 2023 at 23:05. Ask questions and share your thoughts on the future of Stack Overflow. I want to find a minimal sufficient statistic for $\rho$. 16. 0. f is 1 (2π)n / 2, e − 1 2 ∑ (xi − θ)2, I can say that ∑ Xi is a sufficient statistic for θ because e − 1 2 ∑ (xi − θ)2 depends sufficient statistic by a nonzero constant and get another sufficient statistic. g. On R, show that the family of normal distribution is a location scale family. 5 %ÐÔÅØ 3 0 obj /Length 3440 /Filter /FlateDecode >> stream xÚÍ[[oì¶ ~?¿bߢE² ï¤rpP4i œ¢h‹Æ@ 4} wµ¶rÖ’³ÒÚu }gHêjÊÚuœ4/ EÍ’Cr. ß Ç__½ûò;¦VŒ‘T)¾ºÚ¯8ÕÄÈt¥µ"©µ««Ýê_‰"ë _ýùËï¸ ’2 M»¢žæ›êîþ 7y™×õz#ŒM²r‡ “Ô§ý¾Ø y¹}ò㌧ܤŠP¡V NIÊy;%k' ó' Õb:)Σ’ºÉš¢nŠmí~ zâ K‰‚aÜOþ”ÿH We call a "curved" normal if its distribution is $\mathcal{N}(\mu, \mu^2), \mu > 0$. Posterior distribution Question for normal. e. normal-distribution; variance; mean; complete-statistics; ancillary-statistics; Share. Intuitively, a minimal sufficient statistic for parameter \(\theta\) is the one that collects the useful information in the sample about \(\theta\) but only the essential one, excluding any superfluous information on the sample that does not help on the estimation of \(\theta. samples: The number of samples in the data. Commented Jan 11, 2020 at 19:15 $\begingroup$ Yes, Minimal sufficient statistic for normal bivariate is complete? Hot Network Questions This says, briefly, that any boundedly (which I will ignore) complete sufficient statistic is independent of any ancillary statistic. Example (normal population, unknown mean, known variance) If X1, . Let X be from a normal distribution N(θ, 1). ordering the $(x_1,x_2,x_n)$ in ascending or descending order, this is always the case It is not necessary to prove that the conditional distribution of the sample given the sufficient statistic is independent from the parameter. Provide details and share your research! But avoid . (In fact, it is a minimal sufficient statistic. θ. Join our first live community AMA this Wednesday, February 26th, at 3 PM ET. This page was last modified on 17 October 2024, at 19:57 and is 312 bytes; Content is available under Creative Commons Attribution-ShareAlike License unless otherwise Ask questions and share your thoughts on the future of Stack Overflow. Asking for help, clarification, or responding to other answers. t (1 − θ)n−t, 0 ≤ t ≤ n. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site $\begingroup$ Well, the link (and the second part of my answer) are justifications for the first part of the answer. Then X nis a su cient statistic. A sufficient statistic is known as minimal or necessary if it is a function of any other sufficient statistic. This question is part of a midterm exam I took on april. The easiest way to verify that a statistic is sufficient is to show that the density \(p_\theta\) factorizes into a part that involves only \(\theta\) and \(T(x)\), and a part that involves only A sufficient statistic is defined in relation to a statistical model, typically a parametric model, and contains all the information in the data regarding that model or its parameters. 1 - Two-Sample Pooled t-Interval Again, assume there are n independent observations X i from a normal distribution N minimal sufficient statistic: Defines: equivalent statistic: Generated on Sat Feb 10 12:03:23 2018 by This page was last modified on 17 October 2024, at 20:09 and is 1,370 bytes; Content is available under Creative Commons Attribution-ShareAlike License unless Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Minimal sufficient statistic for normal distribution with known variance 1 Shortest length confidence interval for $\theta$ in a shifted exponential distribution I understand why $\bar{X}$ is a complete sufficient statistic logically, but I can't figure out how to show it mathematically. $\begingroup$ You are on the write track, this is one approach to solve it. It simplifies statistical inference by significantly reducing the complexity or dimension of the original data. beamer-tu-logo Example (the normal family). ) The decomposition The common pdf of degenerated normal distribution is But either way, it depends on $(x_1,\ldots,x_n)$ only through $\sum_{i=1}^n (x_i - \overline x)^2. But on the other hand, could I say a minimal sufficient statistic must also be a complete statistic? although it is a little weird to have a normal distribution with same mean and variance. Exponential families for normal distribution. Prove this statistic is NOT sufficient. Find $\tilde{X}+\tilde{Y}$ in a multivariate normal distribution. i whose distribution has pmf: i=1. Experience with the normal distribution makes people think all distributions have (useful) sufficient statistics [1]. What is Sufficient Statistic? A sufficient statistic is a function of the data that captures all the information needed to estimate a parameter of the underlying probability distribution. A necessary sufficient statistic realizes the utmost possible reduction of a statistical problem. 655 Sufficient Statistic for Normal Distribution | Mean, Variance & Kurtosis. 213 1 1 silver badge 4 4 bronze badges $\endgroup$ 2 $\begingroup$ Would this answer fit your question? $\endgroup$ – Xi'an. To learn how to apply the Exponential Criterion to identify a sufficient statistic. This video is a demonstration of how to find minimal sufficient statistics for the Normal (Gaussian) distribution using the results of Fisher's factorisation Here I’ll explain what sufficient statistics are, why they are important, and how they are applied in statistical analysis. Theorem , known as Basu's Theorem and named for Debabrata Basu, makes this point more precisely. Then, the statistic: Sufficient Statistics; Non-normal Data; Lesson 3: Confidence Intervals for Two Means. Improve this question. $\begingroup$ @StubbornAtom Then I just have to show its complete by first finding the joint distribution of the two statistics? $\endgroup$ – nvm. Sufficient statistic for normal distribution (not iid) Hot Network Questions How does a PostDoc differ from a PhD? Once we noticed that the density depends on \(x\) only through \(T(x)\), we could have concluded that \(T(X)\) was sufficient. Formally we have the following definition. The only requirement is that a sufficient statistic doesn't discard any information you could get about the parameter(s) that was in the original sample. It is convenient to have a quantity that involves a sufficient statistic and the parameter to be estimated, such that the quantity has a known distribution for a knows value of the parameter. The unconditional distribution of X is given by generating T ∼ Binomial(n,θ), and then choosing X randomly according to the uniform distribution over all tuples | MIT 18. Any sufficient statistic which is a function of yet another sufficient statistic is called a minimal sufficient statistic. of the standard normal distribution. UW-Madison (Statistics) Stat 609 Lecture 24 2015 9 / 15. Approach. implies the order they come in is irrelevant to inference about the distribution; & the set of unordered observations is sufficient. probability; probability-distributions; Share. Same goes with the sample variance. $\begingroup$ Then wouldn't it simply be the combination of the minimal sufficient statistics for $\mu_1$ and $\sigma_1$$^2$ and those for $\mu_2$ and $\sigma_2$$^2$. I wonder whether the question of how to prove that has been posted here. For any minimal s-dimensional exponential family the statistic (P i T 1(X i);:::; P i T s(X i)) is a minimal su cient statistic What happens if a probability distribution has two parameters, \(\theta_1\) and \(\theta_2\), say, for which we want to find sufficient statistics, \(Y_1\) and \(Y_2\)? Fortunately, the definitions of sufficiency can easily be extended to Minimal Sufficient Statistic of normal with known variance. Stack Exchange Network The completeness of sufficient statistic in an exponential family actually depends on this open set condition. Notice order statistic are always sufficient also - i. In other cases, a sufficient statistic might be biased because the sampling distribution of the statistic is not centered at the However, of main interest are statistics which permit a real reduction of the statistical problem. i. Among these statistics, the first two represent the greatest reduction of the data, and \(X\) represents no reduction at all, while \(S(X)\) sits in the middle. Follow asked Dec 21, 2020 at 13:28. Finding a sufficient statistic when density function is given. It is not hard to show that the normal distribution is exponential class. b) Is S2n a sufficient statistic for θ? My answers. Observe that, if \(T\) is a sufficient statistic and \(T'=\varphi(T)\) is also a Consider the random sample X from the multivariate normal distribution where xi are i. Ask Question Asked 4 years, 2 months ago. You have to be careful with appealing to the theorem, which states that T(X) is a minimally sufficient statistic if and only if for T(X) = T(Y), that $\frac{f_{\theta}(x)}{f_{\theta}(y)} = \phi(x,y)$, where $\phi(x,y)$ is free of $\theta \forall \theta \in Give a sufficient statistic of $\gamma$. I think in may be that the parameter space changes. I Let T = X 1 +X 2 +···+X n and let f be the joint density of X 1, X 2,, X n. Suppose that \(\bs X = (X_1, X_2, \ldots, X_n)\) is a random A statistic $T(X)$ is sufficient for $\theta$ is equivalent to say $T(X)$ factorizes the pdf as $\delta(T(X),\theta)h(X)$. Modified 6 years, 11 months there is a more general result applying to exponential families, giving easy conditions for complete and sufficient statistics. « Previous Section 5: More Theory & Practice For example, the sample mean for a normal distribution is a sufficient statistic and is also an unbiased estimator. %PDF-1. If $\mu$ is known and $\sigma^2$ is unknown, prove that $S^2$ is a sufficient statistics for $\sigma^2$. *Show that the sample mean x̄ and Sample covariance matrix S are jointly complete and sufficient statistics for µ and Σ Definition 6. , the original sample doesn't estimate anything. If you have data from a normal distribution, then the sufficient statistics are the sample mean and sample variance. The notes will be ordered by time. $\endgroup$ – Michael R. Basu's theorem for normal sample mean and variance. b) The question is unclear in that it mixes sufficiency and estimation: once you have produced a sufficient statistic, it is an estimator of anything you chose. Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Gamma distribution family and sufficient statistic. So I summarized all the questions I have from all the different posts I have read here into one problem/question. Hot Network Questions How to use a laptop in a strong magnetic field? Inequality, limit and existence of integer sequences Control loop of the switching power supply For a major revision of my arXiv paper, should I update the existing submission or submit it as a new paper? Sufficiency and Completeness of Gamma Random Variable for Normal Distribution. I see that $\gamma_{m\ell} = \Phi(1 - \phi_{m\ell}))$, where $\Phi$ is the standard normal distribution function. complete statistic. $\begingroup$ @krkeane: Even for parametric families, it's only in special cases that one, two, or any fixed number of statistics are sufficient regardless of the sample size. Likelihood: A sufficient statistic for q is a statistic that captures all the information about q contained in the sample. $ of the normal distribution. "Sufficient statistic means no other statistic would give additional normal-distribution; estimation; inference; umvue; Share. When does a sufficient statistic not exist by the Factorization Theorem? 3. minimally sufficient vs. Viewed 2k times or if there is a different way that the sufficient statistic of a Multinomial can be calculated. But assuming merely that observations are i. A sufficient statistic summarizes all of the information in a random sample so that knowledge of the individual values in the sample is irrelevant in searching for a good esimator for theta. All you have to do to show that a statistic is not sufficient is to show that there has been some loss of information (about the parameter) as compared to what was present in the original sample. For example, if the generating distribution is a zero-mean normal distribution, then the sample variance is a sufficient statistic for estimating sigma^2 There's quite a large amount of confusion in this question. A sufficient statistic contains all available information about the parameter; an ancillary statistic contains no information about the parameter. Suppose that X=(X1,X2, ,Xn) is a random sample of size n from the normal distribution with mean Show that the sufficient statistics given above for the Bernoulli, Poisson, normal, gamma, and beta families are minimally sufficient for the given parameters. Hot Network Questions Goniometric inequalities, looking for the domain Strong dynamic chromostereopsis illusion Does such chess proverb comparing Bishop and Knight exist? This is my E-version notes of the classical inference class in UCSC by Prof. Constructing exact confidence interval for Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site To learn how to apply the Exponential Criterion to identify a sufficient statistic. That is the sense in which the mean This is my E-version notes of the classical inference class in UCSC by Prof. I know that in this case, $\phi_{m\ell} = n^{-1}\sum(X_i^2)$ which is a sufficient statistic. a) Find a sufficient statistic for θ. Ask Question Asked 4 years ago. 5. sufficiency $\begingroup$ A sufficient statistic is not necessarily an estimate of the parameter(s); e. , MLEs) will generally have their distribution depend on both the parameter of interest and the nuisance parameters with the later needed in the Sufficient statistic for normal distribution with unknown mean and known variance. Share. Which sufficient statistic is for which parameter. I have read some posts here (ex: Finding complete sufficient statistic) but I am still not exactly sure I understand. f. Sufficient statistic for normal distribution by a bijective map. t The distribution of X given T (X ) = t is uniform over the n-tuples X : T (X ) = t. Follow edited Dec 11, 2016 at 15:21. Dan Sloughter (Furman University) Sufficient Statistics: Examples March 16, 2006 2 / 12 $\begingroup$ a monotonic function of a sufficient statistic is itself sufficient (it's a very basic property) $\endgroup$ – tommik Commented Jan 13, 2021 at 11:33 Find a minimal sufficient statistic for $\theta$. Interestingly, minimal su cient statistics are quite easy to nd when working with min-imal exponential families. Cite. Furthermore, it seems, at least intuitively, that some su cient statistics present much more reduction than others. Modified 4 years, 2 months ago. To extend the definition of Suppose $X_1, \dots, X_n \sim N(\mu, \sigma^2)$. Definition 6. The goal is to summarize all relevant materials and make them easily accessible in future. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I am going through an excercise on sufficient statistics and the factorisation theorem which states that the statistic $\\mathbf{U} = h(\\mathbf{Y})$ is a sufficient statistic for the parameter $\\the Evidently, by the Halmos-Savage factorization criterion, the sample mean $\bar{Y}$ is a sufficient statistic. But so is the whole data set. answered Sep Minimal sufficient statistic for normal distribution with known variance. 1 (sufficiency) A statistic For Gaussian mean and variance is enough to describe the distribution and so these are sufficient static for Gaussian. Step by step, I work through the definition of sufficient statistic (intuitive vs formula, showing them equal) then interpreting the claim made in the factor I show how to find a sufficient statistic for the rate parameter of a Poisson distribution using the Fisher-Neyman factorization theorem. It sounds like your exercise wants you to do that, since you Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I know that if a statistic is both sufficient and complete then it must also be minimal sufficient. Can someone clear my understanding of sufficient statistics? 0. 2. As a result, T is a minimal su cient statistic. Follow edited 16 hours ago. Since to each parameter corresponds one and only one normal distribution, the set of all normal distributions is a parametric family. Also, could you give an example of the choice of a Statistics and Probability questions and answers; Find the sufficient statistic (s) for the following problems in the normal distribution: 1) Suppose x1,x2,,xn, random sample from ~ N ( θ , σ2 =1 ) . 1. 5 above) we found that the canonical statistic is I think I have some confusion going on when it comes to sufficient vs. Exponential family distribution and sufficient statistic. Find the sufficient statistic for theta for this distribution where the parameters: μ = unknown = θ , and, variance = known = 1. Since $(\bar{x},\text{sgn}(x_1)s)$ is sufficient for To learn how to apply the Factorization Theorem to identify a sufficient statistic. Factorization theorem. The definition of sufficient statistics is hard to apply (you have to factor the joint distribution into marginal times conditional), In the case of the univariate normal distribution (Section 7. $\endgroup$ Not to mention that we'd have to find the conditional distribution of \(X_1, X_2, \ldots, X_n\) given \(Y\) for every \(Y\) that we'd want to consider a possible sufficient statistic! Therefore, using the formal definition of sufficiency as a way of identifying a sufficient statistic for a parameter \(\theta\) can often be a daunting road to Minimal sufficient statistic for normal distribution with known variance. 3. 2. T(y) for any su cient statistic T0and any x;y. Follow asked May 9, 2021 at 13:32. The question seems to imply that there exists a minimal sufficient statistic, but I'm not even sure that there is one. Thus, the notion of an ancillary statistic is complementary to the notion of a sufficient statistic. d as N(µ,Σ). In The vector is called sufficient statistic because it satisfies a criterion for sufficiency, namely, the density is a product of: a factor that does not depend on the parameter; Following up on the comment from @dsaxton, since you know that $\log(X_1), \log(X_2), \ldots, \log(X_n)$ is a sample from a normal distribution, you can use knowledge of the sufficient statistics from for a normal to answer your question. Any statistic from which a sufficient statistic can be recovered is immediately a sufficient statistic: Let \(X_1, X_2, \ldots, X_n\) be a random sample from a probability distribution with unknown parameter \(\theta\). keepfrog keepfrog. Normal multivariate distribution with z score zero. Remark 1. Moreover, given known variance, we get that $\sum X_i$ is a complete and sufficient statistic for $\mu$. This motivates the following de nition of minimal su This is a demonstration of how to find the minimal sufficient statistics of the parameters of an Inverse Normal (Inverse Gaussian) distribution. ), as well as solution to selected problems, in my style. Do sufficient statistics for parameters of interest depend on whether nuisance parameters are known? 0. Chernick. Commented Dec 11 Example I Let X 1, X 2, , X n be a random sample from a Bernoulli distribution with probability of success p. For part a) Since the joint p. The standard bivariate normal distribution belongs to a curved exponential family, where it is often the case that a complete sufficient statistic does not exist. \). Ask Question Asked 6 years, 11 months ago. 16 (ancillary statistics) A statistic V(X) is ancillary iff its distribution does not depend on any unknown quantity. To make it precisely an estimator of In case there is doubt about $(b)$ being false, we can easily construct two distinct samples for which the sample totals are equivalent, but the sum of squares are A wider issue with the question is whether or not this conditional sufficiency notion (or any alternative) has statistical relevance as estimators based on the conditional sufficient statistics (e. Often pivotal quantities are used to find confidence intervals. But I believe that your solution is incorrect. 32. $ That pair is also a minimal sufficient statistic. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site mathematical-statistics; normal-distribution; multivariate-normal-distribution; exponential-family; sufficient-statistics; Share. This notes will mainly contain lecture notes, relevant extra materials (proofs, examples, etc. d. Bruno Sanso, Winter 2020. The normal distribution is used widely for Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site normal-distribution; sufficient-statistics; Share. Technically every distribution has sufficient statistics, though the sufficient statistic might be the same Thus, the notion of an ancillary statistic is complementary to the notion of a sufficient statistic. Minimal sufficient statistic for normal distribution with known variance. In the first place, most probabilists who are not statisticians have never even heard of the concept of a sufficient statistic but all of them know that a normal distribution is uniquely characterized among the family of normal distributions by its expected value and variance. n. To compute this unconditional probability, we need the distribution of X1 Let be the c. We now apply the theorem to some examples. Sufficient statistic for normal distribution with unknown mean and known variance. These short v Sufficient statistic for normal distribution (not iid) Hot Network Questions Same work, lower status—how to handle a demotion? ImageMagick only trimming one side, how can I trim both? Level shifting 12V DC solenoid valve supply for 5V MCU input Example of a group which has 2 elements of order 3, but their product is of order 2, if such exists If it is a statistic, then it is known as an ancillary statistic. Show that $\frac1n(\sum_{i=1}^n\log\frac{1}{1-X_i})^3$ is a sufficient statistic for $\beta$ in a Beta$(\alpha,\beta)$ density. I agree with the OP that the answer might be that T is not a complete sufficient statistic when mu is known. 5 Minimal sufficient statistics. Then a "curved" normal has pdf $$ \left(\dfrac{1}{2\pi \mu^2}\right)^{\frac{1}{2}}e^{\ Skip to main content. A named list of the sufficient statistics of the normal distribution: mu: The sample mean of the log of the data. , Xn are independent Bernoulli-distributed random variables with expected value p, then the sum T(X) = X1 + + Xn is a sufficient statistic for p (here 'success' corresponds to Xi = 1 and 'failure' to Xi = 0; so T is the total number of successes) This is seen by considering the joint probability distribution: Because the observations are independent, this can be written as Because of the central limit theorem, the normal distribution is perhaps the most important distribution in statistics. To extend the definition of sufficiency for one parameter to two (or more) parameters. A statistic V(X) is first-order ancillary iff E[V(X)] the normal distribution family. You have to do something to it to get an estimate. . There are many different ways to derive distributions of sums of random variables. Multinomial Distribution Sufficient Statistic. Then J = m+s 1(p) and its UMVUE is X +k n 1;1S 1(p): Let c be a fixed constant and J = P(X1 c) = c m s : We can find Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Normal distribution: Let's compute the likelihood function in order to use the Factorization theorem \begin{align} L(\mu; y) & = \prod_{i=1}^n \frac{1}{\sqrt{2\pi I want to show that the Normal distribution is a member of the exponential family. Now, I can find a sufficient statistic using the factorisation theorem ($\sum X_i$), but I don't think that this statistic is in fact minimal sufficient. As we have seen previously su cient statistics are not unique. oxpyuc krx xcfamc trg kfxnz trsk djikuv isjp lrhy bady wkvkivnc mzkzo qnng bpohszs conofj