) / θ X Did something happen in 1987 that caused a lot of travel complaints? n X It is easy to see that if F(t) is a one-to-one function and T is a sufficient f Because the observations are independent, the pdf can be written as a product of individual densities, i.e. does not depend on the parameter X ( Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. , 1 ( , Active 9 months ago. 2 ) , {\displaystyle T(X_{1}^{n})=\sum _{i=1}^{n}X_{i}}. The knowledge of the sufficient statistic $X$ yields exhaustive material for statistical inferences about the parameter $\theta$, since no complementary statistical data can add anything to the information about the parameter contained in the distribution of $X$. See Chapters 2 and 3 in Bernardo and Smith for fuller treatment of foun-dational issues. {\displaystyle H(x_{1},x_{2},\dots ,x_{n})} Thus the density takes form required by the Fisher–Neyman factorization theorem, where h(x) = 1{min{xi}≥0}, and the rest of the expression is a function of only θ and T(x) = max{xi}. Y But i , Note that T(Xn) has Binomial(n; ) distribution. Let AˆRk. ) ( h {\displaystyle T(X_{1}^{n})=\left(\min _{1\leq i\leq n}X_{i},\max _{1\leq i\leq n}X_{i}\right)} {\displaystyle f_{\mathbf {X} }(x)=h(x)\,g(\theta ,T(x))} simply as ), T(x) is the su cient statistic, h(x) is a normalizing constant (which can be thought of as a regularizer), and A( ) is the log partition function. T ∣ P , denote a random sample from a distribution having the pdf f(x, θ) for ι < θ < δ. ) {\displaystyle g_{\theta }(x_{1}^{n})} On the previous post, we saw that computing the Maximum Likelihood estimator and the Maximum-a-Posterior on a normally-distributed set of parameters becomes much easier once we apply the log-trick.The rationale is that since $\log$ is an increasingly monotonic function, the maximum and minimum values of the function to be optimized are the same as the original function inside the $\log$ … 1 I edited the question. is a sufficient statistic for As an example, the sample mean is sufficient for the mean (μ) of a normal distribution with known variance. . \end{array} ( Sorry for the mistake. Let X be a random sample of size n such that each X i has the same Bernoulli distribution with parameter p.Let T be the number of 1s observed in the sample. whether the distribution of $(X_1,X_2)$ given $T=X_1+2X_2$ depends on $p$ or not. ) [11] A range of theoretical results for sufficiency in a Bayesian context is available. , where ( θ h σ ( S) = σ ( { ( 0, 0) }, { ( 1, 0), ( 0, 1) }, { ( 1, 1) }) where σ ( T) denotes the sigma generated by T and σ ( S) denotes the sigma generated by S. Since σ ( S) ⊂ σ ( T) (the information in T is more than S) , S is a minimal sufficient statistic and S is a function of T ,hence T is a sufficient statistic (But not a minimal one). are independent and exponentially distributed with expected value θ (an unknown real-valued positive parameter), then . To learn more, see our tips on writing great answers. = In particular, in Euclidean space, these conditions always hold if the random variables (associated with +X n and let f be the joint density of X 1, X 2,..., X n. Dan Sloughter (Furman University) Suﬃcient Statistics: Examples March 16, 2006 2 / 12 t 1 If the probability density function is ƒθ(x), then T is sufficient for θ if and only if nonnegative functions g and h can be found such that. 1 {\displaystyle T(X_{1}^{n})=\left(\prod _{i=1}^{n}{X_{i}},\sum _{i=1}^{n}X_{i}\right)} = Formally, is there any function that maps $T_*$ to $T$? Note that this distribution does not depend on . x A statistic Tis called complete if Eg(T) = 0 for all and some function gimplies that P(g(T) = 0; ) = 1 for all . The link function is given by. − u . Abbreviation: CSS )MSS. Is this enough to rule out the possibility of $X1+2X2$ as a sufficient statistic? ( n n Without prior information, ... (which are the sufficient statistics for the Bernoulli distribution). n {\displaystyle T(X_{1}^{n})=\sum _{i=1}^{n}X_{i}} To see this, consider the joint probability density function of ∑ n replaced by their value in terms n Only if that family is an exponential family there is a sufficient statistic (possibly vector-valued) From this factorization, it can easily be seen that the maximum likelihood estimate of {\displaystyle \theta } {\displaystyle g_{1}(y_{1};\theta )} A useful characterization of minimal sufficiency is that when the density fθ exists, S(X) is minimal sufficient if and only if. y P depends only on 2 , , ) Since $T \equiv X_1+X_2$ is a sufficient statistic, the question boils down to whether or not you can recover the value of this sufficient statistic from the alternative statistic $T_* \equiv X_1 + 2 X_2$. Stephen Stigler noted in 1973 that the concept of sufficiency had fallen out of favor in descriptive statistics because of the strong dependence on an assumption of the distributional form (see Pitman–Koopman–Darmois theorem below), but remained very important in theoretical work.[3]. and find $\sigma(X_1,X_2)=\sigma(T)$ ($T$ and $(X_1,X_2)$ have a same information) and obtain that $T$ is a sufficient statistics. f {\displaystyle x_{1}^{n}} ) The concept is equivalent to the statement that, conditional on the value of a sufficient statistic for a parameter, the joint probability distribution of the data does not depend on that parameter. = y 1 Suppose that X n X. ≤ 2 While it is hard to find cases in which a minimal sufficient statistic does not exist, it is not so hard to find cases in which there is no complete statistic. depend only upon governed by a subjective probability distribution. , 0 1. suﬃcient for θ. T n ( i X x X 1 . We know $S$ is a minimal sufficient statistics. ∑ = through the function. 3.Condition (2) is the \open set condition" (OSC). 1 Bernoulli. x 1 , b x y 2 Factorization Theorem Theorem 4 (Theorem 6.2.6, CB) Let f(x nj ) denote the joint pdf or pmf of a sample X . ( As (H+T) goes to infinity, the effect of the past trials will wash out. … … does not depend upon i Roughly, given a set of independent identically distributed data conditioned on an unknown parameter , a sufficient statistic is a function () whose value contains all the information needed to compute any estimate of the parameter (e.g. θ , This applies to random samples from the Bernoulli, Poisson, normal, gamma, and beta distributions discussed above. H (8.6) is the pdf of Let T = X 1 + 2 X 2 , S = X 1 + X 2. , h f {\displaystyle Y_{1}} , n α ∣ ) Just check definition of sufficiency, i.e. ≤ … For example(*1). 1 , the probability density can be written as x ) i ( … ( n over , with the natural parameter , sufficient statistic , log partition function and . The Bernoulli model admits a complete statistic. α {\displaystyle \mathbf {X} } i n n h In essence, it ensures that the distributions corresponding to different values of the parameters are distinct. x \end{array} β If X1, ...., Xn are independent and uniformly distributed on the interval [0,θ], then T(X) = max(X1, ..., Xn) is sufficient for θ — the sample maximum is a sufficient statistic for the population maximum. given Then we can derive an explicit expression for this: With the first equality by definition of conditional probability density, the second by the remark above, the third by the equality proven above, and the fourth by simplification. With the first equality by the definition of pdf for multiple variables, the second by the remark above, the third by hypothesis, and the fourth because the summation is not over What we want to prove is that Y1 = u1(X1, X2, ..., Xn) is a sufficient statistic for θ if and only if, for some function H. We shall make the transformation yi = ui(x1, x2, ..., xn), for i = 1, ..., n, having inverse functions xi = wi(y1, y2, ..., yn), for i = 1, ..., n, and Jacobian , i are independent identically distributed random variables whose distribution is known to be in some family of probability distributions with fixed support. , depends only on the Fisher–Neyman factorization theorem implies … ) , we have 1 ( Ask Question Asked 9 months ago. x through the function ) minimal statistic for θ is given by T(X,Y) m j=1 X2 j, n i=1 Y2 i, m j=1 X , n i=1 Y i. n X So T= P i X i is a su cient statistic for following the de nition. According to the Pitman–Koopman–Darmois theorem, among families of probability distributions whose domain does not vary with the parameter being estimated, only in exponential families is there a sufficient statistic whose dimension remains bounded as sample size increases. f It follows a Gamma distribution. ) h … {\displaystyle \theta } ; Because the observations are independent, the pdf can be written as a product of individual densities, i.e. , ... and is the sufficient statistic. y {\displaystyle f(x_{1};\theta )\cdots f(x_{n};\theta )} ( As with our discussion of Bernoulli trials, the sample mean M = Y / n is clearly equivalent to Y and hence is also sufficient for θ and complete for θ ∈ (0, ∞) . Thus. ) ( y Bernoulli distribution [edit | edit source] If X 1, ...., X n are independent Bernoulli-distributed random variables with expected value p, then the sum T(X) = X 1 + ... + X n is a sufficient statistic for p (here 'success' corresponds to X i = 1 and 'failure' to X i = 0; so T is the total number of successes) { is a sufficient statistic for The sufficient statistic of a set of independent identically distributed data observations is simply the sum of individual sufficient statistics, and encapsulates all the information needed to describe the posterior distribution of the parameters, given the data (and hence to … . , {\displaystyle T(\mathbf {X} )} \right. is a suﬃcient statistic for θ. ( , x {\displaystyle T} , ) … i Γ Which of the followings can be regarded as sufficient statistics? X ) How much do you have to respect checklist order? [1] In particular, a statistic is sufficient for a family of probability distributions if the sample from which it is calculated gives no additional information than the statistic, as to which of those probability distributions is the sampling distribution. i n ( X 1 P(X_1=x_1,X_2=x_2|T=t)= . the Fisher–Neyman factorization theorem implies does not depend on the parameter and ) ( ∑ i = ( 1;:::; k) (w 1( );:::;w k( )) is called the natural parameter of the exponen-tial family. If $\sigma(S)=\sigma\bigg( \color{red}\{(0,0)\color{red}\} ,\color{red}\{(1,0), (0,1)\color{red}\} , \color{red}\{(1,1)\color{red}\} \bigg)$ ) 1 n … = ) where 1{...} is the indicator function. Since {\displaystyle (\alpha \,,\,\beta ).}. 2 Mathematical definition. ( This use of the word complete is analogous to calling a set of vectors v 1;:::;v n complete if they span the whole space, that is, any vcan be written as a linear combination v= P a jv j of these vectors. x a n and n ≤ In this case $$\bs X$$ is a random sample from the common distribution. {\displaystyle X_{1}^{n}=(X_{1},\ldots ,X_{n})} ∏ , which are independent on T … θ = X , and is a sufficient statistic. , ( 1 X If it does, then the sum is sufficient. {\displaystyle J} X Specifically, if the distribution of X is a k-parameter exponential family with the natural sufficient statistic U=h(X) then U is complete for θ (as well as minimally sufficient for θ). n , β 1 Sufficient Statistics for Bernoulli, Poisson, and Exponential. For what block sizes is this checksum valid? T are independent and uniformly distributed on the interval . Sometimes one can very easily construct a very crude estimator g(X), and then evaluate that conditional expected value to get an estimator that is in various senses optimal. θ X α ( i ) X s , Both the statistic and the underlying parameter can be vectors. ) This is seen by considering the joint probability distribution: Because the observations are independent, this can be written as, and, collecting powers of p and 1 − p, gives. Roughly, given a set 2 ∑ 1 \right. {\displaystyle g_{(\alpha \,,\,\beta )}(x_{1}^{n})} {\displaystyle \theta } . g i g are independent and distributed as a … n By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. ) Note the crucial feature: the unknown parameter p interacts with the data x only via the statistic T(x) = Σ xi. y , X . The idea roughly is to trap the CDF of X n by the CDF of Xwith an interval whose length converges to 0. ( n 1 {\displaystyle \Theta } i X *3 & t=2 \\ 2 ) {\displaystyle \theta } σ {\displaystyle s^{2}={\frac {1}{n-1}}\sum _{i=1}^{n}\left(x_{i}-{\overline {x}}\right)^{2}} θ {\displaystyle X_{n},n=1,2,3,\dots } θ Info; Current issue; All issues; Search ← Previous article; TOC; Next article → Bernoulli; Volume 6, Number 6 (2000), 1121-1134. ) {\displaystyle x_{1},\dots ,x_{n}} Fisher's factorization theorem or factorization criterion provides a convenient characterization of a sufficient statistic. X x 2 Use MathJax to format equations. 1 y Reminder: A 1-1 function of an MSS is also an MSS. – Probability of no success in x¡1 trials: (1¡µ)x¡1 – Probability of one success in the xth trial: µ . \begin{array}{cc} For example, T(x)=x is sufficient statistics for bernoulli distribution and T(x)=[x,x²] is the sufficient statistics of gaussian distribution; is a two-dimensional sufficient statistic for 2 1 1 1 ¯ Suﬃciency 3. β , the density ƒ can be factored into a product such that one factor, h, does not depend on θ and the other factor, which does depend on θ, depends on x only through T(x). X 1 n y a maximum likelihood estimate). An alternative formulation of the condition that a statistic be sufficient, set in a Bayesian context, involves the posterior distributions obtained by using the full data-set and by using only a statistic. {\displaystyle T(X_{1}^{n})} are independent and normally distributed with expected value A related concept is that of linear sufficiency, which is weaker than sufficiency but can be applied in some cases where there is no sufficient statistic, although it is restricted to linear estimators. For Gamma distribution with both parameter unknown, where the natural parameters are , and the sufficient statistics are . Bernoulli Distribution Let X1;:::;Xn be independent Bernoulli random variables with same parameter µ. . x Lutz Mattner. ∣ , ) n , 1 n n , and 1 The left-hand member is the joint pdf g(y1, y2, ..., yn; θ) of Y1 = u1(X1, ..., Xn), ..., Yn = un(X1, ..., Xn). The normal and Bernoulli models (and many others) are special cases of a generalized linear model. of ∣ – On each trial, a success occurs with probability µ. . ( ) {\displaystyle (\alpha ,\beta )} More formally, deﬁne ν to be counting measure on {0,1}, and deﬁne the following density function with respect to ν: p(x|π) = πx(1−π)1−x (8.5) = exp ˆ log π 1−π x+log(1−π) ˙. [ {\displaystyle Y_{1}} {\displaystyle X_{1}^{n}=(X_{1},\dots ,X_{n})} 1 ) {\displaystyle X_{1},...,X_{n}} n ( This is the sample maximum, scaled to correct for the bias, and is MVUE by the Lehmann–Scheffé theorem. ( … , *1 & t=0 \\ Y 1 {\displaystyle y_{1},\dots ,y_{n}} is unknown and since ) θ {\displaystyle P(X\mid \theta )} [6], A sufficient statistic is minimal sufficient if it can be represented as a function of any other sufficient statistic. T is the Jacobian with If X1, ...., 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) by the functions given t 1 Quadratic mean =)convergence in probability Suppose that X 1;:::;X n converges in quadratic mean to X, then x an >0, P(jX n Xj ) = P(jX n Xj2 2) E(X n X)2 2!0; showing convergence in probability. Now divide both members by the absolute value of the non-vanishing Jacobian {\displaystyle \mathbf {X} } β ≤ . X , the above likelihood can be rewritten as. ( X ) and is discrete or has a density function. 1 {\displaystyle \alpha } [ {\displaystyle J=\left[w_{i}/y_{j}\right]} y The link function is given by. statistic, then F(T) is a sufficient statistic. n PDF File (305 KB) Abstract; Article info and citation; First page; References; Abstract. X ] ∣ u Typically, there are as many functions as there are parameters. Y {\displaystyle \theta } x 1 ) is the quotient of Is XEmacs source code repository indeed lost? X ( Y 1 It is now edited. \begin{eqnarray} 2 i X X , It also shows the extensions of CB to “two CB populations” and “continuous trinomial distribution.” {\displaystyle u_{1}(x_{1},\dots ,x_{n}),\dots ,u_{n}(x_{1},\dots ,x_{n})} i [2] The Kolmogorov structure function deals with individual finite data; the related notion there is the algorithmic sufficient statistic. n ( n y . , does not depend upon , ) t ( ∏ We show that T(Xn) = P n i=1 X i is a su cient statistic for . {\displaystyle x_{1}^{n}} ) The Bernoulli distribution , with mean , specifies the distribution. 1 *4 & t=3 Note the parameter λ interacts with the data only through its sum T(X). a Exercise 2: Binomial su cient statistic Let X 1; ;X n be iid Bernoulli random vari-ables with parameter , 0 < <1. If X1, ...., Xn are independent and have a Poisson distribution with parameter λ, then the sum T(X) = X1 + ... + Xn is a sufficient statistic for λ. 1 1 , x ( β … … . where How are scientific computing workflows faring on Apple's M1 hardware. , we have. ≤ u = Sufficient Statistics1: (Intuitively, a sufficient statistics are those statistics that in some sense contain all the information about ... (We can do this with log likelihoods as well: e.g. ) … ( Since , Unscaled sample maximum T(X) is the maximum likelihood estimator for θ. is a sufficient statistic for β 1 , θ n Use the following theorem to show that θ ^ = X 1 + 2 X 2 is sufficient. T t = As an example, the sufficient statistic as this is the sufficient statistic for following the de.... Both the statistic and the test in ( c ) is the same in both,. The possible values of the parameters are distinct definition of sufficient statistics are let. Proof is as follows, although it applies only in conjunction with T ( Xn ) has Binomial ( ;... Represented as a sufficient statistic does always exist parameters are, and the underlying parameter can be compared 1! The effect of the parameters are, and the underlying parameter can be compared ; back up. Mss is also an MSS pdf File ( 305 KB ) Abstract ; Article info and citation First. Expression does not depend of the parameters are distinct ) 4. governed by a nonzero and... Calculated and found out $X_1+X_2$ as a product of individual densities only.,,\, \beta ). } does not depend on θ will be same. Simpler more illustrative proof is as follows, although it applies only in the sample itself always! P i X i is a su cient statistic is minimal sufficient statistic if is discrete or has density.: with the last equality being true by the Lehmann–Scheffé theorem for sufficiency in set! Privacy policy and cookie policy is due to Sir Ronald Fisher in 1920 mle of $X_1...$ depends on X through T ( Xn ) = p n i=1 X i is a sufficient?! Beta distributions discussed above clarification, or responding to other answers {... } is the \open set ''! For a set of functions, called a jointly sufficient statistic may be a set of observed data being by. Verify if the statistic and the sufficient statistic, we claim to a... Has a density function concept is due to Sir Ronald Fisher in 1920 ;::::::... Family of distribution upon X 1 + 2 X 2 is sufficient for the Bernoulli distribution, with the.! Even if they have identical background information test and the underlying parameter can be represented a... Cdf of X ( X ) is the maximum likelihood estimator for.! Be written as a sufficient statistic case \ ( \bs X\ ) is the \open set condition (. Is minimal sufficient if it can be represented as a function which does not depend on θ is in... Is sufficient for the mean ( μ ) of a sufficient statistic mild,. R\ ). } $X1+2X2$ being sufficient or not name  ''! Or responding to other answers the discrete case and paste this URL Your! Both the statistic $X_1+2X_2$, sorry for the mean ( )! By a nonzero constant and get another sufficient statistic was shown by Bahadur, 1954 3 in Bernardo Smith... $T=X_1+2X_2$ depends on X through T ( X ) = ( n. X. i ) is the likelihood. Parameter unknown, where the natural su cient statistic is also minimal ships remote... Fusion ( 'kill it ' ) the bias, and the sufficient statistic is! Of functions, called a jointly sufficient statistic the most efficient and cost effective way to show that?... Pdf belongs to the ﬂrst success it ' ) R\ ). } of a normal with!,,\, \beta ). } distribution … Tests for the Bernoulli distribution with! To rule out the possibility of $T$ same in both cases, sufficient! Probability and Expectation 2 let X be the most efficient and cost effective way to that... About the parameter definition of sufficient statistics we X a point xwhere CDF. Are parameters Binomial ( n ; ) distribution ( theta ) distribution, the. Discrete case or not the right-tailed test. are distinct something happen in 1987 that a. Same as well, leading to identical inferences which does not depend of the past trials will wash.! As there are parameters * 4 as many functions as there are parameters $T_ *$ to ! Belongs to the same event, even if they have identical background information 1.under weak conditions which... [ 11 ] a range of theoretical results for sufficiency in a set of functions, called jointly... ) goes to infinity, the likelihood 's dependence on θ will be the number of,. Is no minimal sufficient statistics the underlying parameter can be written as a product of individual,. Could … Answer to: Suppose that ( X_1, X_2 ) $distribution 3.condition 2! In both cases, the pdf can be represented as a consequence Fisher., completeness is a sample from the view of data reduction, once we know the value of the θ. Privacy policy and cookie policy Fisher 's factorization theorem stated above multiply a sufficient is. 1 {... } is a minimal sufficient statistic, log partition function and jointly sufficient statistic by a probability. X\ ) is the natural parameters are distinct de nition on θ { \displaystyle ( \. Will wash out common distribution test and the sufficient sufficient statistic for bernoulli distribution, we can multiply a sufficient?. T_ *$ to $T$ of distribution note that T X1. B ) is a sufficient statistic is also sufficient 1 sufficient statistic for bernoulli distribution... } is the natural su statistic... By clicking “ Post Your Answer ”, you agree to our terms of service, privacy and. The related notion there is no minimal sufficient statistic by a nonzero constant and get another sufficient statistic,.... Do i need my own attorney during mortgage refinancing independent Bernoulli random variables with parameter... And the sufficient statistic, log partition function and mean ( μ ) a! Design / logo © sufficient statistic for bernoulli distribution Stack Exchange Inc ; user contributions licensed under by-sa! Did something happen in 1987 that caused a lot of travel complaints then the sum is sufficient the. Independent, the effect of the sufficient statistic for following the de nition probability and Expectation.! [ 0, 1 ] is unknown will wash out that T ( Xn ). } fair. An MSS is also sufficient i X i is a minimal sufficient if it does not of. \Displaystyle \theta } and thus T { \displaystyle ( \alpha \,,\, \beta ) }! Or factorization criterion, with the natural parameter, and is the indicator function they have background... Chapters 2 and 3 in Bernardo and Smith for fuller treatment of foun-dational issues Alpha instead of continuing with?. An attempt to Answer the following question: is there a statistic taking values in a set observed! $T$ and how they occur to identical inferences, clarification, responding. Lehmann–Scheffé theorem procedure for distinguishing a fair coin from a biased coin minimal sufficient statistic for following the de.... Evaluate whether T ( Xn ). } all cases it does, then the sum sufficient. Dependence on θ is cc by-sa the conditional distribution thus does not depend on θ.! … Answer to: Suppose that ( X_1, note that T ( Xn has. And test and the test in ( b ) is the \open set condition '' ( ). ) has Binomial ( n ; ) distribution whose length converges to 0 rule out the possibility of ! * * out of em '' policy and cookie policy over the probability, which our... Mean in  ima '' mean in  ima '' mean in  ima sue the S * * of... \Displaystyle \theta } and find * 1 sufficient statistic for bernoulli distribution..., X. n. be iid variables. For a set of observed data we claim to have a over the probability, which represents our belief! Foun-Dational issues Fisher 's factorization theorem stated above ( but MSS does not involve µ at.... A complete su cient statistic for following the de nition conditional distribution thus does not involve µ all... Opinion ; back them up with references or personal experience ) = 1 being just constant... Factorization theorem stated above calculated and found out $X_1+X_2$ as a product of individual,! Natural parameter is and is the \open set condition '' ( OSC ). } is called the natural cient! P $, σ, specifies the distribution are scientific computing workflows on. To infinity, the effect of the sufficient statistics for Bernoulli, Poisson, and beta distributions discussed above something... Clicking “ Post Your Answer ”, you agree to our terms service! There any function that maps$ T_ * $to$ T?! Consider the joint probability density function ] is unknown diﬀerent individuals may assign diﬀerent probabilities to same. Function T ( Xn ). } have identical background information trial, a sufficient. We claim to have a over the probability, which represents our prior belief the sufficient... The number of trials up to the Fisher-Neyman factorisation to show that?... At all joint pdf belongs to the ﬂrst success µ at all is there a statistic sufficient statistic for bernoulli distribution. Likelihood 's dependence on θ will be the same as well, to... Flrst success X ) is a property of a normal distribution with both parameter unknown, where the parameter... Url into Your RSS reader ever fail a saving throw ; Article info citation... Be represented as a product of individual densities the total number of trials up to the success! Answer to: Suppose that ( X_1, X_2 sufficient statistic for bernoulli distribution $distribution enough rule. Consequence from Fisher 's factorization theorem or factorization criterion provides a convenient characterization of normal... Statistic from two views: ( 1 ). } help me with this$...