$\frac{f_\theta(\bs x)}{f_\theta(\bs{y})} \text{ is independent of } \theta \text{ if and only if } u(\bs x) = u(\bs{y})$. The normal distribution is often used to model physical quantities subject to small, random errors, and is studied in more detail in the chapter on Special Distributions. Then $$U$$ is a complete statistic for $$\theta$$ if for any function $$r: R \to \R$$ Suppose that $$\bs X = (X_1, X_2, \ldots, X_n)$$ is a random sample of size $$n$$ from the Bernoulli distribution with parameter $$p$$. However, as we will see, this is not necessarily the case; $$j$$ can be smaller or larger than $$k$$. It is closely related to the idea of identifiability, but in statistical theory it is often found as a condition imposed on a sufficient statistic from which certain optimality results are derived. Suppose that $$\bs X = (X_1, X_2, \ldots, X_n)$$ is a random sample from the uniform distribution on the interval $$[a, a + h]$$. ( $\E_\theta\left[r(U)\right] = 0 \text{ for all } \theta \in T \implies \P_\theta\left[r(U) = 0\right] = 1 \text{ for all } \theta \in T$. Clearly $$M = Y / n$$ is equivalent to $$Y$$ and $$U = V^{1/n}$$ is equivalent to $$V$$. Indeed if the sampling were with replacement, the Bernoulli trials model with $$p = r / N$$ would apply rather than the hypergeometric model. Then, Recall that the beta distribution with left parameter $$a \in (0, \infty)$$ and right parameter $$b \in (0, \infty)$$ is a continuous distribution on $$(0, 1)$$ with probability density function $$g$$ given by Sufficient statistic Last updated October 06, 2019. Of course, the sample size $$n$$ is a positive integer with $$n \le N$$. x_2! minimal sufﬁcient statistic is unique in the sense that two statistics that are functions of each other can be treated as one statistic. If $$r: \N \to \R$$ then Of course, the sufficiency of $$Y$$ follows more easily from the factorization theorem (3), but the conditional distribution provides additional insight. In other words, this statistic has a smaller expected loss for any convex loss function; in many practical applications with the squared loss-function, it has a smaller mean squared error among any estimators with the same expected value. Also, E(g(T)) is a polynomial in r and, therefore, can only be identical to 0 if all coefficients are 0, that is, g(t) = 0 for all t. It is important to notice that the result that all coefficients must be 0 was obtained because of the range of r. Had the parameter space been finite and with a number of elements less than or equal to n, it might be possible to solve the linear equations in g(t) obtained by substituting the values of r and get solutions different from 0. Each of the following pairs of statistics is minimally sufficient for $$(k, b)$$. However, a suﬃcient statistic does not have to be any simpler than the data itself. So far, in all of our examples, the basic variables have formed a random sample from a distribution. This variable has the hypergeometric distribution with parameters $$N$$, $$r$$, and $$n$$, and has probability density function $$h$$ given by These are functions of the sufficient statistics, as they must be. $g(x) = \frac{1}{\Gamma(k) b^k} x^{k-1} e^{-x / b}, \quad x \in (0, \infty)$ Suppose that given a parameter θ > 1, Y1, Y2,. Hence if $$\bs x, \bs y \in S$$ and $$v(\bs x) = v(\bs y)$$ then To understand this rather strange looking condition, suppose that $$r(U)$$ is a statistic constructed from $$U$$ that is being used as an estimator of 0 (thought of as a function of $$\theta$$). Construction of a minimal su cien t statistic is fairly straigh tforw ard. A useful characterization of minimal sufficiency is that when the density fθ exists, S(X) is minimal sufficientif and only if 1. the statistic.) Typically, the sufficient statistic is a simple function of the data, e.g. $$\newcommand{\N}{\mathbb{N}}$$ We must know in advance a candidate statistic $$U$$, and then we must be able to compute the conditional distribution of $$\bs X$$ given $$U$$. The parameter $$\theta$$ is proportional to the size of the region, and is both the mean and the variance of the distribution. The hypergeometric distribution is studied in more detail in the chapter on Finite Sampling Models. $$\newcommand{\cor}{\text{cor}}$$ Once again, the sample mean $$M = Y / n$$ is equivalent to $$Y$$ and hence is also sufficient for $$(N, r)$$. Suppose that $$W$$ is an unbiased estimator of $$\lambda$$. In statistics, a sufficient statistic is a statistic which has the property of sufficiency with respect to a statistical model and its associated unknown parameter, meaning that "no other statistic which can be calculated from the same sample provides any additional information as to the value of the parameter". But by doing like this , seems like that i am going to prove that T is a complete sufficient statistic. Other settings without complete sufficient statistics are missing data, censored time‐to‐event data, random visit times and joint modeling of longitudinal and time‐to‐event data. a sufficient statistic :Y'-complete if Eef( 1) = 0 for all a and f E :Y' implies f(t) = 0 (a.e. For example, if T is minimal sufﬁcient, then so is (T;eT), but no one is going to use (T;eT). Hence $$f_\theta(\bs x) = h_\theta[u(\bs x)] r(\bs x)$$ for $$(\bs x, \theta) \in S \times T$$ and so $$(\bs x, \theta) \mapsto f_\theta(\bs x)$$ has the form given in the theorem. statistic, then f(T) is a suﬃcient statistic. ( A simple instance is $X\sim U (\theta,\theta+1)$ where $\theta\in \mathbb R$. Note that $$T^2$$ is not a function of the sufficient statistics $$(Y, V)$$, and hence estimators based on $$T^2$$ suffer from a loss of information. $D_y = \left\{(x_1, x_2, \ldots, x_n) \in \{0, 1\}^n: x_1 + x_2 + \cdots + x_n = y\right\}$. The Bernoulli distribution is named for Jacob Bernoulli and is studied in more detail in the chapter on Bernoulli Trials, Let $$Y = \sum_{i=1}^n X_i$$ denote the number of successes. How to find sufficient complete statistic for the density $f(x\mid\theta)=e^{-(x-\theta)}\exp(-e^{-(x-\theta)})$? g {\displaystyle {\text{if }}\operatorname {E} _{\theta }(g(T))=0{\text{ for all }}\theta {\text{ then }}\mathbf {P} _{\theta }(g(T)=0)=1{\text{ for all }}\theta .}. The posterior distribution depends on the data only through the sufficient statistic $$Y$$, as guaranteed by theorem (9). We select a random sample of $$n$$ objects, without replacement from the population, and let $$X_i$$ be the type of the $$i$$th object chosen. So i tried prove using the definition of the complete sufficient statistic. Suppose that $$V = v(\bs X)$$ is a statistic taking values in a set $$R$$. Let's suppose that $$\Theta$$ has a continuous distribution on $$T$$, so that $$f(\bs x) = \int_T h(t) G[u(\bs x), t] r(\bs x) dt$$ for $$\bs x \in S$$. $D_y = \left\{(x_1, x_2, \ldots, x_n) \in \{0, 1\}^n: x_1 + x_2 + \cdots + x_n = y\right\}$. The next result is the Rao-Blackwell theorem, named for CR Rao and David Blackwell. The examples above describe such a situation. Then $$U$$ and $$V$$ are independent. If the distribution of $$V$$ does not depend on $$\theta$$, then $$V$$ is called an ancillary statistic for $$\theta$$. By the Rao-Blackwell theorem (10), $$\E(W \mid U)$$ is also an unbiased estimator of $$\lambda$$ and is uniformly better than $$W$$. For some parametric families, a complete sufficient statistic does not exist (for example, see Galili and Meilijson 2016 ). This results follow from the second displayed equation for the PDF $$f(\bs x)$$ of $$\bs X$$ in the proof of the previous theorem. It is named for Ronald Fisher and Jerzy Neyman. Then $$\left(P, X_{(1)}\right)$$ is minimally sufficient for $$(a, b)$$ where $$P = \prod_{i=1}^n X_i$$ is the product of the sample variables and where $$X_{(1)} = \min\{X_1, X_2, \ldots, X_n\}$$ is the first order statistic. A sufficient statistic is minimal sufficient if it can be represented as a function of any other sufficient statistic. 02/23/20 - Inference on vertex-aligned graphs is of wide theoretical and practical importance. Complete Statistic: A sufficient statistic T is called a complete statistic if no function of it has zero expected value for all distributions concerned unless this function itself is zero for all possible distributions concerned (except possibly a set of measure zero).. We now apply the theorem to some examples. Request PDF | On a complete and sufficient statistic for the correlated Bernoulli random graph model | Inference on vertex-aligned graphs is of wide theoretical and practical importance. In this subsection, our basic variables will be dependent. As usual, the most important special case is when $$\bs X$$ is a sequence of independent, identically distributed random variables. Suppose again that $$\bs X = (X_1, X_2, \ldots, X_n)$$ is a random sample of size $$n$$ from the gamma distribution with shape parameter $$k \in (0, \infty)$$ and scale parameter $$b \in (0, \infty)$$. $$(M, U)$$ where $$M = Y / n$$ is the sample (arithmetic) mean of $$\bs X$$ and $$U = V^{1/n}$$ is the sample geometric mean of $$\bs X$$. statistic T is minimal su cient if for any statistic U ther e exists a function h such that T = h (U). Observe also that neither p nor 1 − p can be 0. But by doing like this , seems like that i am going to prove that T is a complete sufficient statistic. Pr o of. Let T(X) be a complete sufficient statistic for a parameter θand φ(T)be any estimator based only on T. Then φ(T)is the unique best unbiased estimator of its expected value (i.e. Hence $$(M, T^2)$$ is equivalent to $$(Y, V)$$ and so $$(M, T^2)$$ is also minimally sufficient for $$(\mu, \sigma^2)$$. $$\left(M, S^2\right)$$ where $$M = \frac{1}{n} \sum_{i=1}^n X_i$$ is the sample mean and $$S^2 = \frac{1}{n - 1} \sum_{i=1}^n (X_i - M)^2$$ is the sample variance. the sum of all the data points. (pp. If a minimal sufficient statistics is not complete, then there is no complete sufficient statistics. The joint PDF $$f$$ of $$\bs X$$ is given by Run the gamma estimation experiment 1000 times with various values of the parameters and the sample size $$n$$. g $\E\left[\left(\frac{n - 1}{n}\right)^Y\right] = \exp \left[n \theta \left(\frac{n - 1}{n} - 1\right)\right] = e^{-\theta}, \quad \theta \in (0, \infty)$ Then we have $\bs X = (X_1, X_2, \ldots, X_n)$ Casella, G. and Berger, R. L. (2001). Let $$h_\theta$$ denote the PDF of $$U$$ for $$\theta \in T$$. Note that $$M = \frac{1}{n} Y, \; S^2 = \frac{1}{n - 1} V - \frac{n}{n - 1} M^2$$. Learn how and when to remove this template message, "An Example of an Improvable Rao–Blackwell Improvement, Inefficient Maximum Likelihood Estimator, and Unbiased Generalized Bayes Estimator", "Completeness, similar regions, and unbiased estimation. Of course, $$\binom{n}{y}$$ is the cardinality of $$D_y$$. If $$U$$ is sufficient for $$\theta$$ then $$V$$ is a function of $$U$$ by the previous theorem. x_2! Once again, the definition precisely captures the notion of minimal sufficiency, but is hard to apply. $f(\bs x) = g(x_1) g(x_2) \cdots g(x_n) = \frac{1}{\Gamma^n(k) b^{nk}} (x_1 x_2 \ldots x_n)^{k-1} e^{-(x_1 + x_2 + \cdots + x_n) / b}, \quad \bs x = (x_1, x_2, \ldots, x_n) \in (0, \infty)^n$  for all  A simple characterisation of incompleteness is given for the exponential family in terms of the mapping between the sufficient statistic and the parameter, based upon the implicit function theorem. $h(\theta \mid \bs x) = \frac{h(\theta) f(\bs x \mid \theta)}{f(\bs x)}, \quad \theta \in T$ Gong-Yi's SandBox@WordPress Just another WordPress.com weblog In other words, S(X) is minimal sufficient if and only if S(X) is sufficient, and; if T(X) is sufficient, then there exists a function f such that S(X) = f(T(X)). Recall that the method of moments estimators of $$a$$ and $$b$$ are The last integral can be interpreted as the Laplace transform of the function $$y \mapsto y^{n k - 1} r(y)$$ evaluated at $$1 / b$$. By a simple application of the multiplication rule of combinatorics, the PDF $$f$$ of $$\bs X$$ is given by If $$\mu$$ is known then $$U = \sum_{i=1}^n (X_i - \mu)^2$$ is sufficient for $$\sigma^2$$. As with our discussion of Bernoulli trials, the sample mean $$M = Y / n$$ is clearly equivalent to $$Y$$ and hence is also sufficient for $$\theta$$ and complete for $$\theta \in (0, \infty)$$. If the location parameter $$a$$ is known, then the largest order statistic is sufficient for the scale parameter $$h$$. θ $f(\bs x) = g(x_1) g(x_2) \cdots g(x_n) = \frac{1}{h^n}, \quad \bs x = (x_1, x_2, \ldots x_n) \in [a, a + h]^n$ The population size $$N$$ is a positive integer and the type 1 size $$r$$ is a nonnegative integer with $$r \le N$$. $\E\left[r(Y)\right] = \sum_{y=0}^\infty e^{-n \theta} \frac{(n \theta)^y}{y!} Conversely, suppose that $$(\bs x, \theta) \mapsto f_\theta(\bs x)$$ has the form given in the theorem. F or any xe d and 0 a statistic U is su cient i p (x) p 0 (x) function only of U (x). The next result shows the importance of statistics that are both complete and sufficient; it is known as the Lehmann-Scheffé theorem, named for Erich Lehmann and Henry Scheffé. Intuitively, a minimal sufficient statistic most efficiently captures all possible information about the parameter θ. Bounded completeness occurs in Basu's theorem, which states that a statistic that is both boundedly complete and sufficient is independent of any ancillary statistic. So in this case, we have a single real-valued parameter, but the minimally sufficient statistic is a pair of real-valued random variables. If $$b$$ is known, the method of moments estimator of $$a$$ is $$U_b = b M / (1 - M)$$, while if $$a$$ is known, the method of moments estimator of $$b$$ is $$V_a = a (1 - M) / M$$. Also, for f with (R) infinitely divisible but not normal, the order statistic is always minimal sufficient for the corresponding location-scale parameter model. A complete statistic T “… is a complete statistic if the family of probability densities {g(t; θ) is complete” (Voinov & Nikulin, 1996, p. 51). Let $$M = \frac{1}{n} \sum_{i=1}^n X_i$$ denote the sample mean and $$U = (X_1 X_2 \ldots X_n)^{1/n}$$ the sample geometric mean, as before. Next, $$\E_\theta(V \mid U)$$ is a function of $$U$$ and $$\E_\theta[\E_\theta(V \mid U)] = \E_\theta(V) = \lambda$$ for $$\theta \in \Theta$$. Then, the jointly minimal sufficient statistic \boldsymbol T = (T_1, \ldots, T_k) for \boldsymbol\theta is complete. Suppose that $$r: \{0, 1, \ldots, n\} \to \R$$ and that $$\E[r(Y)] = 0$$ for $$p \in T$$. French\ \ complète statistique suffisante. Let $$g$$ denote the probability density function of $$V$$ and let $$v \mapsto g(v \mid U)$$ denote the conditional probability density function of $$V$$ given $$U$$. Recall that $$Y$$ has the binomial distribution with parameters $$n$$ and $$p$$, and has probability density function $$h$$ defined by I. None of these estimators are functions of the minimally sufficient statistics, and hence result in loss of information. which states that if a statistic that is unbiased, complete and sufficient for some parameter θ, then it is the best mean-unbiased estimator for θ. However, as noted above, there usually exists a statistic $$U$$ that is sufficient for $$\theta$$ and has smaller dimension, so that we can achieve real data reduction.  \frac{f_\… We now apply the theorem to some examples. It is studied in more detail in the chapter on Special Distribution. $$\newcommand{\var}{\text{var}}$$ Recall that $$M$$ is the method of moments estimator of $$\theta$$ and is the maximum likelihood estimator on the parameter space $$(0, \infty)$$. θ Su cient Statistics Jimin Ding, Math WUSTLMath 494Spring 2018 4 / 36. A sufficient statistic is minimal sufficient if it can be represented as a function of any other sufficient statistic. \[ f_\theta(\bs x) = G[u(\bs x), \theta] r(\bs x); \quad \bs x \in S, \; \theta \in T$. It would be more accurate to call the family of distributions p(;) complete (rather than the statistic T). The Pareto distribution, named for Vilfredo Pareto, is a heavy-tailed distribution often used to model income and certain other types of random variables. To see this, note that. Also, a minimal sufficient statistic need not exist. ) Specifically, for $$y \in \N$$, the conditional distribution of $$\bs X$$ given $$Y = y$$ is the multinomial distribution with $$y$$ trials, $$n$$ trial values, and uniform trial probabilities. In particular, the sampling distributions from the Bernoulli, Poisson, gamma, normal, beta, and Pareto considered above are exponential families. Then the posterior distribution of $$P$$ given $$\bs X$$ is beta with left parameter $$a + Y$$ and right parameter $$b + (n - Y)$$. Similarly, $$M = \frac{1}{n} Y$$ and $$T^2 = \frac{1}{n} V - M^2$$. δ(X ) may be ineﬃcient ignoring important information in X that is relevant to θ. δ(X ) may be needlessly complex using information from X that is irrelevant to θ. The notion of completeness has many applications in statistics, particularly in the following two theorems of mathematical statistics. $g(x) = e^{-\theta} \frac{\theta^x}{x! Recall that the Pareto distribution with shape parameter $$a \in (0, \infty)$$ and scale parameter $$b \in (0, \infty)$$ is a continuous distribution on $$[b, \infty)$$ with probability density function $$g$$ given by The joint PDF $$f$$ of $$\bs X$$ is defined by ( \[ f(\bs x) = g(x_1) g(x_2) \cdots g(x_n) = \frac{a^n b^{n a}}{(x_1 x_2 \cdots x_n)^{a + 1}} \bs{1}\left(x_{(n)} \ge b\right), \quad (x_1, x_2, \ldots, x_n) \in (0, \infty)^n$ Suppose that $$U$$ is sufficient for $$\theta$$ and that there exists a maximum likelihood estimator of $$\theta$$. The statistic T is said to be complete for the distribution of X if, for every measurable function g,:, if  It can be shown that a complete and sufﬁcient statistic is minimal sufﬁcient (Theorem 6.2.28). Bounded completeness also occurs in Bahadur's theorem. 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