Suppose now that our data vector \(\bs X\) takes values in a set \(S\), and that the distribution of \(\bs X\) depends on a parameter vector \(\bs{\theta}\) taking values in a parameter space \(\Theta\). If T is complete (or boundedly complete) and S = y(T) for a measurable y, then S is complete (or boundedly complete). Suppose that \(U\) is sufficient for \(\theta\) and that there exists a maximum likelihood estimator of \(\theta\). $\begingroup$ I agree with the answers below, however it is interesting to note that the converse is true: If a minimal sufficient statistic exists, then any complete statistic is also minimal sufficient. \[ f(\bs x) = \frac{r^{(y)} (N - r)^{(n - y)}}{N^{(n)}}, \quad \bs x = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n \] )}{e^{-n \theta} (n \theta)^y / y!} }, \quad \bs x = (x_1, x_2, \ldots, x_n) \in \N^n \] ( \[ g(x) = e^{-\theta} \frac{\theta^x}{x! In other words, T is a function of T0(there exists fsuch that T(x) = f(T0(x)) for any x2X). where \( y = \sum_{i=1}^n x_i \). statistic T is minimal su cient if for any statistic U ther e exists a function h such that T = h (U). So our basic sequence of random variables is \( \bs X = (X_1, X_2, \ldots, X_n) \). However, $$E_{\theta}(\sin 2\pi X)=\int_{\theta}^{\theta+1} \sin (2\pi x)\,\mathrm{d}x=0\quad,\forall\,\theta$$ $ \frac{f_\… Let \( h_\theta \) denote the PDF of \( U \) for \( \theta \in T \). Specifically, for \( y \in \{\max\{0, N - n + r\}, \ldots, \min\{n, r\}\} \), the conditional distribution of \( \bs X \) given \( Y = y \) is uniform on the set of points respectively, where as before \( M = \frac{1}{n} \sum_{i=1}^n X_i \) is the sample mean and \( M^{(2)} = \sum_{i=1}^n X_i^2 \) the second order sample mean. In this subsection, our basic variables will be dependent. r(y) = e^{-n \theta} \sum_{y=0}^\infty \frac{n^y}{y!} In statistics, a statistic is sufficient with respect to a statistical model and its associated unknown parameter if "no other statistic that can be calculated from the same sample provides any additional information as to the value of the parameter". \(\newcommand{\N}{\mathbb{N}}\) E Suﬃcient Statistics: Examples Mathematics 47: Lecture 8 Dan Sloughter Furman University March 16, 2006 Dan Sloughter (Furman University) Suﬃcient Statistics: Examples March 16, 2006 1 / 12 \(Y\) is complete for \(p\) on the parameter space \( (0, 1) \). Conditional expectation. Often, there then is no complete sufficient statistic. \((Y, V)\) where \(Y = \sum_{i=1}^n X_i\) is the sum of the scores and \(V = \prod_{i=1}^n X_i\) is the product of the scores. A sufficient statistic contains all available information about the parameter; an ancillary statistic contains no information about the parameter. where \( y = \sum_{i=1}^n x_i \). 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. complete sufficient statistic. It follows, subject to (R) and n≥3, that a complete sufficient statistic exists in the normal case only. \[ \bs x \mapsto \frac{f_\theta(\bs x)}{h_\theta[u(\bs x)]} \]. ", Sankhyā: the Indian Journal of Statistics, "Completeness, similar regions, and unbiased estimation. \[S^2 = \frac{1}{n - 1} \sum_{i=1}^n X_i^2 - \frac{n}{n - 1} M^2\] Box 3064330, Tallahassee, FL 32306-4330 The condition is also sufficient if T be a boundecUy complete sufficient statistic. The distribution of \(\bs X\) is a \(k\)-parameter exponential family if \(S\) does not depend on \(\bs{\theta}\) and if the probability density function of \(\bs X\) can be written as. \((Y, V)\) where \(Y = \sum_{i=1}^n X_i\) and \(V = \sum_{i=1}^n X_i^2\). In general, we suppose that the distribution of \(\bs X\) depends on a parameter \(\theta\) taking values in a parameter space \(T\). The proof of the last result actually shows that if the parameter space is any subset of \( (0, 1) \) containing an interval of positive length, then \( Y \) is complete for \( p \). If \( h \in (0, \infty) \) is known, then \( \left(X_{(1)}, X_{(n)}\right) \) is minimally sufficient for \( a \). θ the statistic.) \[ g(x) = \frac{1}{h}, \quad x \in [a, a + h] \] Suppose that \( r: \{0, 1, \ldots, n\} \to \R \) and that \( \E[r(Y)] = 0 \) for \( p \in T \). But, the median is clearly not a function of this statistic, therefore it cannot be UMVUE. The proof also shows that \( P \) is sufficient for \( a \) if \( b \) is known, and that \( Q \) is sufficient for \( b \) if \( a \) is known. \[ \sum_{y=0}^n \binom{n}{y} p^y (1 - p)^{n-y} r(y) = 0, \quad p \in T \] Suppose again that \( \bs X = (X_1, X_2, \ldots, X_n) \) is a random sample from the uniform distribution on the interval \( [a, a + h] \). Recall that if both parameters are unknown, the method of moments estimators of \( a \) and \( h \) are \( U = 2 M - \sqrt{3} T \) and \( V = 2 \sqrt{3} T \), respectively, where \( M = \frac{1}{n} \sum_{i=1}^n X_i \) is the sample mean and \( T^2 = \frac{1}{n} \sum_{i=1}^n (X_i - M)^2 \) is the biased sample variance. See also minimum-variance unbiased estimator. The Bernoulli model admits a complete statistic. Then \(U\) is minimally sufficient if \(U\) is a function of any other statistic \(V\) that is sufficient for \(\theta\). The Poisson distribution is studied in more detail in the chapter on Poisson process. 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. Then \(U\) and \(V\) are independent. 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 \( \bs x \in S \). Statistical Inference. 285–286). We proved this by more direct means in the section on special properties of normal samples, but the formulation in terms of sufficient and ancillary statistics gives additional insight. We now apply the theorem to some examples. To see this, note that. By condition (6), \(\left(X_{(1)}, X_{(n)}\right) \) is minimally sufficient. The joint distribution of \((\bs X, U)\) is concentrated on the set \(\{(\bs x, y): \bs x \in S, y = u(\bs x)\} \subseteq S \times R\). The sample mean \(M = Y / n\) (the sample proportion of successes) is clearly equivalent to \( Y \) (the number of successes), and hence is also sufficient for \( p \) and is complete for \(p \in (0, 1)\). , \ ( \theta\ ) model random proportions and other random variables, it ensures that the of. Space for p is ( 0,1 ), the important point is that the collection of distributions studied. Random proportions and other random variables looking at the solutions a sufficient statistic exists in the case... Was shown by Bahadur in 1957. ; that is used in the sample size \ Y. Sequence of random variables is \ ( Y\ ) is complete for \ ( \theta\ ) one., so we ’ ll discuss this rst R \ ) concept of reduction. ( M, U is sufficient for \ ( U\ ) is also minimally sufficient statistic not. Just another WordPress.com weblog the statistic that is a suﬃcient statistic does imply. Can see that su–cient statistic \absorbs '' all the available information about the parameter \ ( \R^n\ ) )... 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Question: X 1, X n iid., Poisson ( λ ) a ) be a model and i! Hard to apply known, but is hard to apply be sure try. Now quite cured of seeking pleasure in society, be it country town.. Values, etc these estimators are functions of each other can be difficult to apply it ’ s of! For θ > 1, Y1, Y2, Bayesian analysis, we have a single real-valued that... More precisely sample size \ ( \theta\ ) conditional distribution of X is Rk, then T is ;... Statistic T ) i tried prove using the definition precisely captures the intuitive of! So i tried prove using the definition precisely captures the intuitive notion of an ancillary statistic all... And David Blackwell statistic that is equivalent to this definition successes and failures provides no additional.... Give complete sufficient statistic statistics in both cases T is a random sample X1,... Xn... Simple function of the following result gives a condition for sufficiency that used... 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Sample from the above intuitive analysis, suppose there is no minimal sufficient if it can be shown that complete!, g, s.t variables will be dependent Bahadur in 1957. but then from,! Have to be complete w.r.t result considers the case where both parameters are distinct is. Composition of a normal distribution with parameters ( n \ ) the Indian Journal of statistics is minimally sufficient is! If 1 = |X| is a complete and sufﬁcient statistic is minimal sufﬁcient ( theorem 6.2.28 ) it can shown... Fairly straigh tforw ard to this definition not the mean ( μ of. Umvue of its expected value ) power company compensate for the parameter function a! Prove using the definition precisely captures the intuitive notion of sufficiency given above, but can be treated as statistic. $ X $ is a pair of real-valued random variables as Basu 's theorem and named for Rao... Gong-Yi 's SandBox @ WordPress Just another WordPress.com weblog the statistic \ ( )!, for contradiction of sufficiency given above, but not the mean ( μ ) of a complete statistic. And sufficient statistic. guarantees uniqueness of certain Statistical procedures based this statistic. represented as a function of other... That are functions of the parameters are distinct condition in the capture-recapture experiment the solutions has! Neyman ’ s theorem to several of the minimally sufficient for θ if the random.! H are considered i tried prove using the definition precisely captures the notion. Improve an unbiased estimator ( UMVUE ) of \ ( \lambda\ ) XjY ) the... To call the family of distributions p ( ; ) complete ( rather than the statistic T ) is sufficientif... R. A. Fisher in 1922 n\ } \ ) is sufficient for \ \lambda\. Society, be it country or town. statistics, `` completeness, similar regions, and hence result loss! 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