If. $f(\bs x) = g(x_1) g(x_2) \cdots g(x_n) = p^y (1 - p)^{n-y}, \quad \bs x = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n$ Our rst central result on su cient statistics will depend on the notion of conditional expectation, so we’ll discuss this rst. In this case, the outcome variable has the form Let $$U = u(\bs X)$$ be a statistic taking values in $$R$$, and let $$f_\theta$$ and $$h_\theta$$ denote the probability density functions of $$\bs X$$ and $$U$$ respectively. Here is the formal definition: A statistic $$U$$ is sufficient for $$\theta$$ if the conditional distribution of $$\bs X$$ given $$U$$ does not depend on $$\theta \in T$$. Suppose that $$r: \{0, 1, \ldots, n\} \to \R$$ and that $$\E[r(Y)] = 0$$ for $$p \in T$$. 0 Compare the method of moments estimates of the parameters with the maximum likelihood estimates in terms of the empirical bias and mean square error. $\E\left[r(Y)\right] = \int_0^\infty \frac{1}{\Gamma(n k) b^{n k}} y^{n k-1} e^{-y/b} r(y) \, dy = \frac{1}{\Gamma(n k) b^{n k}} \int_0^\infty y^{n k - 1} r(y) e^{-y / b} \, dy$ The statistic T is said to be boundedly complete for the distribution of X if this implication holds for every measurable function g that is also bounded. $\endgroup$ – knrumsey Jul 6 '18 at 22:24 Suppose that $$U = u(\bs X)$$ is a statistic taking values in a set $$R$$. θ if for all functions h E θ{h(T)} = 0 for all θ =⇒ h(t) = 0 a.s. 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 $$p$$. None of these estimators are functions of the minimally sufficient statistics, and hence result in loss of information. Question: X 1, X 2, ..., X n iid., Poisson(λ). Answer of Complete Sufficient Statistic. If $$\sigma^2$$ is known then $$Y = \sum_{i=1}^n X_i$$ is minimally sufficient for $$\mu$$. $\bs x \mapsto \frac{f_\theta(\bs x)}{h_\theta[u(\bs x)]}$. Alternatively, T 1 and T 2 can be used to construct reasonable estimates (read: maximum likelihood estimator) of μ and σ, respectively. When a wind turbine does not produce enough electricity how does the power company compensate for the loss? the statistic.) Compare the estimates of the parameters in terms of bias and mean square error. The Pareto distribution, named for Vilfredo Pareto, is a heavy-tailed distribution often used to model income and certain other types of random variables. The condition is also sufficient if T be a boundecUy complete sufficient statistic. Then $$U$$ is minimally sufficient for $$\theta$$ if the following condition holds: for $$\bs x \in S$$ and $$\bs y \in S$$ Casella, G. and Berger, R. L. (2001). Then, the jointly minimal sufficient statistic $\boldsymbol T = (T_1, \ldots, T_k)$ for $\boldsymbol\theta$ is complete. It is studied in more detail in the chapter on Special Distribution. (A case in which there is no minimal sufficient statistic was shown by Bahadur in 1957. V is rst-or der ancil lary if the exp e ctation E [(X)] do es not dep end on (i.e., E [V (X)] is c onstant). $D_y = \left\{(x_1, x_2, \ldots, x_n) \in \{0, 1\}^n: x_1 + x_2 + \cdots + x_n = y\right\}$. The joint PDF $$f$$ of $$\bs X$$ is given by We now apply the theorem to some examples. E $D_y = \left\{(x_1, x_2, \ldots, x_n) \in \{0, 1\}^n: x_1 + x_2 + \cdots + x_n = y\right\}$. The statistic $$Y$$ is sufficient for $$\theta$$. $h(y) = \binom{n}{y} p^y (1 - p)^{n-y}, \quad y \in \{0, 1, \ldots, n\}$, $$Y$$ is sufficient for $$p$$. Also, for f with (R) infinitely divisible but not normal, the order statistic is always minimal sufficient for the corresponding location-scale parameter model. In other words, T is a function of T0(there exists fsuch that T(x) = f(T0(x)) for any x2X). Sufficient statistics for three example models . In particular, suppose that $$V$$ is the unique maximum likelihood estimator of $$\theta$$ and that $$V$$ is sufficient for $$\theta$$. Suppose again that $$\bs X = (X_1, X_2, \ldots, X_n)$$ is a random sample from the uniform distribution on the interval $$[a, a + h]$$. of the (same) complete statistic are equal almost everywhere (i.e. German\ \ vollständige suffiziente Statistik. Suppose that $$\bs X = (X_1, X_2, \ldots, X_n)$$ is a random sample from the beta distribution with left parameter $$a$$ and right parameter $$b$$. These are functions of the sufficient statistics, as they must be. So the result follows from the factorization theorem (3). Compare the estimates of the parameters. Let $$f_\theta$$ denote the probability density function of $$\bs X$$ and suppose that $$U = u(\bs X)$$ is a statistic taking values in $$R$$. Thus $$\E_\theta(V \mid U)$$ is an unbiased estimator of $$\lambda$$. Lehman-Scheffe: (An unbiased estimator that is a function of a complete and sufficient statistic is the unique UMVUE!) We want to de ne E(XjY), the conditional expectation of X, given Y. Run the uniform estimation experiment 1000 times with various values of the parameter. \) for $$y \in \N$$. ) The following result gives an equivalent condition. Also, for f with (R) infinitely divisible but not normal, the order statistic is always minimal sufficient for the corresponding location-scale parameter model. Of course, the sample size $$n$$ is a positive integer with $$n \le N$$. [2] Let X be a random sample of size n such that each Xi has the same Bernoulli distribution with parameter p. Let T be the number of 1s observed in the sample. On the other hand, if $$b = 1$$, the maximum likelihood estimator of $$a$$ on the interval $$(0, \infty)$$ is $$W = -n / \sum_{i=1}^n \ln X_i$$, which is a function of $$P$$ (as it must be). If $$y \in \{\max\{0, N - n + r\}, \ldots, \min\{n, r\}\}$$, the conditional distribution of $$\bs X$$ given $$Y = y$$ is concentrated on $$D_y$$ and Sufficient Statistics Let U=h(X) be a statistic taking values in a set T. Intuitively, U is sufficient for θ if U contains all of the information about θ that is available in the entire data variable X. $$\newcommand{\E}{\mathbb{E}}$$ After some algebra, this can be written as From the factorization theorem (3), it follows that $$(U, V)$$ is sufficient for $$(a, b)$$. ( Statistical Inference. Moreover, in part (a), $$M$$ is complete for $$\mu$$ on the parameter space $$\R$$ and the sample variance $$S^2$$ is ancillary for $$\mu$$ (Recall that $$(n - 1) S^2 / \sigma^2$$ has the chi-square distribution with $$n - 1$$ degrees of freedom.) Given $$Y = y$$, $$\bs X$$ is concentrated on $$D_y$$ and A statistic T is said to be complete w.r.t. 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)). Because of the central limit theorem, the normal distribution is perhaps the most important distribution in statistics. Minimal sufficiency follows from condition (6). It is named for Ronald Fisher and Jerzy Neyman. It is easy to see that iff(t) is a one to one function andTis a suﬃcient statistic, thenf(T) is a suﬃcient statistic. Recall that the gamma distribution with shape parameter $$k \in (0, \infty)$$ and scale parameter $$b \in (0, \infty)$$ is a continuous distribution on $$(0, \infty)$$ with probability density function $$g$$ given by Then $$\left(X_{(1)}, X_{(n)}\right)$$ is minimally sufficient for $$(a, h)$$, where $$X_{(1)} = \min\{X_1, X_2, \ldots, X_n\}$$ is the first order statistic and $$X_{(n)} = \max\{X_1, X_2, \ldots, X_n\}$$ is the last order statistic. = The parameter $$\theta$$ is proportional to the size of the region, and is both the mean and the variance of the distribution. where $$y = \sum_{i=1}^n x_i$$. As always, be sure to try the problems yourself before looking at the solutions. A bit of though t will lead us to the idea that a su cien statistic T pro vides the most e cien t degree of data compression will ha v e the prop ert y … Let be the order statistics of a random sample from a We can rewrite the PDF as So our basic sequence of random variables is $$\bs X = (X_1, X_2, \ldots, X_n)$$. Moreover, The concept is perhaps best understood in terms of the Lehmann-Scheffé theorem “…if a sufficient statistic is boundedly complete it is minimal sufficient. Let $$h$$ denote the prior PDF of $$\Theta$$ and $$f(\cdot \mid \theta)$$ the conditional PDF of $$\bs X$$ given $$\Theta = \theta \in T$$. to denote the dependence on $$\theta$$. = $g(x) = e^{-\theta} \frac{\theta^x}{x! ) Lehmann-Scheffé Theorem. 2.A one-to-one function of a CSS is also a CSS (See later remarks). A sufficient statistic is minimal sufficient if it can be represented as a function of any other sufficient statistic. \[ h(\theta \mid \bs x) = \frac{h(\theta) G[u(\bs x), \theta]}{\int_T h(t) G[u(\bs x), t] dt}$ In part (v) you need to nd an unbiased estimator of 2 that is a function of the complete su cient statistic. Of course, the important point is that the conditional distribution does not depend on $$\theta$$. 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. It would be more accurate to call the family of distributions p(;) complete (rather than the statistic T). Department of Statistics, Florida State University 214 Rogers Building (OSB), 117 N. Woodward Ave. P.O. As an example, the sample mean is sufficient for the mean (μ) of a normal distribution with known variance. Compare the estimates of the parameters in terms of bias and mean square error. Construction of a minimal su cien t statistic is fairly straigh tforw ard. Exercise. r(y) \theta^y\] $f(\bs x) = g(x_1) g(x_2) \cdots g(x_n) = \frac{1}{(2 \pi)^{n/2} \sigma^n} \exp\left[-\frac{1}{2 \sigma^2} \sum_{i=1}^n (x_i - \mu)^2\right], \quad \bs x = (x_1, x_2 \ldots, x_n) \in \R^n$ 0 A statistic V is ancil lary if its distribution do es not dep end on . 1. Let X;Y be random variables. Then there exists a positive constant $$C$$ such that $$h_\theta(y) = C G(y, \theta)$$ for $$\theta \in T$$ and $$y \in R$$. Run the normal estimation experiment 1000 times with various values of the parameters. A useful characterization of minimal sufficiency is that when the density fθ exists, S(X) is minimal sufficientif and only if 1. List elements digit difference sort Conservation of Mass and Energy Could you please stop shuffling the deck and play already? ON STATISTICS INDEPENDENT OF A COMPLETE SUFFICIENT STATISTIC By D. BASU Indian Statistical Institute, Calcutta, 1. Given $$Y = y \in \N$$, random vector $$\bs X$$ takes values in the set $$D_y = \left\{\bs x = (x_1, x_2, \ldots, x_n) \in \N^n: \sum_{i=1}^n x_i = y\right\}$$. A statistic T(X) is complete iff for any function g not depending on q, Eq[g(T)] = 0 for all q 2 implies Pq(g(T) = 0) = 1 for all q 2 . It would be more precise to say the family of densities of T F T = {f T(t;θ),θ ∈ Θ} One of the most famous results in statistics, known as Basu's theorem (see Basu, 1955), says that a complete sufficient statistic and any ancillary statistic are stochastically independent. 1 Lehman-Scheffe: (An unbiased estimator that is a function of a complete and sufficient statistic is the unique UMVUE! x_2! ( θ Examples exists that when the minimal sufficient statistic is not complete then several alternative statistics exist for unbiased estimation of θ, while some of them have lower variance than others.[5]. Hence if $$\bs x, \bs y \in S$$ and $$v(\bs x) = v(\bs y)$$ then $f(\bs x) = g(x_1) g(x_2) \cdots g(x_n) = \frac{1}{B^n(a, b)} (x_1 x_2 \cdots x_n)^{a - 1} [(1 - x_1) (1 - x_2) \cdots (1 - x_n)]^{b-1}, \quad \bs x = (x_1, x_2, \ldots, x_n) \in (0, 1)^n$ }, \quad \bs x = (x_1, x_2, \ldots, x_n) \in \N^n \] Continuous uniform distributions are studied in more detail in the chapter on Special Distributions. However, $$E_{\theta}(\sin 2\pi X)=\int_{\theta}^{\theta+1} \sin (2\pi x)\,\mathrm{d}x=0\quad,\forall\,\theta$$ Suppose now that $$\bs X = (X_1, X_2, \ldots, X_n)$$ is a random sample of size $$n$$ from the Poisson distribution with parameter $$\theta$$. $\E[r(Y)] = \sum_{y=0}^n r(y) \binom{n}{k} p^y (1 - p)^{n-y} = (1 - p)^n \sum_{y=0}^n r(y) \binom{n}{y} \left(\frac{p}{1 - p}\right)^y$ Recall that the method of moments estimators of $$a$$ and $$b$$ are Casella, G. and Berger, R. L. (2001). This concept was introduced by R. A. Fisher in 1922. Once again, the definition precisely captures the notion of minimal sufficiency, but is hard to apply. The estimator of $$r$$ is the one that is used in the capture-recapture experiment. Let $$h_\theta$$ denote the PDF of $$U$$ for $$\theta \in T$$. In the continuous case, a similar inter- pretation works. Here i have attached my work so far. In general, $$S^2$$ is an unbiased estimator of the distribution variance $$\sigma^2$$. This result follows from the first displayed equation for the PDF $$f(\bs x)$$ of $$\bs X$$ in the proof of the previous theorem. Suppose that $$V = v(\bs X)$$ is a statistic taking values in a set $$R$$. Recall that $$M$$ is the method of moments estimator of $$\theta$$ and is the maximum likelihood estimator on the parameter space $$(0, \infty)$$. 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 $$b \in (0, \infty)$$. $U = \left( \frac{n-1}{n} \right)^Y$. 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. In particular, these conditions always hold if the random variables (associated with Pθ ) are all discrete or are all continuous. De nition 3. Often, there then is no complete sufficient statistic. For some parametric families, a complete sufficient statistic does not exist (for example, see Galili and Meilijson 2016 [3]). . Reminder: A 1-1 }, \quad x \in \N \] Example 6.2.15. $\E_\theta\left[r(U)\right] = 0 \text{ for all } \theta \in T \implies \P_\theta\left[r(U) = 0\right] = 1 \text{ for all } \theta \in T$. $$(Y, V)$$ where $$Y = \sum_{i=1}^n X_i$$ and $$V = \sum_{i=1}^n X_i^2$$. 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. But $$X_i^2 = X_i$$ since $$X_i$$ is an indicator variable, and $$M = Y / n$$. F or any xe d and 0 a statistic U is su cient i p (x) p 0 (x) function only of U (x). 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 In other words, S(X) is minimal sufficientif and only if 1. Equivalently, $$\bs X$$ is a sequence of Bernoulli trials, so that in the usual langauage of reliability, $$X_i = 1$$ if trial $$i$$ is a success, and $$X_i = 0$$ if trial $$i$$ is a failure. Abbreviation: CSS )MSS. An UMVUE of the parameter $$\P(X = 0) = e^{-\theta}$$ for $$\theta \in (0, \infty)$$ is 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. This results follow from the second displayed equation for the PDF $$f(\bs x)$$ of $$\bs X$$ in the proof of the previous theorem. The Poisson distribution is named for Simeon Poisson and is used to model the number of random points in region of time or space, under certain ideal conditions. θ So far, in all of our examples, the basic variables have formed a random sample from a distribution. $f(\bs x) = \frac{1}{h^n} \bs{1}[x_{(1)} \ge a] \bs{1}[x_{(n)} \le a + h], \quad \bs x = (x_1, x_2, \ldots, x_n) \in \R^n$ Probability density functions, called a JOINTLY sufficient statistic, therefore it can be! 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