$$ Ben Lambert . As for the first question, I found a reference - Theorem 2.29, Mark J. Schervish, Theory of Statistics, 1995. = For example, if T is minimal sufcient, then so is (T;eT), but no one is going to use (T;eT). Stack Overflow for Teams is moving to its own domain! New Orleans: (985) 781-9190 | New York City: (646) 820-9084 But: 2. question: how to show that if the ratio is constant as a function of $\theta$ then $(x_{(1)},x_{(n)})=(y_{(1)},y_{(n)})$? &= \prod_{i=1}^n \text{U}(x_i| -\theta, \theta) \\[6pt] These short videos. Please take a moment to review my steps and let me know where exactly I went wrong because the correct answer is shown as max of $|Xi|$. This use of the word complete is analogous to calling a set of vectors v 1;:::;v n complete if they span the whole space, that is, any vcan be written as a linear combination v= P a jv j of . Since $|X|$ follows $U(0,\theta)$, I will have $\frac{|X|}{\theta}$ following $U(0,1)$ which means that it is ancillary. The joint density of the sample can be written as: $f(x_1, x_2,..,x_n|\theta) = (\frac{1}{2\theta})^n,-\theta-Y_1 \land \theta>Y_n$$, $$\mathbf 1_{(-\theta,\theta)}(Y_1) \cdot \mathbf 1_{(-\theta,\theta)}(Y_n) = \mathbf 1_{(-\theta,\theta)}(Y^*)$$, $$K_1(Y_1,Y_2;\theta)=(\frac{1}{2})^n \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_1) \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_n)$$, $$\theta-1\theta \land Y_n-1<\theta$$, [Math] Minimal Sufficient statistic for Uniform($\theta, \theta+1$), [Math] Degree of the minimal sufficient statistic for $\theta$ in $U(\theta-1,\theta+1)$ distribution. To demonstrate sufficiency formally, we note that the likelihood function reduces to: $$\begin{align} In the examples discussed above the obtained sufficient statistics are also necessary. Minimal sufficiency follows from the fact that there is no sufficient statistic from which this statistic cannot be obtained. i). \mathcal{L}(\theta)=\frac{1}{(2\pi)^{n/2}}\exp\{-\sum_{i=1}^nX_i^2/2 +\bar{X}_n \theta -n\theta^2/2\} It is easy to show that $T(X) = (X_{(1)},X_{(n)})$ is a sufficient statistic for $\theta$ where $X_{(1)}$ and $X_{(n)}$ stands for the minimum and the maximum from the sample $X_1,\dots,X_n$ respectively. Your intuition should show that $(X_{(1)}, X_{(n)})$ is not minimal, because the support is symmetric: if you observed $\boldsymbol x = (3, -1, 4, -5, 1, 0)$, you immediately know that $\theta \ge 5$, and that this is the best you can do in the sense that the other data contribute no additional information about the parameter. How do planetarium apps and software calculate positions? $\frac{0}{0}$)? When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. $(\theta,c)$ unknown, Doubt Regarding the Sufficient Statistic Problem. And the latter, the pair, cannot be a minimal sufficient statistic if the former is sufficient because if you know the maximum you don't have enough information to find the pair, and the pair is sufficient. Theorem 6.1 (Basu Theorem) If T (X) T ( X) is a complete and minimal sufficient statistic, then T (X) T ( X) is independent of every ancillary statistic. A necessary sufficient statistic realizes the utmost possible reduction of a statistical problem. More than a million books are available now via BitTorrent. $$ What is the difference between an "odor-free" bully stick vs a "regular" bully stick? On the other hand, we have also shown that $Y^*=max\{Y_1,Y_n\}$ is the single-dimensional and (thus) minimal sufficient sufficient statistic for $\theta$ for a symmetric Uniform distribution. The likelihood function is given by $$ L(\mathbf{x}| \theta) = \prod \mathbf{1} [ \theta < x_i < \theta+1] = \mathbf{1} [\min (\mathbf{x}) > \theta] \mathbf{1} [\max (\mathbf{x}) < \theta+1]$$. Starting with the unbiased estimate To = X1, derive the MVUE of theta. In order to skirt any indeterminacy problems, we can take the first condition to be $f_\theta (x) = k(x,y) f_\theta (y)$. giacomomaraglino Asks: Different Correlation Coefficents with different Time Ranges I built a Time-Series that displays the price of the Electricty Price in South Italy and two of their most important commodities (commodities, gas) used to produce the eletrical energy. I once read that if a function of sufficient statistic is ancillary, then it cannot be complete. Refer to the lecture notes here on page 5. It is easy to show that $T(X) = (X_{(1)},X_{(n)})$ is a sufficient statistic for $\theta$ where $X_{(1)}$ and $X_{(n)}$ stands for the minimum and the maximum from the sample $X_1,\dots,X_n$ respectively. Counting from the 21st century forward, what is the last place on Earth that will get to experience a total solar eclipse? Does English have an equivalent to the Aramaic idiom "ashes on my head"? +X n and let f be the joint density of X 1, X 2, . With lambda ~ 0, critical density is 0.5 x 10**29 g /cm**3 . Connect and share knowledge within a single location that is structured and easy to search. A statistic Tis called complete if Eg(T) = 0 for all and some function gimplies that P(g(T) = 0; ) = 1 for all . I Let T = X (n) and let f be the joint density of X 1, X Let $X_1,\dots,X_n$ be a sample from uniform distribution on $(-\theta,\theta)$ with parameter $\theta>0$. Neither $X_{(n)}$ nor $(X_{(1)},X_{(n)})$ is minimal sufficient: Specially the last line made everything clear to me. L_\mathbf{x}(\theta) Find a minimal sufficient statistic for $\theta$. By the factorization theorem, it is easy to verify that the vector $\mathbf Y = (Y_1,Y_2)$ where $Y_1 = X_\left(1\right)$ and $Y_2=X_\left(n\right)$ is a joint sufficient vector of degree two for $\theta$, with $$K_1(Y_1,Y_2;\theta)=(\frac{1}{2})^n \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_1) \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_n)$$, From the two indicator functions and from the definition of order statistics, we have that $$\theta-1\theta \land Y_n-1<\theta$$. Joint density of the sample $ X=(X_1,X_2,\ldots,X_n)$ for $\theta\in\mathbb R$ is as you say $$f_{\theta}( x)=\mathbf1_{\theta
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