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Let $X_1, \ldots, X_n$ be iid with common pdf $$ f(x;\theta,\gamma) = \frac{\theta x^{\theta-1}}{\gamma^{\theta}}1(0<x<\gamma);\,\, \theta, \gamma >0$$ and let $T_1, \ldots, T_n$ be the corresponding order statistics. Further let $\displaystyle (S,T) = \left(\prod_{i=1}^{n-1}T_i,T_n\right)$ Show $(S,T)$ is sufficient for this model.

I have two questions:

  1. Solving this:

the joint pdf is \begin{align} & f(\vec{t};\theta\gamma) \\[8pt] = {} & n!\frac{\theta t_1^{\theta-1}}{\gamma^{\theta}}1(0<t_1<\gamma)\cdots \frac{\theta t_n^{\theta-1}}{\gamma^{\theta}}1(0<t_n<\gamma) \\[8pt] = {} & n!\frac{\theta^n(t_1\cdots t_n)^\theta}{\gamma^{n\theta}(t_1\cdots t_n)}1(0<t_1<\cdots<t_n<\gamma) \\[8pt] = {} & n!\frac{\theta^n(s\cdot t)^\theta}{\gamma^{n\theta}(s \cdot t)}1(0<t_1<\cdots<t_n<\gamma) \\[8pt] = {} & \frac{n!}{s\cdot t}\cdot \frac{\theta^n(s\cdot t)^\theta}{\gamma^{n\theta}}1(0<t_1<\cdots<t_n<\gamma) \\[8pt] = {} & h(\vec{t})g(s,t; \theta,\gamma) \end{align}

This shows the sufficiency by the Factorization theorem, correct?

  1. My second question has to do with conditional probability--specifically when working with a continuous distribution so I will use the above. By definition we achieve sufficiency if we can show the conditional distribution doesn't depend on the given parameters. I always have difficulty computing a conditional distribution in these instances. Namely, I know that $\displaystyle f(\vec{t} ; (s,t) ) = \frac{f_{\vec{T},(S,T)}(\vec{t},(s,t))}{f_{(S,T)}(s,t)} $

But can someone show me to how to actually derive this in this case, I know that $f_{(S,T)}(s,t)$ will follow from doing a transformation of random variable method. The issue I run into is trying to compute what $f_{\vec{T},(S,T)}(\vec{t},(s,t))$ would be.

Michael Hardy
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oliverjones
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  • @StubbornAtom $X_1,...,X_n$ are the observations and $T_1,...T_n$ are the corresponding order statistics. – oliverjones Aug 06 '20 at 20:02
  • Then you should use the density of $X_1,\ldots,X_n$. – StubbornAtom Aug 06 '20 at 20:04
  • @StubbornAtom I need the order statistics to be able to factor in (1) above so I don't immediately see how to do it with the density of $X_1,...,X_n$ aside from maybe changing the $n!$ to $n$ because there would still be $n$ possible choices for $T=T_n$. – oliverjones Aug 06 '20 at 20:09
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    Actually it doesn't really matter. Your approach is fine. – StubbornAtom Aug 06 '20 at 21:54
  • Where you wrote $$ \frac{n!}{s\cdot t}\cdot \frac{\theta^n(s\cdot t)^\theta}{\gamma^{n\theta}}1(0<t_1<\cdots<t_n<\gamma), $$ you should note that this is the same as $$ \frac{n!}{s\cdot t}\cdot \frac{\theta^n(s\cdot t)^\theta}{\gamma^{n\theta}}1(t_n<\gamma). $$ Thus there is no dependence on $t_1,\ldots,t_{n-1}$ except through $s. \qquad$ – Michael Hardy Aug 07 '20 at 18:46

1 Answers1

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You are supposed to use the joint density of the data you observe, which is just the sample $\boldsymbol X=(X_1,X_2,\ldots,X_n)$. Suppose $(X_{(1)},X_{(2)},\ldots,X_{(n)})$ are the corresponding order statistics.

For $\theta,\gamma>0$, density of $\boldsymbol X$ is

\begin{align} f_{\theta,\gamma}(x_1,x_2,\ldots,x_n)&=\left(\frac{\theta}{\gamma^\theta}\right)^n \left(\prod_{i=1}^n x_i\right)^{\theta-1} \mathbf1_{0<x_1,\ldots,x_n<\gamma} \\&=\left(\frac{\theta}{\gamma^\theta}\right)^n \left(\prod_{i=1}^n x_{(i)}\right)^{\theta-1} \mathbf1_{0<x_{(n)}<\gamma} \\&=\left(\frac{\theta}{\gamma^\theta}\right)^n \left(\prod_{i=1}^{n-1} x_{(i)}\right)^{\theta-1} x_{(n)}^{\theta-1}\,\mathbf1_{0<x_{(n)}<\gamma} \end{align}

Here we used $\prod\limits_{i=1}^n x_i=\prod\limits_{i=1}^n x_{(i)}$, since this is just a rearrangement of the product of $n$ observations.

Hence by Factorization theorem, $\boldsymbol T=\left(\prod\limits_{i=1}^{n-1} X_{(i)},X_{(n)}\right)$ is a sufficient statistic for $(\theta,\gamma)$.

This means that the distribution of the data $\boldsymbol X$ given $\boldsymbol T$ is independent of $(\theta,\gamma)$. To show this rigorously from definition for continuous distributions is somewhat technical as far as I am aware, depending on the notion of a sufficient partition. That's why we usually stick to Factorization theorem in these situations.

StubbornAtom
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