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Given a random sample $X_1,\ldots,X_n$ distributed according to:

$$f(x_i\mid\theta) = e^{-(x_i-\theta)}$$ Show that the minimal sufficient statistic $X_{(1)}$ is independent of the set of ancillary statistics: $$(Y_1,\ldots,Y_{n-1})=(X_{(n)}-X_{(1)},\ldots,X_{(n)}-X_{(n-1)})$$ where $X_{(i)}$ denotes the $i^\text{th}$ order statistic of a random sample of size $n$.

My first thought is to perform a simple transformation using the joint order statistic distribution of $X$: $$f_{X_{(1)},\ldots,X_{(n)}}(x_1,\ldots,x_n)=n!f_X(x_1)\cdots f_X(x_n)$$ If the order statistic inequalities condition is met $x_1<\cdots<x_n$. Then confirm independence using factorization of the joint distribution.

Using the definitions of $Y_i$ we perform the multivariate transformation, I defined $Y_n=X_{(1)}$ for simplicity. However I quickly run into a difficult jacobian matrix computation for the determinant. Each inverse mapping from $Y\to X$ is of form: $$X_i=Y_n+Y_1-Y_i$$ The jacobian of this takes on a form I'm not familiar with computing. Not sure how to proceed here or if I should take a look at another approach!

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  • $\begingroup$ $X_{(1)}$ is also a complete statistic, so this follows from Basu's theorem. $\endgroup$ May 13 '21 at 20:15
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Your approach will work. The actual value of the Jacobian is immaterial, since you obtain the vector $Y$ by multiplying a matrix with constant coefficients onto the vector $X$, and so the Jacobian contributes a constant to the joint density of $Y$. The joint density of $Y$ is then proportional to $$ \left(\prod_{i=1}^n f(x_i)\right) I(\theta<x_1<x_2<\cdots<x_n) =e^{n\theta -\sum x_i} I(\theta <x_1<x_2<\cdots<x_n). $$ Now substitute the formulae you've obtained for $x_i$ in terms of $y_j$: $$ x_i=\begin{cases} y_n+y_1-y_i&\text{if $i<n$}\\ y_n+y_1&\text{if $i=n$} \end{cases}\tag1 $$ to find $$\sum x_i=ny_n +ny_1-\sum_{i=1}^{n-1}y_i. $$ The final step to demonstrate the factorizability of the joint density of $Y$ is to use (1) to convert the constraint $\theta <x_1<x_2<\cdots<x_n$ into a statement about $y_1,\ldots,y_n$. You should find that the result is $$y_1>y_2>\cdots>y_{n-1}>0\quad \text{and}\quad y_n>\theta.$$

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