Using Basu theorem

Question

Let $X_1,X_2...,X_n$ a random sample with pdf

$$f(x|\theta)= \exp(-x+\alpha), \ \ \alpha < x < \infty$$

Using Basu Theorem prove that $T(X)=\min X_i$ and $T(X)=\frac{1}{n-1}\sum ^n_i(X_i-\bar{X})^2$ are independent.

I have already a proof that $T(X)=\min X_i$ is sufficient and complete. I dont know how to prove $T(X)=\frac{1}{n-1}\sum ^n_i(X_i-\bar{X})^2$ is ancillary.

Answer 1

The density $f(x|\theta)=\exp(-x+\alpha)=\exp(-(x-\alpha))$ is from the location family (LF) of densities $f(x|\theta)=\psi(x-\theta), -\infty<\theta<\infty$ with $\psi$ being any density. Here $\psi(x)=\exp(-x)$ is the exponential density. For location family densities, if $T(\textbf X)$ is a location-invariant function ($T(\textbf X+c\textbf 1)=T(\textbf X)$), then $T(\textbf X)$ is ancillary.

$$\begin{split}T(\textbf X+c\textbf 1)&=\frac{1}{n-1}\sum_{i=1}^n \left(X_i+c-\frac{\sum (X_i+c)}n\right)^2\\ &=\frac 1 {n-1}\sum _i\left(X_i+c-\frac {\sum_iX_i}{n}-\frac{nc}{n}\right)^2\\ &=\frac 1 {n-1}\sum _i\left(X_i-\frac {\sum_iX_i}{n}\right)^2\\ &=\frac 1 {n-1}\sum _i\left(X_i-\bar X\right)^2=T(\textbf X)\end{split}$$

Thus $T(\textbf X)$ is location-invariant, and ancillary. We can now apply Basu's theorem to conclude that $T(\textbf X)$ and $T(\textbf X)$ are independent.

Answer 2

Note that

\begin{align} \sum_{i=1}^n (X_i-\overline X)^2&=\sum_{i=1}^n\{(X_i-\alpha)-(\overline X-\alpha)\}^2 \\&=\sum_{i=1}^n(X_i-\alpha)^2 - n(\overline X-\alpha)^2 \end{align}

Now distribution of $X_i-\alpha$ is clearly free of $\alpha$ for every $i=1,\ldots,n$. This implies the distribution of $\frac1n\sum\limits_{i=1}^n (X_i-\alpha)=\overline X-\alpha$ is also free of $\alpha$. Hence the overall distribution of the sample variance is independent of $\alpha$, making it ancillary.