1
$\begingroup$

Let $X_{1}, X_{2}, ..., X_{n}$ be a random sample from $\text{Uniform}(\theta,\theta+1)$ population with $-\infty<\theta<\theta+1< \infty$ show that $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$. Also, show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic.

For the first part I did the following $$f(x|\theta,\theta+1)=I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})$$ then $$\frac{f(x|\theta,\theta+1)}{f(y|\theta,\theta+1)}=\frac{I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})}{I_{(\theta,\theta+1)}(y_{(1)},y_{(n)})}$$

This is a constant function in $\theta$ iff $x_{(1)}=y_{(1)}$ and $x_{(n)}=y_{(n)}$ s.t. $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$.

However, I am not sure how to proceed to show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic. Can some help me with this?

$\endgroup$
1
  • $\begingroup$ You might want to see this, which is nearly identical, and this, which is similar. $\endgroup$ – Jacob Maibach 2 days ago
1
$\begingroup$

Any invertible function of a minimal sufficient statistic is also minimal sufficient. Such a function is given by $$M(x,y) = (y-x, (x+y)/2)$$ which has the inverse $$M^{-1}(x,y) = (y - x/2, y + x/2).$$ This can also be conceptualized as a $2 \times 2$ matrix: $$M = \begin{bmatrix} -1 & 1 \\ 1/2 & 1/2 \end{bmatrix}.$$

$\endgroup$
1
  • $\begingroup$ please keep posting! I learn a lot from your answers. $\endgroup$ – Matthew Pilling 2 days ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.