Question

Let \(X_1,X_2\) be a random sample from the following probability density function
\[ f_\alpha(x)= \begin{cases} \alpha x^{\alpha-1}e^{-x^\alpha} & \text{if } x>0 \\ 0 & \text{otherwise,} \end{cases} \]
where \(\alpha\in(0,\infty)\) is an unknown parameter. Let
\[ X_{(1)}=\min\{X_1,X_2\} \quad\text{and}\quad X_{(2)}=\max\{X_1,X_2\}. \]
Then which of the following statements is correct?

• A. \(\left(X_1+X_2,\ X_{(1)}\right)\) is a minimal sufficient statistic
• B. \(\left(X_1,\ X_{(2)}\right)\) is a minimal sufficient statistic
• C. \(\left(X_1-X_2,\ X_{(1)}\right)\) is a sufficient statistic but not a minimal sufficient statistic
• D. \(\left(X_1-X_2,\ X_{(2)}\right)\) is a complete sufficient statistic

Hint:

Use the Lehmann-Scheffe likelihood ratio method: the ratio \(f(x_1,x_2)/f(y_1,y_2)\) stays free of \(\alpha\) only when \(\{x_1,x_2\}=\{y_1,y_2\}\), which is exactly the order statistics \((X_{(1)},X_{(2)})\), the same information as \((X_1+X_2,X_{(1)})\).

Answer 1

The Correct Option is A

Step 1: Write down the joint density.
For $X_1,X_2$ iid with density $f_\alpha(x)=\alpha x^{\alpha-1}e^{-x^\alpha}$ for $x>0$, the joint density is \[ f_\alpha(x_1,x_2)=\alpha^2(x_1x_2)^{\alpha-1}e^{-(x_1^\alpha+x_2^\alpha)} \]

Step 2: Try to factor this using the Fisher-Neyman theorem.
The factorization theorem says $T(X_1,X_2)$ is sufficient if we can write the joint density as $g(T;\alpha)\cdot h(x_1,x_2)$, where $h$ does not involve $\alpha$. Here the whole expression depends on $(x_1,x_2)$ only through $x_1x_2$ and $x_1^\alpha+x_2^\alpha$, and both of these only depend on which two numbers $x_1,x_2$ are, not which one is labeled first. So the natural candidate for a sufficient statistic is the unordered pair, the order statistics $(X_{(1)},X_{(2)})$.

Step 3: Show no further reduction is possible.
Could we get away with just $X_1+X_2$ alone? No, because $\alpha$ shows up as an exponent, and this Weibull-type family is not an exponential family with a fixed sufficient statistic of dimension one. Two different pairs with the same sum but different individual values, like $(1,3)$ and $(2,2)$, give different values of $x_1^\alpha+x_2^\alpha$ for general $\alpha$.

Step 4: Translate $(X_{(1)},X_{(2)})$ into the form given in option (A).
The sum of the two observations does not depend on order: $X_1+X_2=X_{(1)}+X_{(2)}$. So if we know $X_1+X_2$ and $X_{(1)}$ together, we can solve for the other order statistic: \[ X_{(2)}=(X_1+X_2)-X_{(1)} \] This shows $(X_1+X_2,X_{(1)})$ carries exactly the same information as $(X_{(1)},X_{(2)})$.

Step 5: Test option (B) with a concrete pair.
Take $X_1=5,X_2=2$. Then $X_{(2)}=\max=5=X_1$. The pair $(X_1,X_{(2)})=(5,5)$ gives no way to recover $X_2=2$. So this statistic does not preserve enough information, and option (B) fails.

Step 6: Test option (C) with the same idea.
$(X_1-X_2,X_{(1)})$ can be inverted back to the exact ordered pair $(X_1,X_2)$, more detailed than the order statistics actually require.

Step 7: Test option (D) the same way.
$(X_1-X_2,X_{(2)})$ similarly recovers the full labeled sample $(X_1,X_2)$, so it also fails to be minimal.

Step 8: Conclude.
\[ \boxed{\left(X_1+X_2,\ X_{(1)}\right)\text{ is a minimal sufficient statistic}} \]