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 the joint density of the sample.
\(X_1,X_2\) are iid with density \[ f_\alpha(x)=\alpha x^{\alpha-1}e^{-x^\alpha}, \qquad x>0. \] The joint density is \[ f_\alpha(x_1,x_2)=\alpha^2(x_1x_2)^{\alpha-1}\exp\!\left[-(x_1^\alpha+x_2^\alpha)\right]. \]

Step 2: Apply the Lehmann-Scheffe method for minimal sufficiency.
A statistic \(T(X_1,X_2)\) is minimal sufficient if, for two sample points \((x_1,x_2)\) and \((y_1,y_2)\), the ratio \[ \frac{f_\alpha(x_1,x_2)}{f_\alpha(y_1,y_2)} \] is free of \(\alpha\) if and only if \(T(x_1,x_2)=T(y_1,y_2)\).

Step 3: Form the likelihood ratio.
\[ \frac{f_\alpha(x_1,x_2)}{f_\alpha(y_1,y_2)}=\left(\frac{x_1x_2}{y_1y_2}\right)^{\alpha-1}\exp\!\left[-(x_1^\alpha+x_2^\alpha)+(y_1^\alpha+y_2^\alpha)\right]. \] For this to stay constant as \(\alpha\) ranges over all of \((0,\infty)\), the power sums must match, \[ x_1^\alpha+x_2^\alpha=y_1^\alpha+y_2^\alpha \quad\text{for every }\alpha, \] and then automatically \(x_1x_2=y_1y_2\) as well. Letting \(\alpha\to\infty\) forces the larger of \(x_1,x_2\) to equal the larger of \(y_1,y_2\), then the smaller values must match too. So this happens exactly when \(\{x_1,x_2\}\) and \(\{y_1,y_2\}\) are the same pair of numbers, regardless of order.

Step 4: State the minimal sufficient statistic.
So the ratio is free of \(\alpha\) exactly when the unordered pair \(\{x_1,x_2\}\) matches, which means the pair of order statistics \[ \left(X_{(1)},X_{(2)}\right)=\big(\min\{X_1,X_2\},\ \max\{X_1,X_2\}\big) \] is the minimal sufficient statistic.

Step 5: Rewrite \(\left(X_1+X_2,X_{(1)}\right)\) in terms of the order statistics.
Since \(X_1+X_2=X_{(1)}+X_{(2)}\), knowing \(X_1+X_2\) together with \(X_{(1)}\) tells us \[ X_{(2)}=(X_1+X_2)-X_{(1)}. \] So \(\left(X_1+X_2,X_{(1)}\right)\) determines \(\left(X_{(1)},X_{(2)}\right)\) exactly, and vice versa. The two statistics carry identical information, so \(\left(X_1+X_2,X_{(1)}\right)\) is also a minimal sufficient statistic. This matches option (A).

Step 6: Check option (B).
\(\left(X_1,X_{(2)}\right)\) keeps the raw, unsorted value \(X_1\) together with the maximum. If \(X_1\) is the smaller of the two, \(X_{(2)}=X_2\) and we recover \(X_2\) directly; but if \(X_1\) is the larger one, \(X_{(2)}=X_1\) tells us nothing new, and the value of \(X_2\) has been lost. Because the information retained changes depending on which observation happens to be bigger, this is not the minimal sufficient statistic.

Step 7: Check option (C).
Knowing \(X_1-X_2\) together with \(X_{(1)}\) lets us recover both \(X_1\) and \(X_2\) individually, since the sign of \(X_1-X_2\) tells us which one is the minimum. This is strictly more detailed than the order statistics \(\left(X_{(1)},X_{(2)}\right)\), since it also remembers which original observation was which. This statistic is not the coarsest possible sufficient statistic.

Step 8: Check option (D).
The same issue applies to \(\left(X_1-X_2,X_{(2)}\right)\): it still keeps track of which observation was which, so it does not reduce to the minimal sufficient statistic, and a statistic that is not minimal cannot be complete either.

Final Answer:
Among the options, only \(\left(X_1+X_2,X_{(1)}\right)\) carries exactly the same information as the order statistics, no more and no less. \[ \boxed{\left(X_1+X_2,\ X_{(1)}\right)\text{ is a minimal sufficient statistic}} \]