Question:

Let \(X_1,X_2,X_3\) be a random sample from a distribution having probability mass function
\[ f_{\theta}(x)=\begin{cases}\theta & \text{if } x=1\\ 1-\theta & \text{if } x=2\\ 0, & \text{otherwise},\end{cases} \]
where \(\theta\in\Theta=(0,1)\). Let \(\underline{X}=(X_1,X_2,X_3)\). Then which of the following is NOT a sufficient statistic for \(\theta\)?

Show Hint

A statistic is sufficient only if it lets you recover the count of \(1\)s among \(X_1,X_2,X_3\); check whether two samples with different counts can give the same statistic value.
Updated On: Aug 3, 2026
  • \(T_1(\underline{X})=(X_1-X_2,\ X_1+X_2,\ X_1+X_3)\)
  • \(T_2(\underline{X})=(X_1+X_2,\ X_1-X_3,\ X_2+X_3)\)
  • \(T_3(\underline{X})=(X_1-X_2,\ X_2-X_3,\ X_3)\)
  • \(T_4(\underline{X})=(X_1+X_2,\ X_3)\)
Show Solution
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The Correct Option is B

Solution and Explanation

Step 1: Find the sufficient statistic.
With \(n_1\) = count of \(X_i=1\), joint pmf = \(\theta^{n_1}(1-\theta)^{3-n_1}\). So \(n_1\) (equivalently \(\sum X_i\)) is sufficient.

Step 2: Check (A).
$T_1$ recovers \(X_1,X_2,X_3\) fully (invertible linear system), so sufficient.

Step 3: Check (C).
$T_3$ also recovers \(X_1,X_2,X_3\) fully, sufficient.

Step 4: Check (D).
$T_4$'s sum gives \(X_1+X_2+X_3\), determines \(n_1\), sufficient.

Step 5: Check (B) with two samples.
$(1,2,1)$: \(n_1=2\), \(T_2=(3,0,3)\). $(2,1,2)$: \(n_1=1\), \(T_2=(3,0,3)\). Same \(T_2\) but different \(n_1\) and different likelihoods, so NOT sufficient.

Final Answer:
\[ \boxed{T_2(\underline{X})=(X_1+X_2,\ X_1-X_3,\ X_2+X_3)} \]
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