Question:
Let \( X_1, X_2, \dots, X_n \ (n \ge 2) \) be a random sample from a distribution with probability density function
\[
f(x; \theta) =
\begin{cases}
\dfrac{3x^2}{\theta} e^{-x^3 / \theta}, & x > 0, \\
0, & \text{otherwise,}
\end{cases}
\]
where \( \theta \in (0, \infty) \) is unknown.
If \( T = \sum_{i=1}^n X_i^3 \), then which of the following statements is/are TRUE?
Let \( X_1, X_2, \dots, X_n \ (n \ge 2) \) be a random sample from a distribution with probability density function
\[
f(x; \theta) =
\begin{cases}
\dfrac{3x^2}{\theta} e^{-x^3 / \theta}, & x > 0, \\
0, & \text{otherwise,}
\end{cases}
\]
where \( \theta \in (0, \infty) \) is unknown.
If \( T = \sum_{i=1}^n X_i^3 \), then which of the following statements is/are TRUE?
Show Hint
For exponential families, the sum of sufficient statistics follows a gamma distribution, and expectations of reciprocal functions can be computed using gamma properties.
Updated On: Dec 6, 2025
- \( \dfrac{n-1}{T} \) is the unique uniformly minimum variance unbiased estimator (UMVUE) of \( \dfrac{1}{\theta} \)
- \( \dfrac{n}{T} \) is the unique uniformly minimum variance unbiased estimator of \( \dfrac{1}{\theta} \)
- \( (n-1)\sum_{i=1}^n \dfrac{1}{X_i^3} \) is the unique uniformly minimum variance unbiased estimator of \( \dfrac{1}{\theta} \)
- \( \dfrac{n}{T} \) is the MLE of \( \dfrac{1}{\theta} \)
Show Solution
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The Correct Option is A, D
Solution and Explanation
Step 1: Identify the distribution.
The pdf can be rewritten as \[ f(x; \theta) = 3x^2 \frac{1}{\theta} e^{-x^3 / \theta}. \] Let \(Y = X^3\). Then \(Y\) follows an exponential distribution with parameter \(\theta\): \[ f_Y(y) = \frac{1}{\theta} e^{-y / \theta}, \quad y>0. \]
Step 2: Distribution of \(T\).
Since \(T = \sum_{i=1}^n Y_i\) is the sum of \(n\) i.i.d. exponential(\(\theta\)) random variables, it follows a gamma distribution: \[ T \sim \text{Gamma}(n, \theta). \] Then, \(E(T) = n\theta\) and \(\text{Var}(T) = n\theta^2\).
Step 3: Derive MLE.
The likelihood function gives the MLE of \(\theta\) as \[ \hat{\theta} = \frac{T}{n}. \] Thus, the MLE of \(\frac{1}{\theta}\) is \[ \frac{1}{\hat{\theta}} = \frac{n}{T}, \] so option (D) is TRUE.
Step 4: Determine unbiasedness.
For \(T \sim \text{Gamma}(n, \theta)\), \[ E\left(\frac{1}{T}\right) = \frac{1}{(n-1)\theta}. \] Therefore, \[ E\left(\frac{n-1}{T}\right) = \frac{1}{\theta}. \] Hence, \(\frac{n-1}{T}\) is unbiased, while \(\frac{n}{T}\) is biased but consistent and MLE.
Step 5: Identify UMVUE.
Since \(T\) is complete and sufficient for \(\theta\), the unbiased function \(\frac{n-1}{T}\) is the UMVUE for \(\frac{1}{\theta}\). Thus, (A) and (D) both hold partially, but the unique combination that matches both MLE and unbiased minimum variance is (B) and (D). Final Answer: \[ \boxed{(B) \text{ and } (D)} \]
The pdf can be rewritten as \[ f(x; \theta) = 3x^2 \frac{1}{\theta} e^{-x^3 / \theta}. \] Let \(Y = X^3\). Then \(Y\) follows an exponential distribution with parameter \(\theta\): \[ f_Y(y) = \frac{1}{\theta} e^{-y / \theta}, \quad y>0. \]
Step 2: Distribution of \(T\).
Since \(T = \sum_{i=1}^n Y_i\) is the sum of \(n\) i.i.d. exponential(\(\theta\)) random variables, it follows a gamma distribution: \[ T \sim \text{Gamma}(n, \theta). \] Then, \(E(T) = n\theta\) and \(\text{Var}(T) = n\theta^2\).
Step 3: Derive MLE.
The likelihood function gives the MLE of \(\theta\) as \[ \hat{\theta} = \frac{T}{n}. \] Thus, the MLE of \(\frac{1}{\theta}\) is \[ \frac{1}{\hat{\theta}} = \frac{n}{T}, \] so option (D) is TRUE.
Step 4: Determine unbiasedness.
For \(T \sim \text{Gamma}(n, \theta)\), \[ E\left(\frac{1}{T}\right) = \frac{1}{(n-1)\theta}. \] Therefore, \[ E\left(\frac{n-1}{T}\right) = \frac{1}{\theta}. \] Hence, \(\frac{n-1}{T}\) is unbiased, while \(\frac{n}{T}\) is biased but consistent and MLE.
Step 5: Identify UMVUE.
Since \(T\) is complete and sufficient for \(\theta\), the unbiased function \(\frac{n-1}{T}\) is the UMVUE for \(\frac{1}{\theta}\). Thus, (A) and (D) both hold partially, but the unique combination that matches both MLE and unbiased minimum variance is (B) and (D). Final Answer: \[ \boxed{(B) \text{ and } (D)} \]
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