Question

Let \(X_1,X_2,\ldots,X_n\) be a random sample of size \(n\;(n\geq 1)\) from a distribution with the probability density function

• A. \(\dfrac{3}{n}\sum_{i=1}^{n}e^{X_i}\)
• B. \(\dfrac{1}{n}\sum_{i=1}^{n}X_i\)
• C. \(\dfrac{1}{3n}\sum_{i=1}^{n}e^{X_i}\)
• D. \(\dfrac{1}{n}\sum_{i=1}^{n}X_i^3\)

Hint:

For maximum likelihood estimation, first form the likelihood, then take log, differentiate with respect to the unknown parameter, and solve the likelihood equation.

Answer 1

The Correct Option is C

Step 1: Write the log likelihood.
With density $\frac1{2\beta^3}e^{3x-e^x/\beta}$, the log likelihood is $-3n\ln\beta+3\sum X_i-\frac1\beta\sum e^{X_i}+\text{const}$.

Step 2: Differentiate in $\beta$.
\[ \frac{d\ell}{d\beta}=-\frac{3n}\beta+\frac1{\beta^2}\sum e^{X_i}. \]

Step 3: Solve the equation.
Setting it to $0$ and multiplying by $\beta^2$ gives $3n\beta=\sum e^{X_i}$, so $\hat\beta=\frac1{3n}\sum e^{X_i}$.

Step 4: Confirm a maximum.
The second derivative at $\hat\beta$ works out to $-\frac{3n}{\hat\beta^2}<0$, so it is indeed a maximum.

Step 5: Conclude.
The MLE is option (C).
\[ \boxed{\frac1{3n}\sum_{i=1}^n e^{X_i}} \]