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 likelihood function.
For the random sample \(X_1,X_2,\ldots,X_n\), the likelihood function is
\[ L(\beta)=\prod_{i=1}^{n}\frac{1}{2\beta^3}e^{3X_i-\frac{e^{X_i}}{\beta}} \] \[ L(\beta)=\left(\frac{1}{2\beta^3}\right)^n e^{3\sum_{i=1}^{n}X_i-\frac{1}{\beta}\sum_{i=1}^{n}e^{X_i}} \]

Step 2: Write the log-likelihood function.
Taking logarithm,
\[ \ell(\beta)=\log L(\beta) \] \[ \ell(\beta) = -n\log 2-3n\log \beta +3\sum_{i=1}^{n}X_i -\frac{1}{\beta}\sum_{i=1}^{n}e^{X_i} \]

Step 3: Differentiate with respect to \(\beta\).
\[ \frac{d\ell}{d\beta} = -\frac{3n}{\beta} + \frac{1}{\beta^2}\sum_{i=1}^{n}e^{X_i} \] For maximum likelihood estimator, put
\[ \frac{d\ell}{d\beta}=0 \] Thus,
\[ -\frac{3n}{\beta} + \frac{1}{\beta^2}\sum_{i=1}^{n}e^{X_i}=0 \] Multiplying by \(\beta^2\),
\[ -3n\beta+\sum_{i=1}^{n}e^{X_i}=0 \] Therefore,
\[ 3n\beta=\sum_{i=1}^{n}e^{X_i} \] \[ \hat{\beta}=\frac{1}{3n}\sum_{i=1}^{n}e^{X_i} \]

Step 4: Verify maximum.
The second derivative is
\[ \frac{d^2\ell}{d\beta^2} = \frac{3n}{\beta^2} - \frac{2}{\beta^3}\sum_{i=1}^{n}e^{X_i} \] At
\[ \hat{\beta}=\frac{1}{3n}\sum_{i=1}^{n}e^{X_i}, \] we have
\[ \sum_{i=1}^{n}e^{X_i}=3n\hat{\beta} \] So,
\[ \frac{d^2\ell}{d\beta^2} = \frac{3n}{\hat{\beta}^2} - \frac{2(3n\hat{\beta})}{\hat{\beta}^3} \] \[ = \frac{3n}{\hat{\beta}^2} - \frac{6n}{\hat{\beta}^2} \] \[ = -\frac{3n}{\hat{\beta}^2}<0 \] Hence, this gives a maximum.

Step 5: Final conclusion.
The maximum likelihood estimator of \(\beta\) is
\[ \boxed{\hat{\beta}=\frac{1}{3n}\sum_{i=1}^{n}e^{X_i}} \] Therefore, the correct option is
\[ \boxed{(C)} \]