Question

Let the sample mean based on a random sample from Exp($\lambda$) distribution be 3.7. Then the maximum likelihood estimate of $1 - e^{-\lambda}$ equals ........... (round off to two decimal places).
 

Hint:

For the exponential distribution, the MLE of $\lambda$ is the reciprocal of the sample mean, $\hat{\lambda} = 1/\bar{X}$.

Answer 1

Correct Answer: 0.22

Step 1: Recall MLE for exponential distribution.
For $X_1, X_2, \ldots, X_n \sim \text{Exp}(\lambda)$, the probability density function is \[ f(x; \lambda) = \lambda e^{-\lambda x}, x > 0. \] The log-likelihood is \[ \ln L = n \ln \lambda - \lambda \sum X_i. \]

Step 2: Differentiate and set derivative to zero.
\[ \frac{d(\ln L)}{d\lambda} = \frac{n}{\lambda} - \sum X_i = 0 $\Rightarrow$ \hat{\lambda} = \frac{n}{\sum X_i} = \frac{1}{\bar{X}}. \]

Step 3: Substitute given mean.
Given $\bar{X} = 3.7$, \[ \hat{\lambda} = \frac{1}{3.7} = 0.27027. \]

Step 4: Compute required expression.
\[ 1 - e^{-\lambda} = 1 - e^{-0.27027} = 1 - 0.7639 = 0.2361. \]

Step 5: Round off.
\[ \boxed{1 - e^{-\hat{\lambda}} = 0.23.} \]