Question

Let \(X_1,X_2,\ldots,X_{10}\) be a random sample of size \(10\) from an \(Exp(\beta)\) distribution, where \(\beta>0\) is an unknown parameter. Let

• A. \(P\) is correct and \(Q\) is NOT correct
• B. \(P\) is NOT correct and \(Q\) is correct
• C. Both \(P\) and \(Q\) are correct
• D. Neither \(P\) nor \(Q\) is correct

Hint:

For \(X\sim Exp(\beta)\) with scale parameter \(\beta\), the MLE is \(\hat{\beta}=\bar{X}\), and by invariance property, the MLE of any function \(g(\beta)\) is \(g(\hat{\beta})\).

Answer 1

The Correct Option is A

Step 1: Recall the exponential distribution.
For \(X\sim Exp(\beta)\), taking \(\beta\) as the scale parameter, the probability density function is
\[ f(x)=\frac{1}{\beta}e^{-\frac{x}{\beta}},\qquad x>0 \] The maximum likelihood estimator of \(\beta\) is the sample mean.
Since
\[ T=\frac{1}{10}\sum_{i=1}^{10}X_i, \] we get
\[ \hat{\beta}=T \]

Step 2: Check statement \(P\).
The median \(m\) of an exponential distribution satisfies
\[ P(X\leq m)=\frac12 \] Now, the distribution function is
\[ F(x)=1-e^{-\frac{x}{\beta}} \] So,
\[ 1-e^{-\frac{m}{\beta}}=\frac12 \] \[ e^{-\frac{m}{\beta}}=\frac12 \] Taking logarithm,
\[ -\frac{m}{\beta}=\log_e\left(\frac12\right) \] \[ \frac{m}{\beta}=\log_e 2 \] Thus,
\[ m=\beta\log_e 2 \] Using invariance property of MLE, the MLE of the median is
\[ \hat{m}=\hat{\beta}\log_e 2 \] \[ \hat{m}=T\log_e 2 \] Therefore, statement \(P\) is correct.

Step 3: Check statement \(Q\).
We need the MLE of
\[ P(X_1<2) \] For exponential distribution,
\[ P(X_1<2)=F(2) \] \[ F(2)=1-e^{-\frac{2}{\beta}} \] Using \(\hat{\beta}=T\), the MLE of \(P(X_1<2)\) is
\[ 1-e^{-\frac{2}{T}} \] But statement \(Q\) gives
\[ e^{-\frac{2}{T}} \] which is the MLE of
\[ P(X_1>2) \] not of \(P(X_1<2)\).
Therefore, statement \(Q\) is NOT correct.

Step 4: Final conclusion.
Statement \(P\) is correct and statement \(Q\) is NOT correct.
Hence,
\[ \boxed{(A)} \]