Question

If $X_{1}=17, X_{2}=10, X_{3}=32$ and $X_{4}=5$ be the observed values of a random sample from the discrete distribution $P(X=x)=\begin{cases}\frac{\theta^{2x} e^{-\theta^2}}{x!} & \text{if } x=0,1,2,\dots, \theta>0 \\ 0 & \text{otherwise}\end{cases}$. Then MLE is

• A. $16$
• B. $12$
• C. $8$
• D. $4$

Hint:

By the invariance property of MLE, if $\hat{\lambda}$ is the MLE of $\lambda$, then $g(\hat{\lambda})$ is the MLE of $g(\lambda)$. Since $\hat{\lambda} = \overline{x} = 16$ and $\theta = \sqrt{\lambda}$, then $\hat{\theta} = \sqrt{16} = 4$.

Answer 1

The Correct Option is D

The given distribution is a Poisson distribution with parameter $\lambda = \theta^2$. We need to find the Maximum Likelihood Estimate (MLE) for $\theta$.

Step 1: \color{redIdentify the MLE of the Poisson parameter
For a Poisson distribution with parameter $\lambda$, the MLE $\hat{\lambda}$ is the sample mean $\overline{x}$.
Here, $\lambda = \theta^2$, so $\widehat{\theta^2} = \overline{x}$.

Step 2: \color{redCalculate the Sample Mean
The observations are $17, 10, 32, 5$.
$\overline{x} = \frac{17 + 10 + 32 + 5}{4} = \frac{64}{4} = 16$.

Step 3: \color{redSolve for $\theta$
We have $\hat{\theta}^2 = 16$.
Taking the square root (noting $\theta > 0$):
$\hat{\theta} = \sqrt{16} = 4$.
The MLE of $\theta$ is 4.