Question

If $X_{1}, X_{2}, \dots, X_{n}$ denote a random sample of size $n$ from normal population $N(0, \theta^{2})$ then MVUE of $\theta^{2}$ is

• A. $\overline{x}^{2}$
• B. $\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}$
• C. $\sum_{i=1}^{n} \frac{(X_{i}-\overline{X})^{2}}{n-1}$
• D. $\sum_{i=1}^{n} \frac{(X_{i}-\overline{X})^{2}}{n}$

Hint:

When the mean $\mu$ is known (e.g., $\mu=0$), we do not lose a degree of freedom. In such cases, the divisor for the unbiased estimator of variance is $n$, whereas it is $n-1$ when the mean is unknown and estimated by $\overline{X}$.

Answer 1

The Correct Option is B

To find the Minimum Variance Unbiased Estimator (MVUE) of $\theta^2$ for a normal distribution with a known mean, we utilize the properties of sufficiency and unbiasedness.

Step 1: \color{redIdentify the Distribution and Parameters
The sample $X_1, X_2, \dots, X_n$ is drawn from $N(0, \theta^2)$.
The probability density function (PDF) is:
$f(x; \theta^2) = \frac{1}{\sqrt{2\pi\theta^2}} \exp\left(-\frac{x^2}{2\theta^2}\right)$.
Since the mean $\mu$ is known to be $0$, the only unknown parameter is the variance $\theta^2$.

Step 2: \color{redFind the Sufficient Statistic
Using the Factorization Theorem on the joint PDF:
$L(\theta^2) = (2\pi\theta^2)^{-n/2} \exp\left(-\frac{\sum X_i^2}{2\theta^2}\right)$.
The statistic $T = \sum_{i=1}^n X_i^2$ is a complete sufficient statistic for $\theta^2$ because the distribution belongs to the exponential family.

Step 3: \color{redCheck for Unbiasedness
We know that for $X_i \sim N(0, \theta^2)$, the term $\frac{X_i}{\theta} \sim N(0, 1)$.
Thus, $\left(\frac{X_i}{\theta}\right)^2 \sim \chi^2_1$.
Summing these independent variables gives $\sum_{i=1}^n \left(\frac{X_i}{\theta}\right)^2 = \frac{1}{\theta^2} \sum X_i^2 \sim \chi^2_n$.
The expectation of a $\chi^2_n$ variable is $n$:
$E\left[\frac{1}{\theta^2} \sum X_i^2\right] = n \implies E\left[\sum X_i^2\right] = n\theta^2$.

Step 4: \color{redConstruct the MVUE
By the Lehmann-Scheffé Theorem, any function of a complete sufficient statistic that is unbiased for the parameter is the unique MVUE.
Since $E\left[\sum X_i^2\right] = n\theta^2$, then $E\left[\frac{1}{n} \sum X_i^2\right] = \theta^2$.
Therefore, $\frac{1}{n} \sum_{i=1}^n X_i^2$ is the MVUE of $\theta^2$.