Analysis of the $T_n$ Estimator for $\sigma^2$

We are evaluating the statistic $T_n = \frac{1}{10n} \sum_{i=1}^{10} \sum_{j=1}^n X_{ij}^2$ as an estimator for $\sigma^2$. The model is given by $X_{ij} = b_i + \epsilon_{ij}$, where $b_i \sim i.i.d. N(0, \tau^2)$ and $\epsilon_{ij} \sim i.i.d. N(0, \sigma^2)$.

Unbiasedness of $T_n$

An estimator is unbiased if its expected value equals the parameter being estimated. We calculate $E[T_n]$.

First, let's find the expected value of $X_{ij}^2$:

• Since $b_i \sim N(0, \tau^2)$, $E[b_i] = 0$ and $Var(b_i) = E[b_i^2] - (E[b_i])^2 = \tau^2$. Thus, $E[b_i^2] = \tau^2$.
• Similarly, since $\epsilon_{ij} \sim N(0, \sigma^2)$, $E[\epsilon_{ij}] = 0$ and $Var(\epsilon_{ij}) = E[\epsilon_{ij}^2] - (E[\epsilon_{ij}])^2 = \sigma^2$. Thus, $E[\epsilon_{ij}^2] = \sigma^2$.
• $X_{ij}^2 = (b_i + \epsilon_{ij})^2 = b_i^2 + 2b_i\epsilon_{ij} + \epsilon_{ij}^2$.
• Because $b_i$ and $\epsilon_{ij}$ are independent and have zero mean, $E[2b_i\epsilon_{ij}] = 2E[b_i]E[\epsilon_{ij}] = 2(0)(0) = 0$.
• Therefore, $E[X_{ij}^2] = E[b_i^2] + E[2b_i\epsilon_{ij}] + E[\epsilon_{ij}^2] = \tau^2 + 0 + \sigma^2 = \tau^2 + \sigma^2$.

Now, we compute the expected value of $T_n$:

$ E[T_n] = E\left[\frac{1}{10n} \sum_{i=1}^{10} \sum_{j=1}^n X_{ij}^2\right] $

By linearity of expectation:

$ E[T_n] = \frac{1}{10n} \sum_{i=1}^{10} \sum_{j=1}^n E[X_{ij}^2] = \frac{1}{10n} \sum_{i=1}^{10} \sum_{j=1}^n (\tau^2 + \sigma^2) $

The term $(\tau^2 + \sigma^2)$ is summed $10n$ times:

$ E[T_n] = \frac{1}{10n} (10n)(\tau^2 + \sigma^2) = \tau^2 + \sigma^2 $

For $T_n$ to be an unbiased estimator of $\sigma^2$, we need $E[T_n] = \sigma^2$. This requires $\tau^2 + \sigma^2 = \sigma^2$, which implies $\tau^2 = 0$. Thus, $T_n$ is unbiased only when $\tau^2 = 0$.

Consistency of $T_n$

An estimator is consistent if it converges in probability to the true parameter value as the sample size increases. We need to check if $T_n \xrightarrow{p} \sigma^2$ as $n \to \infty$.

We have established that $E[T_n] = \tau^2 + \sigma^2$. This expectation is constant with respect to $n$.

Using the Law of Large Numbers, the average of the $X_{ij}^2$ terms converges in probability to their mean. Since $T_n$ is the average of $10n$ such terms (appropriately weighted), it converges in probability to its expectation:

$ T_n \xrightarrow{p} E[T_n] = \tau^2 + \sigma^2 \quad \text{as } n \to \infty $

For $T_n$ to be a consistent estimator of $\sigma^2$, the limit must be $\sigma^2$. Therefore, we require $\tau^2 + \sigma^2 = \sigma^2$. This condition holds if and only if $\tau^2 = 0$. Thus, $T_n$ is consistent for $\sigma^2$ only when $\tau^2 = 0$.

Conclusion

The analysis shows that the estimator $T_n$ is unbiased for $\sigma^2$ if and only if $\tau^2 = 0$. Similarly, $T_n$ is consistent for $\sigma^2$ if and only if $\tau^2 = 0$. These findings match the conditions described in options C and D.