This is a statistical problem about proving that an estimator $t$ is unbiased for the population variance $\sigma^2$ and calculating its variance. Let me guide you step by step.

Problem Restatement:

Given:

• Independent measurements $x[n]$ that follow a normal distribution $\mathcal{N}(\mu, \sigma^2)$, where $n = 0, 1, \ldots, N-1$.
• An estimator for variance is defined as: $t = \frac{1}{N} \sum_{n=0}^{N-1} \left( x[n] - \mu \right)^2$

We are asked to:

1. Prove that $t$ is an unbiased estimator of $\sigma^2$.
2. Calculate the variance of the estimator $t$.

1. Proving Unbiasedness of $t$:

The estimator $t$ is unbiased if: $\mathbb{E}[t] = \sigma^2$

Steps:

1. Substitute the definition of $t$: [ \mathbb{E}[t] = \mathbb{E}\left[\frac{1}{N} \sum_{n=0}^{N-1} \left( x[n] - \mu \right)^2 \right] ]

2. Use the linearity of expectation: [ \mathbb{E}[t] = \frac{1}{N} \sum_{n=0}^{N-1} \mathbb{E}\left[ \left( x[n] - \mu \right)^2 \right] ]

3. Recall the definition of variance: For $x[n] \sim \mathcal{N}(\mu, \sigma^2)$, [ \mathbb{E}\left[ (x[n] - \mu)^2 \right] = \text{Var}(x[n]) = \sigma^2 ]

4. Substitute this result: $\mathbb{E}[t] = \frac{1}{N} \sum_{n=0}^{N-1} \sigma^2$

5. Simplify the sum: $\mathbb{E}[t] = \frac{1}{N} \cdot N \cdot \sigma^2 = \sigma^2$

Thus, $\mathbb{E}[t] = \sigma^2$, which proves that $t$ is an unbiased estimator of $\sigma^2$.

2. Variance of the Estimator $t$:

The variance of $t$ is defined as: [ \text{Var}(t) = \mathbb{E}[t^2] - (\mathbb{E}[t])^2 ]

Steps:

1. Expand $t$: $t = \frac{1}{N} \sum_{n=0}^{N-1} (x[n] - \mu)^2$

   Denote $y_n = (x[n] - \mu)^2$. Then: $t = \frac{1}{N} \sum_{n=0}^{N-1} y_n$

2. Recall the variance formula for the mean: If $y_n$ are independent and identically distributed random variables, then: $\text{Var}\left( \frac{1}{N} \sum_{n=0}^{N-1} y_n \right) = \frac{1}{N^2} \sum_{n=0}^{N-1} \text{Var}(y_n)$

3. Find $\text{Var}(y_n)$: Since $y_n = (x[n] - \mu)^2$, use the properties of the normal distribution to calculate higher moments. Specifically, for $x[n] \sim \mathcal{N}(\mu, \sigma^2)$, it can be shown that: [ \mathbb{E}[y_n^2] = \mathbb{E}[(x[n] - \mu)^4] = 3\sigma^4 ] Thus: [ \text{Var}(y_n) = \mathbb{E}[y_n^2] - (\mathbb{E}[y_n])^2 = 3\sigma^4 - \sigma^4 = 2\sigma^4 ]

4. Substitute back into the variance formula: $\text{Var}(t) = \frac{1}{N^2} \cdot N \cdot 2\sigma^4 = \frac{2\sigma^4}{N}$

Final Results:

1. Unbiasedness: $\mathbb{E}[t] = \sigma^2$

2. Variance of $t$: $\text{Var}(t) = \frac{2\sigma^4}{N}$

Do you want further clarifications or detailed derivations? Let me know! Here are some additional related questions:

Related Questions:

1. What changes if $\mu$ is unknown and replaced by the sample mean?
2. How is $t$ related to the sample variance $S^2$?
3. Can $t$ be considered a maximum likelihood estimator (MLE)?
4. How would the results change for a non-normal distribution?
5. What are the implications of $N \to \infty$ on the variance of $t$?

Tip:

When working with estimators, always check both unbiasedness and variance to understand the trade-off between accuracy and variability.