Find the mean and variance of $V_n=\frac{1}{n}\sum_{i=1}^n(X_i-u)^2$

Question

Suppose that $X_1,X_2,...,X_n$ is a random sample from a distribution with mean $\mu$ and variance $\sigma^2$. Suppose also that $v:=\mathbb{E}[(X_1-\mu)^4]<\infty$. Find the mean and variance of $V_n=\frac{1}{n}\sum_{i=1}^n(X_i-\mu)^2$

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My attempt

$V_n=\frac{1}{n}\sum_{i=1}^n(X_i-\mu)^2=\frac{1}{n}\sum_{i=1}^n(X_i^2-2\mu X_i+\mu^2)$

Thus,

$\mathbb{E}[V_n]=\mathbb{E}[\frac{1}{n}\sum_{i=1}^n(X_i^2-2\mu X_i+\mu^2)]=\mathbb{E}[\frac{1}{n}\sum_{i=1}^nX_i^2]-\mathbb{E}[\frac{1}{n}\sum_{i=1}^n2\mu X_i]+\mathbb{E}[\frac{1}{n}\sum_{i=1}^n\mu^2]=\mathbb{E}[\frac{1}{n}\sum_{i=1}^nX_i^2]-\frac{1}{n}\sum_{i=1}^n\mathbb{E}[2\mu X_i]+\frac{1}{n}\sum_{i=1}^n\mathbb{E}[\mu^2]=\mathbb{E}[\frac{1}{n}\sum_{i=1}^nX_i^2]-2\mu^2+\mu^2=\mathbb{E}[\frac{1}{n}\sum_{i=1}^nX_i^2]-\mu^2$

This is where I get a little unsure.

$\mathbb{E}[\frac{1}{n}\sum_{i=1}^nX_i^2]=\frac{1}{n}\sum^n_{i=1}\mathbb{E}[X_i^2]=\frac{1}{n}\sum^n_{i=1}\mathbb{E}[X_i^2]=\frac{1}{n}\mathbb{E}[X^2]$

This seems unsatisfactory. How should I proceed?

Answer 1

For the expectation, using linearity of expectation, we have \begin{align} \mathbb{E}[V_n] &= \mathbb{E}[\frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2]\\ &=\frac{1}{n} \sum_{i=1}^n \mathbb{E}[(X_i - \mu)^2]\\ &=\frac{1}{n} \sum_{i=1}^n \sigma^2 = \sigma^2 \end{align} Here, the fact "variance of $X_i$ is $\mathbb{E}[(X_i - \mu)^2]$" is used.

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For the variance, since random variables $X_i - \mu$ are independent, we have \begin{align} Var[V_n] &= Var[\frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2]\\ &= \frac{1}{n^2} \sum_{i=1}^n Var[(X_i - \mu)^2]\\ &= \frac{1}{n^2} \sum_{i=1}^n (\mathbb{E}[(X_i - \mu)^4] - \mathbb{E}[(X_i - \mu)^2]^2) \\ &= \frac{1}{n^2} \sum_{i=1}^n (v - \sigma^4)= \frac{1}{n} (v - \sigma^4) \end{align}

In the third equality, we use $Var[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$.