Let $\{X_i\}_{1\le i\le n}$ be a sample of dependent random variables with $X_i\sim\mathscr N(\mu,\sigma^2)$ and $\mathrm{Cov}[X_i,X_j]=\rho\sigma$ for $i\neq j$. Find the mean and variance of $\bar X=\frac1n\sum\limits_{i=1}^nX_i$.
Computing the mean is easy:
$$\mathrm E\big[\bar X\big]=\mathrm E\bigg[\frac1n\sum_{i=1}^nX_i\bigg]=\frac1n\sum_{i=1}^n\mathrm E\big[X_i\big]=\frac{n\mu}n=\mu$$
Computing the variance, however, I find a bit confusing. I don't think I neglected anything in my work, but I have a nagging feeling that I might have. I know that
$$\mathrm V\big[\bar X\big]=\mathrm E\big[\bar X^2\big]-\mathrm E\big[\bar X\big]^2=\mathrm E\big[\bar X^2\big]-\mu^2$$
so here's what I've done to compute the mean of $\bar X^2$:
$$\begin{align*}
\mathrm E\big[\bar X^2\big]&=\mathrm E\Bigg[\frac1{n^2}\left(\sum_{i=1}^n{X_i}^2+2\sum_{1\le i<j\le n}X_iX_j\right)\Bigg]\\[1ex]
&=\frac1{n^2}\left(\sum_{i=1}^n\mathrm E\big[{X_i}^2\big]+2\mathrm \sum_{1\le i<j\le n}\mathrm E[X_iX_j]\right)\\[1ex]
&=\frac1{n^2}\left(\sum_{i=1}^n\left(\mathrm V[X_i]+\mathrm E[X_i]^2\right)+2\sum_{1\le i<j\le n}\left(\mathrm{Cov}[X_i,X_j]+\mathrm E[X_i]\mathrm E[X_j]\right)\right)\\[1ex]
&=\frac1{n^2}\left(\sum_{i=1}^n(\sigma^2+\mu^2)+2\sum_{1\le i<j\le n}(\rho\sigma+\mu^2)\right)\\[1ex]
&=\frac{n(\sigma^2+\mu^2)+2\frac{n(n-1)}2(\rho\sigma+\mu^2)}{n^2}\\[1ex]
&=\mu^2+\frac{n-1}n\rho\sigma+\frac{\sigma^2}n
\end{align*}$$
Is this derivation correct?
Best Answer
Yes, looks ver much correct to me. However, I think that expanding the formula of variance by using sum of variances and twice covariances would’ve been much technically easier.