# derive the mean and variance of $\bar X$ using means of sums rules

I can't find anywhere what the means of sums rules are so i'm confused with this question

The random variables $X_1......X_5$ are jointly multivariate normal. Their expectations are $E(x)= \mu_i$ and variances are $\sigma_i^2$ for $i=1,2....5$ respectively. The correlations are: $Corr(X_i,X_j)= \frac {E(X_iX_j)-\mu_i\mu_j}{\sigma_i\sigma_j} = \rho$ for all $1 \le i\ne j \le 5$.

a) Derive the mean of $\bar X = \sum_{1 \le i \le 5} \frac {X_i}{n}$ using all the rules for computing the means of sums

b) Derive the variance of $\bar X$ of part a) using all the rules for computing variances of sums

Let $X_1,..., X_n$ be random variables, not necessarily independent.

Then

$$E(\sum_{i=1}^n X_i) = \sum_{i=1}^n E(X_i)$$

$$var(\sum_{i=1}^n X_i) = \sum_{i=1}^n var(X_i) + 2\sum_{i \neq j}cov(X_i, X_j)$$ $$= \sum_{i=1}^n var(X_i) +2\sum_{i \neq j}\sigma_i\sigma_j corr(X_i, X_j).$$

EDIT So in this case

$$var(\overline{X}) = \frac{1}{n^2}var(\sum_{i=1}^n X_i) = \frac{1}{n^2} (\sum_{i=1}^n\sigma_i^2 + 2\rho \sum_{i \neq j}\sigma_i\sigma_j )$$

• So I just did the standard $E(\bar x)=E(\frac {X_1...X_5}{n})$ then $\frac {1}{n} E(X_1)+E(X_2)+...E(X_5)$ then $\frac {1}{n} * n\mu = \mu$ is this correct or is it different because these are multivariate $X_i$ – moku Mar 25 '14 at 19:23
• Yes, that's exactly right (if you put brackets around the sum $E(x_1 +...+X_n)$) – Frank Mar 25 '14 at 19:28
• alright thxs mate. So there is no difference even tho the $X_i$'s represent distributions and not observations from one sample – moku Mar 25 '14 at 19:30
• I'm not sure what you mean, but for any collection of RV's $X_1,...,X_n$, whether they have the same distribution or not, or are independent or not, the above formulas hold. – Frank Mar 25 '14 at 19:32
• so what is the $corr(X_i,X_j)$ simplified further like $Var(X_i)= \frac {\sigma^2}{n}$ – moku Mar 25 '14 at 19:50