Covariance and Correlation in Multinormal random variable

Find the covariance and correlation of $N_i$ and $N_j$, where $N_1, N_2, \ldots,N_r$ are multinormal random variable.

At the beginning, I think that I have: $$P(N_1=n_1,N_2=n_2,\ldots,N_r=n_r)=\frac{1}{n_1!n_2!\cdots n_r!} p_1^{n_1} p_2^{n_2} \cdots p_r^{n_r}$$

Then, I think I should calculate $P(N_i), P(N_j), P(N_i,N_j)$, but I don't know how.

• Might you have meant "multinomial"? At first I thought you meant multivariate normal, but then your question wouldn't make much sense. ${}\qquad{}$ – Michael Hardy Apr 25 '15 at 22:02

The probability that you wrote above should be $$P(N_1=n_1,N_2=n_2,\ldots,N_r=n_r)=\frac{n!}{n_1!n_2!\cdots n_r!} p_1^{n_1} p_2^{n_2} \cdots p_r^{n_r}.$$
You have $n$ independent trials and at each trial you choose one of $r$ alternatives, with probabilities $p_1,\ldots,p_r$. The random variable $N_k$ is how many times the $k$th alternative was chosen.
For $k=1,\ldots,r$, let $M_k$ be the number of times the $k$th alternative was chosen on the first trial, so that $M_k$ must be either $0$ or $1$.
A consequence of the fact that the $n$ trials are independent and identically distributed is that $\cov(N_k,N_j) = n\cov(M_k,M_j)$. Assuming $k\ne j$, we can now write \begin{align} & \cov(M_k,M_j) = \E(M_k M_j) - \E(M_k)\E(M_j) \\[8pt] = {} & \Pr(M_k=1=M_j) - \Pr(M_k=1)\Pr(M_j=1) \\[8pt] = {} & 0 - p_kp_j. \end{align} The correlation is the covariance divided by the product of the two standard deviations. The standard deviation is the square root of the variance. The variance is \begin{align} & \operatorname{var}(M_k) = \E(M_k^2) - (\E(M_k))^2 \\[8pt] = {} & \Pr(M_k=1) - (\Pr(M_k=1))^2 = p_k - p_k^2 = p_k(1-p_k). \end{align} This may be familiar to you from seeing it stated that the variance of a Bernoulli distribution is $pq$, where $p+q=1$. The random variable $M_k$ is Bernoulli-distributed.