covariance of normal distribution Suppose we have two normal distribution $\mathcal{N}(0,s)$, $\mathcal{N}(0,t)$. 
how can I find the covariance if I don't have the distribution function $f(x,y)$ or anything, not given the correlation (but they are correlated); I only know the distributions.
 A: You cannot find the covariance of possibly correlated random variables $X$ and $Y$ if all you know is the individual distributions of $X$ and $Y$.  If you know the joint distribution, then you can find the covariance.  In particular, in your case if you knew the joint density function $f(x,y)$ of $X$ and $Y$, you could compute the covariance.  You don't really need to know $f(x,y)$ fully to find the covariance. For example, if you know $E(XY)$, you can find the covariance, and there are other sorts of information from which you could find the covariance. But you do need more information than the individual distributions. 
Examples: Let $X$ and $Y$ be independent and normally distributed. Then $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(X)E(Y)-E(X)E(Y)=0$.
Let $X$ be normally distributed with mean $0$ and variance $1$. Let $Y=2X$.
Then  $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(2X^2)-E(X)E(2X)$. But $E(2X^2)=2E(X^2)$. And since $\text{Var}(X)=E(X^2)-(E(X))^2$, we have $E(X^2)=1$. It follows that $\text{Cov}(X,Y)=2$.
Comment: Let $\sigma$ and $\tau$ be any two positive numbers, and let $c$ be any number such that $|c|\le \sigma\tau$. Then there are normally distributed random variables $X$ and $Y$, with variances $\sigma^2$ and $\tau^2$ respectively, such that $\text{Cov}(X,Y)=c$. (The means of $X$ and $Y$ are irrelevant, and can be taken to be $0$.)  So apart from the condition 
$$-\sigma\tau \le \text{Cov}(X,Y) \le \sigma\tau,$$
the covariance of $X$ and $Y$ can be "anything."  Thus, in a very strong sense, if we know the variances of $X$ and $Y$, we know very little about $\text{Cov}(X,Y)$, only an inequality. 
A: If the correlation is $\rho,$ then the covariance is $\rho\sqrt{st\,}$ (provided what you meant is that $s$ and $t$ are the variances.
