Question: I just want to confirm whether this is correct or not:

> 
> The covariance of $X$ and $Y$ is equal to the standard deviation of $X$ times the standard deviation of $Y$.
> 
> 
> 

or, in mathematical notation,

$$
\mathrm{Cov}(X,Y) = \sqrt{\mathrm{Var(X)}}\sqrt{\mathrm{Var}(Y)}
$$

Answer: 
This is false in general. In fact, it is true **if and only if** $X=aY+b$ (almost surely) for some fixed constants $a \geq 0$ and $b \in \mathbb{R}$. That is $X$ and $Y$ must be positively linearly related for this to hold. [Wikipedia](http://en.wikipedia.org/wiki/Covariance) also has a decent page on this.

For a counterexample to your statement, consider *any* two **independent** random variables $X$ and $Y$ each with strictly positive variance. Then,
$
\mathrm{Cov}(X,Y) = 0 \>,
$
but,
$
\sqrt{\mathrm{Var}(X)} \sqrt{\mathrm{Var}(Y)} > 0 \>.
$

---

A quick proof (and a slick one, I think; it's not my own) of the *if and only if* assertion uses the Cauchy–Schwarz inequality and goes as follows. Let $U$ and $V$ be random variables such that $\newcommand{\e}{\mathbb{E}}\e U^2 < \infty$ and $\e V^2 < \infty$. Then, $|\e UV|^2 < \e U^2 \e V^2$ if and only if $\e (t U - V)^2 > 0$ for all $t$. But, if $\e(t\_0 U - V)^2 = 0$ for some $t\_0$, then $t\_0 U = V$ almost surely. Now, set $U = X - \e X$ and $V = Y - \e Y$ to get the desired conclusion.
While cardinal has answered the question, I wanted to add a little something extra.

The Cauchy-Schwarz inequality shows that $\left|\operatorname{Cov}(X,Y)\right|\leq \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}$, with equality if and only if $X$ and $Y$ are linearly dependent. In general, you can take the ratio $\rho=\frac{\operatorname{Cov}(X,Y)}{ \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}}$ which will be between $-1$ and $1$. This gives the correlation between $X$ and $Y$. It is $1$ when the variables are perfectly correlated (one goes up exactly when the other does), $-1$ when they are perfectly anti-correlated (one goes up exactly when the other goes down), $0$ when they are independent (knowing that one goes up tells you nothing about the behavior of the other), and in general gives a measure of the behavior when things are somewhere in between.

One of the first places where this comes up is [least squares approximation.](http://en.wikipedia.org/wiki/Least_squares) . While you can always do a least squares approximation to get a line of best fit for your data points, the correlation coefficient tells you whether your line of best fit is actually a good fit. It will be near $0$ for data that is essentially random, and small in magnitude for data which is non-linear.