I'm trying to prove that the covariance of a bivariate normal distribution, $Cov(X,Y)$, is equal to $\rho\sigma_X\sigma_Y$. I'm getting stuck.

The approach I have to use is to apply a change of variables to standardize the variables X and Y, and apply the formula for covariance $Cov(X,Y) = E[(X-\mu_X)(Y-\mu_Y)]$, which is equal to:


Where $f(x,y)$ is the nasty pdf for a bivariate normal. Then, when you standardize and do a change of variable ... and complete the square... you get:

$$Cov(X,Y)=\sigma_X\sigma_Y\int_{-\infty}^\infty\int_{-\infty}^\infty \frac{uv}{2\pi\sqrt{1-\rho^2}}\exp\left(-\frac{(v-\rho u)^2}{2(1-\rho^2)} - \frac{u^2}{2}\right) dudv$$

So, this can be separated into two integrals that look like the expected value of normal random variables...

$$Cov(X,Y)=\sigma_X\sigma_Y\int_{-\infty}^\infty \frac{u}{\sqrt{2\pi}} \exp{\left(\frac{-u^2}{2}\right)} \int_{-\infty}^\infty\frac{v}{\sqrt{2\pi}\sqrt{1-\rho^2}}\exp\left( \frac{-(v-\rho u)^2}{2(1-\rho^2)} \right)dvdu$$

But that looks like $\sigma_X\sigma_Y*E[N(0,1)]*E[N(\rho u, 1-\rho^2)]$ which is ... zero! Because the expected value of $N(0,1)$ is 0. But it really should come out to $\rho\sigma_X\sigma_Y$ What am I missing?


1 Answer 1


But that looks like $σ_X σ_Y \cdot \mathsf E[N(0,1)]\cdot \mathsf E[N(ρu,1−ρ^2 )]$ which is ... zero! Because the expected value of $N(0,1)$ is $0$. But it really should come out to $ρσ_X σ_Y$.

What am I missing?

You don't have a product of two separate integrals.

$$\int_\Bbb R g(u) \left(\int_\Bbb R h(u,v)\operatorname d v\right)\operatorname d u \quad\neq\quad \left(\int_\Bbb R g(u) \operatorname d u \right)\cdot \left(\int_\Bbb R\int_\Bbb R h(u,v)\operatorname d v\operatorname d u \right)$$

  • $\begingroup$ Oh man, after looking at it again and evaluating it properly, it all falls into place now! Thank you! $\endgroup$
    – user19650
    Commented Apr 21, 2016 at 7:03

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