1. ## correlation

Suppose $(X_1, X_2)$ has the bivariate normal distribution with density function
$f(x_1, x_2) = \frac{1}{2\pi \sqrt{1-p^2}} exp[- \frac{x_1^2 - 2p x_1 x_2 + x_2^2}{2(1-p^2)}]$

Show that $Corr(X_1^2, X_2^2) = p^2$

2. Originally Posted by graticcio
Suppose $(X_1, X_2)$ has the bivariate normal distribution with density function
$f(x_1, x_2) = \frac{1}{2\pi \sqrt{1-p^2}} exp[- \frac{x_1^2 - 2p x_1 x_2 + x_2^2}{2(1-p^2)}]$

Show that $Corr(X_1^2, X_2^2) = p^2$
First you should recognise from the joint pdf that $E(X_1) = E(X_2) = 0$ and $Var(X_1) = Var(X_2) = 1$.

Therefore $\text{Corr} (X_1^2, X_2^2) = \frac{\text{Cov} (X_1^2, X_2^2)}{\sigma_1 \sigma_2} = \frac{E(X_1^2 \cdot X_2^2) - E(X_1^2) \cdot E(X_2^2)}{\sigma_1 \sigma_2} = E(X_1^2 \cdot X_2^2) - 1$. Mr F Edit: The mistake I mention below is here. See post #3.

Note: $Var(X_i) = E(X_i^2) - [E(X_i)]^2 \Rightarrow 1 = E(X_i^2) - 0^2 \Rightarrow E(X_i^2) = 1$.

So the task is to calculate $E(X_1^2 \cdot X_2^2)$:

$E(X_1^2 \cdot X_2^2) = \int_{x_1 = -\infty}^{+\infty} \int_{x_2 = -\infty}^{+\infty} x_1^2 \, x_2^2 \, \frac{1}{2 \pi \sqrt{1 - \rho^2}} \exp \left( -\frac{(x_1^2 - 2 \rho x_1 x_2 + x_2^2)}{2 (1 - \rho^2)}\right) \, dx_2 \, dx_1$

Now complete the square and re-arrange:

$= \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^2 \, \exp \left( - \frac{x_1^2}{2} \right) \left[ \int_{x_2 = -\infty}^{+\infty} x_2^2 \, g(x_1, x_2) \, dx_2 \right] \, dx_1$

where $g(x_1, x_2) = \frac{1}{\sqrt{2 \pi (1 - \rho^2)}} \exp \left( -\frac{(x_2 - \rho x_1)^2}{2 (1 - \rho^2)}\right)$ is the pdf of $N(\mu = \rho x_1, \, \sigma^2 = 1 - \rho^2)$.

Now note that $\int_{y = -\infty}^{+\infty} y^2 \, g(x_1, y) \, dy$ represents $E(Y^2)$ and so:

$Var(Y) = E(Y^2) - [E(Y)]^2 \Rightarrow 1 - \rho^2 = E(Y^2) - \rho^2 x_1^2 \Rightarrow E(Y^2) = (1 - \rho^2) + \rho^2 x_1^2$.

Therefore:

$E(X_1^2 \cdot X_2^2) = \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^2 \, \exp \left( - \frac{x_1^2}{2} \right) [(1 - \rho^2) + \rho^2 x_1^2] \, dx_1$

$= (1 - \rho^2) \, \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^2 \, \exp \left( - \frac{x_1^2}{2} \right) \, dx_1 + \rho^2 \, \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^4 \, \exp \left( - \frac{x_1^2}{2} \right) \, dx_1$

$= (1 - \rho^2) E(X_1^2) + \rho^2 E(X_1^4)$

where the moments are for a standard normal distribution. Cheating and refering to a table of moments (see Normal distribution - Wikipedia, the free encyclopedia):

$= (1 - \rho^2) + 3 \rho^2 = 1 + 2 \rho^2$.

Therefore $\text{Corr} (X_1^2, X_2^2) = E(X_1^2 \cdot X_2^2) - 1 = 1 + 2 \rho^2 - 1 = 2 \rho^2$.

This is different from the result that was given to be proved so there's a mistake somewhere (that I don't have time to look for). Nevertheless, I'm sure you get the idea.

3. Originally Posted by mr fantastic
First you should recognise from the joint pdf that $E(X_1) = E(X_2) = 0$ and $Var(X_1) = Var(X_2) = 1$.

Therefore $\text{Corr} (X_1^2, X_2^2) = \frac{\text{Cov} (X_1^2, X_2^2)}{\sigma_1 \sigma_2} = \frac{E(X_1^2 \cdot X_2^2) - E(X_1^2) \cdot E(X_2^2)}{\sigma_1 \sigma_2} = E(X_1^2 \cdot X_2^2) - 1$.

[snip]
The above is wrong. It should be:

Therefore $\text{Corr} (X_1^2, X_2^2) = \frac{\text{Cov} (X_1^2, X_2^2)}{ sd(X_1^2) \, sd(X_2^2) } = \frac{E(X_1^2 \cdot X_2^2) - E(X_1^2) \cdot E(X_2^2)}{sd(X_1^2) \, sd(X_2^2)} = \frac{E(X_1^2 \cdot X_2^2) - 1}{sd(X_1^2) \, sd(X_2^2)}$.

I leave it for you to calculate the required standard deviations. The technique is in post #2.

Edit: $E(X_i^4) = 3$ and $E(X_i^2) = 1$ therefore $Var(X_i^2) = 2$ and all is well.