# Quadratic form of a bivariate normal

This is a homework problem. Let $(X,Y)\sim N(\mu_1,\mu_2,\sigma^2_1,\sigma^2_2,\rho)$. Show that if $\sigma_1,\sigma_2 >0,|\rho|<1$, then $$\dfrac{1}{1-\rho^2}\left\{\dfrac{(X-\mu_1)^2}{\sigma^2_1}-2\rho\dfrac{(X-\mu_1)(Y-\mu_2)}{\sigma_1\sigma_2}+\dfrac{(Y-\mu_2)^2}{\sigma^2_2}\right\}$$ has a $\chi^2_2$ distribution.

EDIT:

There is a generalization of this: $(X-\mu)^T\Sigma^{-1}(X-\mu) \sim \chi^2_n$ where $X=(X_1,\ldots,X_n)^T$. The proof is theorem 7 here (http://www2.econ.iastate.edu/classes/econ671/Hallam/documents/QUAD_NORM.pdf)

Can anyone think of a better way to show this?

• Can you solve this for the special case where $\mu_1=\mu_2=0$ and $\sigma_1=\sigma_2=1$? (Hint: what distribution does $Y-\rho X$ have?) If you can, you're done, because the general case easily reduces to this special case upon standardizing the variables. – whuber Sep 10 '13 at 1:12
• I don't see the connection: $Y-\rho X\sim N(-\rho, \sigma^2_y-\rho^2\sigma^2_x-2\rho^2\sigma_x\sigma_y)$. – bdeonovic Sep 10 '13 at 15:05
• I think you may have miscalculated, Benjamin. After all, consider the case of unit variance variables ($\sigma_x=\sigma_y=1$): you are asserting the variance of $Y-\rho X$ is $1-\rho^2-2\rho^2$ = $1-3\rho^2$, but this is negative whenever $\rho^2 \gt 1/3$, which is certainly a possibility. Review your computation of the negative signs :-). Please notice, too, that I did not suggest you look at $Y-\rho X$ in general: look at this combination only for standardized variables. – whuber Sep 10 '13 at 15:09
• Ah, yes I see my mistake. It should be $Y-\rho X\sim N(-\rho, 1-\rho^2)$. So I can make it look like $(1-\rho^2)^{-1}(Y(Y-\rho X)+X(X-\rho Y))$ but I do not know what product of correlated normals is. Perhaps this is not what you intended. – bdeonovic Sep 10 '13 at 15:28
• When $X$ and $Y$ are standardized, the variance of $Y-\rho X$ expands into the sum Var$(Y)$ - $2\rho$Cov$(Y,X)$ + Var$(X)$ = $1 - 2\rho\rho + 1$ = $2-2\rho^2$. This is a constant. Moreover, the mean of $Y-\rho X$ obviously is zero because both $Y$ and $X$ have zero means. Yet, because $Y-\rho X$ is a linear combination of Normal variables, it must have a Normal distribution and we have found that it has to be $\sqrt{2}\sqrt{1-\rho^2}$ times a standard Normal distribution. – whuber Sep 10 '13 at 15:32

As pointed out in the comments, we can translate and scale in order to consider the case $\mu_1=\mu_2=0$ and $\sigma_1=\sigma_2$. Call $Z$ the random variable we want to show it has a $\chi^2$ distribution. Then $$Z=\frac 1{1-\rho^2}\left(\left(X-\rho Y\right)^2+(1-\rho^2)Y^2\right).$$ Notice that $X-\rho Y$ and $Y$ are non-correlated Gaussian distributions.