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.


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?

  • 2
    $\begingroup$ 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. $\endgroup$ – whuber Sep 10 '13 at 1:12
  • $\begingroup$ I don't see the connection: $Y-\rho X\sim N(-\rho, \sigma^2_y-\rho^2\sigma^2_x-2\rho^2\sigma_x\sigma_y)$. $\endgroup$ – bdeonovic Sep 10 '13 at 15:05
  • $\begingroup$ 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. $\endgroup$ – whuber Sep 10 '13 at 15:09
  • $\begingroup$ 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. $\endgroup$ – bdeonovic Sep 10 '13 at 15:28
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.