# Mean and variance of $Y'\Sigma^{-1}Y-Y_1^2/\sigma_1^2$ when $Y\sim N_2(0,\Sigma)$

Let $$\underline Y=(Y_1,Y_2)'$$ have the bivariate normal distribution $$N_2(\underline0,\Sigma)$$, where $$\Sigma=\begin{pmatrix}\sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2\end{pmatrix}.$$ Obtain the mean and variance of $$U=\underline Y'\Sigma^{-1}\underline Y-Y_1^2/\sigma_1^2$$.

I could obtain the Mean of $$U$$ easily since $$Y'\Sigma^{-1}Y$$ is distributed as Chi- Square with df-2 and $$Y_1^2/\sigma_1^2$$ with df 1. But I am having trouble obtaining variance since covariance is also present here. Can you please help in finding covariance?

• Hint: $U = Y^\prime(\Sigma^{-1} - T)Y$ where $T=\pmatrix{1/\sigma_1^2&0\\0&0}.$ – whuber Jun 24 at 13:41
• Does the answer help? Actually the problem is small enough to just expand $U$ and show it has a chi-square distribution directly without using any special result. – StubbornAtom Jun 25 at 21:55
• Yes, the answer is really helpful! I actually tried to expand U in terms of Y1 and Y2 but got stuck because after the expansion I had to find covariance between some terms that did not look good. – rick Jun 26 at 20:19
• You would only need the covariance between $Y_1$ and $Y_2$, nothing beyond that. Please try to typeset the question using MathJax whenever possible. I added the self-study tag to your post in line with our policy on homework/exam questions. – StubbornAtom Jun 27 at 7:55

As @whuber mentioned, you have $$U=Y'AY$$ where $$A=\Sigma^{-1}-\begin{pmatrix}1/\sigma_1^2 & 0 \\ 0& 0\end{pmatrix}$$.

Note that $$A$$ is symmetric, so you can use the result in the question here to get the variance:

$$\operatorname{Var}(U)=2\operatorname{tr}((A\Sigma)^2)$$

Also observe that $$A\Sigma=I_2-\begin{pmatrix}1/\sigma_1^2 & 0 \\ 0& 0\end{pmatrix}\Sigma=\begin{pmatrix}0 & -\rho\frac{\sigma_2}{\sigma_1} \\ 0& 1\end{pmatrix}$$

Calculation of the variance would be somewhat simpler if you notice that $$A\Sigma$$ is an idempotent matrix, which means you can also use this theorem for the exact distribution of $$U$$:

Suppose $$Y\sim N(\mathbf 0,\Sigma)$$ where $$\Sigma$$ is positive definite and let $$A$$ be a symmetric matrix. Then $$Y'AY\sim \chi^2_r$$ if and only if $$A\Sigma$$ is idempotent (or equivalently $$A\Sigma A=A$$) and $$\operatorname{rank}(A\Sigma)=r$$.

This is part of a general result proven here. You can find more on distributions of quadratic forms of multivariate normal distribution in standard textbooks (Rao's Linear Statistical Inference and Its Applications and Seber/Lee's Linear Regression Analysis for example).

Indeed $$U\sim \chi^2_1$$ but that is not to say that difference of arbitrary chi-square variables has a chi-square distribution.