# Does the sum of squared "dependent" Gaussian variables still exhibit a Chi-squared distribution?

Suppose we have $$x_1,\cdots,x_n$$, such that $$x_i$$ are i.i.d according to $$x_i\sim N(0,1)$$ for $$i=1,\cdots,n$$. Then we know that $$Y=\sum_{i=1}^nx_i^2$$

$$$$Y\sim \chi^2(n)$$$$

Now suppose $$\tilde{x}_1,\cdots,\tilde{x}_n$$ are not longer independent and follow an AR(1) process - i.e.,

$$$$\tilde{x}_t=\rho \tilde{x}_{t-1}+u_t,\quad u_t \sim N(0,\sigma^2_u)\quad\text{and}\quad \lvert \rho\rvert<1$$$$ thus, implying $$\tilde{x}_t\sim N(0, \sigma^2_u/1-\rho^2)$$. Can anything be said about the distribution of $$\tilde{Y}_t=\sum_{i=1}^n\tilde{x}_i^2$$? For instance

$$$$\frac{\sqrt{1-\rho^2}}{\sigma_u} \tilde{Y_t}\overset{?}{\sim} \chi^2(?)$$$$ with perhaps adjusted degrees of freedom?

## 1 Answer

Suppose we have $$X_1 = u_1$$ and for $$i>1$$ $$X_i = \rho X_{i-1} + u_i$$. I'll set $$\sigma^2_u = 1$$ for simplicity. Then $$\begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 & \dots & 0 \\ \rho & 1 & 0 & 0 & \dots & 0 \\ \rho^2 & \rho & 1 & 0 & \dots & 0 \\ &&&\vdots&&\\ \rho^{n-1} & \rho^{n-2} & &\dots& & 1 \end{bmatrix}\begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n\end{bmatrix}$$ so $$X \sim \mathcal N(\mathbf 0, \Omega_\rho)$$ where $$\Omega_\rho = L_\rho L_\rho^T$$ with $$L_\rho$$ being the above matrix.

This means that $$Y := X^TX = u^TL_\rho^TL_\rho u$$ is a Gaussian quadratic form. If $$L_\rho^TL_\rho$$ were idempotent we'd be able to use Cochran's theorem to get a chi squared distribution, but that won't be true here. Quadratic forms with Gaussian RVs have been studied for years and there are lots of results and papers out there such as "The Distribution of Quadratic Forms of Gaussian Vectors" by Zorin and Iyubimov (1987).

It's a standard result that $$\text E[X^TX] = \text{tr}(\Omega_\rho)$$ and $$\text{Var}[X^TX] = 2 \text{tr}(\Omega_\rho^2).$$ If we are to have $$Y$$ follow a central chi-squared distribution then we'd need $$2 \text E[Y] = \text {Var}[Y]$$ which in this case implies $$\text{tr}(\Omega_\rho) = \text{tr}(\Omega_\rho^2)$$ which is not true in general. This means $$Y$$ cannot have a central chi squared distribution.

We can also use this to show that $$Y$$ does not necessarily have a noncentral chi squared distribution. If $$Y \sim \chi^2_k(\delta)$$ then $$\text E[Y] = k+\delta$$ and $$\text {Var}[Y] = 2(k + 2\delta)$$. Matching the first two moments like this leads to the system $$k + \delta = \text{tr}(\Omega_\rho) \\ k + 2\delta = \text{tr}(\Omega_\rho^2)$$ so $${k \choose \delta} = \begin{bmatrix} 2 & -1 \\ -1 & 1\end{bmatrix}{\text{tr } \Omega_\rho \choose \text{tr }\Omega_\rho^2}$$ i.e. $$k = 2\cdot\text{tr }\Omega - \text{tr }\Omega^2 \\ \delta = \text{tr }\Omega^2 - \text{tr }\Omega.$$

The problem is that this can lead to negative values of $$k$$, so this means that it also cannot be that $$Y$$ is guaranteed to be a noncentral chi squared.

• (+1) This is almost correct, but I note your simplifying assumption for the first error term. To get an exact answer you would take $u_1 \sim \text{N}(0, \sigma_u^2/(1-\rho^2))$ instead of $u_1 \sim \text{N}(0,1)$, so the variance matrix for the error vector isn't the identity matrix (and consequently $\Omega_\rho \neq L_\rho L_\rho^\text{T}$). I'll just leave this comment here to note this approximation, in case anyone wants to do the additional work to get the exact answer.
– Ben
Aug 31, 2021 at 20:07
• @Ben ah thank you! If I have time I'll try to update with the exact answer (unless someone else does it first)
– jld
Aug 31, 2021 at 20:23