Sorry if the formatting is poor, this is my first time asking a question. I'm investigating how squared gaussians behave, using the techniques provided here, which are giving me inconsistent results. Here's the (simplified) setup: I have four random variables, $Q_1, Q_2, Q_3, Q_4$, which are distributed as follows: $$ \left[ \begin{array}{c} Q_1\\ Q_2\\ Q_3\\ Q_4\\ \end{array} \right] \sim N_4\left(\left[\begin{array}{c} 0\\ 0\\ 0\\ 0 \end{array} \right], \left[ \begin{array}{cccc} 1&0&\alpha&0\\ 0&1&0&\beta\\ \alpha&0&1&0\\ 0&\beta&0&1\\ \end{array} \right]\right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \alpha,\beta \in [0,1] $$

I'm dealing with the distribution $BQ_1^2 + CQ_2^2 - 2Q_1Q_2$ (D1) where $B,C$ are constants. Furthermore I want to contrast the behavior of that distribution to distribution $\frac{1}{2}(BQ_1^2 + CQ_2^2 - 2Q_1Q_2 + BQ_3^2 + CQ_4^2 - 2Q_3Q_4)$ (D2), under the circumstances where $\alpha = \beta = 1$. Ideally, and as I understand it, they are distributed the same way under that correlation assumption, as the repeated terms are perfectly correlated. Using the techniques mentioned in the linked question, and an "A" matrix (again, see the question I linked to above) that looks like $$ \left[ \begin{array}{cccc} B&-1&0&0\\ -1&C&0&0\\ 0&0&B&-1\\ 0&0&-1&C\\ \end{array} \right] $$

I get that (D2) as above is distributed according to $$\frac{1}{2}\left(\left(\frac{\sqrt{1 - \alpha}}{2} + \frac{\sqrt{1 + \alpha}}{2}\right)^2 BT_1 + \left(\frac{\sqrt{1 - \beta}}{2} + \frac{\sqrt{1 + \beta}}{2}\right)^2 CT_2 + \left(\frac{\sqrt{1 - \alpha}}{2} + \frac{\sqrt{1 + \alpha}}{2}\right)^2 BT_3 + \left(\frac{\sqrt{1 - \beta}}{2} + \frac{\sqrt{1 + \beta}}{2}\right)^2 CT_4\right)$$ where each $T_i$ is an independent chi square random variable with d.o.f. 1. Under the assumption that $\alpha = \beta = 1$, this becomes $$\frac{1}{2}\left(\frac{1}{2}BT_1 + \frac{1}{2}CT_2 + \frac{1}{2}BT_3 + \frac{1}{2}CT_4\right)$$ However, using a truncated $A$ matrix to discern the distribution of $D1$, $$ \left[ \begin{array}{cc} B&-1\\ -1&C\\ \end{array} \right] $$ And similarly a truncated covariance matrix that is just the 2X2 identity, I get that $(D1)$ is distributed according to $$\frac{1}{2}(B + C - \sqrt{4 + B^2 - 2 B C + C^2})Y_1 + \frac{1}{2}(B + C + \sqrt{4 + B^2 - 2 B C + C^2})Y_2$$ Where $Y_i$ are again independent chi square random variables with d.o.f. 1.

My issue is that $\frac{1}{2}(B + C - \sqrt{4 + B^2 - 2 B C + C^2})Y_1 + \frac{1}{2}(B + C + \sqrt{4 + B^2 - 2 B C + C^2})Y_2$ doesn't particularly look like $\frac{1}{2}\left(\frac{1}{2}BT_1 + \frac{1}{2}CT_2 + \frac{1}{2}BT_3 + \frac{1}{2}CT_4\right)$, even though we would expect that $BQ_1^2 + CQ_2^2 - 2Q_1Q_2 = \frac{1}{2}(BQ_1^2 + CQ_2^2 - 2Q_1Q_2 + BQ_3^2 + CQ_4^2 - 2Q_3Q_4)$ when $Q_1 = Q_3$ and $Q_2 = Q_4$. I've tried massaging them to work out the same to no avail; it turns out the rest of my mathematics not shown here allows me to put these "into" moment generating functions. The relevant term for $(D1)$ becomes $$(1- (B + C - \sqrt{4 + B^2 - 2 B C + C^2}))^\frac{1}{2}(1-(B + C + \sqrt{4 + B^2 - 2 B C + C^2}))^\frac{1}{2}$$ and the relevant term for (D2) becomes $$(1-\frac{1}{2}B)^\frac{1}{2}(1-\frac{1}{2}C)^\frac{1}{2}(1-\frac{1}{2}B)^\frac{1}{2}(1-\frac{1}{2}C)^\frac{1}{2}$$ $$=(1-\frac{1}{2}B)(1-\frac{1}{2}C)$$ which appears to have no hope of being equal just by virtue of dimensionality. What am I doing wrong? I am fairly convinced that these things should be equal in the perfectly correlated case, and yet here I am, and they are not at all equal. Something else I noted was that the distribution for (D2) isn't sensitive to adjustments in the off diagonal values of the $A$ matrix, but the distribution for (D1) is. That is to say, if we replace all the values of $-1$ in the matrices by $-\sqrt{2}$, the chi squared decomposition for (D2) remains the same but is different for (D1).

As I said at the top, I left some things out. I can include them if this picture is not clear enough.


It turns out that I had incorrectly calculated the resulting distributions in Mathematica. I discovered that when I went through by hand.


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