Suppose $\bar{X}$ is distributed $N_3(\bar{0}, \bar{\Sigma})$, where:
$\bar{\Sigma} = \begin{bmatrix} 3 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 3 \end{bmatrix}$
Find $P((X_1 - 2X_2 + X_3)^2 > 15.36)$.
My first idea was to square root both sides and split into disjoint cases $P(X_1 - 2X_2 + X_3 > \sqrt{15.36})$ and $P(X_1 - 2X_2 + X_3 < -\sqrt{15.35})$.
Then I wanted to write $Y = X_1 - 2X_2 + X_3$ and find the distribution of $Y$.
However, because the $X_i$ are not independent (they are correlated), I cannot conclude that $Y$ is normal. So now I am a bit stuck.
My second idea was to perform some transformation of $\bar{X}$ into some random vector of uncorrelated (hence independent) normal random variables (e.g. $\bar{Z} \sim N_3(\bar{0}, I_3)$ where $I_3$ is the $3 \times 3$ identity matrix).
According to my text, if $\bar{Y} = \bar{A} \bar{X} + \bar{b}$ for some $3 \times 3$ matrix $\bar{A}$ and $3 \times 1$ vector $\bar{b}$, then $Y$ has $N_3(\bar{b}, \bar{A} \bar{\Sigma} \bar{A}')$ distribution. So ideally I would find some $\bar{A}$ such that $\bar{A} \bar{\Sigma} \bar{A}' = I_3$. But this is also proving to be difficult.
Even if I were to find such a transformation, I'm not sure where I would proceed from here (how is the transformed random vector related to the probability statement about the $X_i$'s?)
Does anyone have any ideas? My text (Hogg Introduction to Mathematical Statistics 7th ed.) is not offering much in the way of hints or examples.
I was able to find this answer thanks to the comments of user @Andrew:
Write $Y = \sum_{i=1}^3 a_i X_i$, where $a_1 = a_3 = 1$ and $a_2 = 2$.
It is easy to show in general that:
$E[Y] = \sum_{i=1}^3 a_i E[X_i]$
and
$Var[Y] = \sum_{i=1}^3 a_i^2 Var[X_i] + 2 \sum_{i<j} a_i a_j Cov(X_i, X_j)$
In our case:
$Var[Y] = 1^2 Var[X_1] + (-2)^2 Var[X_2] + 1^2 Var[X_3] + 2(1)(-2) Cov(X_1, X_2) + 2(1)(1) Cov(X_1, X_3) + 2(-2)(1) Cov(X_2, X_3)$
Plugging in the terms of the matrix:
$Var[Y] = 3 + 4 \times 2 + 3 - 4 \times 2 + 2 \times 1 - 4 \times 1$
$ \implies Var[Y] = 4$
Furthermore each $X_i$ has mean $0$, and so we have that $Y \sim N(0, 4)$.
The original problem reduces to:
$P(Y^2 > 15.36) = P(Y < -\sqrt{15.36}) + P(Y > \sqrt{15.36})$
$= 1 - P(Y < \sqrt{15.36}) + 1 - P(Y < \sqrt{15.35})$
$ = 2 - 2P(Y < \sqrt{15.36})$
$= 2(1 - P(Z < \frac{\sqrt{15.36}}{2} )) = 0.05$
where $Z \sim N(0,1)$. Consulting a normal distribution table provides us the final answer.