Relative closeness of Gaussian vector This is a problem about random vectors being close (in terms of being parallel) to a given vector.
Let $X\sim\mathcal{N}(\mu_X, \sigma^2I)$ and $\mu_Y$ be a fixed vector such that $\langle\mu_X, \mu_Y\rangle=0$ and $\|\mu_X\|_2=\|\mu_Y\|_2=\mu$. I'm trying to find the probability of the event $|\langle X,\mu_X\rangle|\ge |\langle X,\mu_Y\rangle|$.
My attempt is the following. Write $X=\mu_X + \sigma g$ where $g\sim\mathcal{N}(0, I)$ (the $0$ here denotes the zero vector). Then we have
\begin{align}
P(|\langle X,\mu_X\rangle|\ge |\langle X,\mu_Y\rangle|) &= P(|\|\mu_X\|^2 + \sigma\langle g,\mu_X\rangle|\ge |\langle \mu_x,\mu_Y\rangle + \sigma\langle g,\mu_Y\rangle|)\\
&= P(|\|\mu_X\|^2 + \sigma\langle g,\mu_X\rangle|\ge |\sigma\langle g,\mu_Y\rangle|)\\
&= P\left(\left|\frac{\mu^2}{\sigma} + \langle g,\mu_X\rangle\right|\ge |\langle g,\mu_Y\rangle|\right).
\end{align}
Here I'm stuck and don't see how to obtain a closed-form expression. If there was no absolute value, then this would be simply $P(\langle g,\mu_Y-\mu_X\rangle\le\mu^2/\sigma) = \Phi\left(\frac{\mu}{\sqrt{2}\sigma}\right)$ since $\|\mu_Y-\mu_X\|_2 = \sqrt{2}\mu$.
However, splitting into cases based on the absolute function didn't seem to help here. Even a "good" upper bound will be quite helpful. Could you provide a hint on how to proceed from here? Thanks!
 A: If $X\sim\operatorname{\mathcal N}(\mu,\Sigma),$ where $\mu\in\mathbb R^{n\times1}$ and where $\Sigma\in\mathbb R^{n\times n}$ is nonnegative-definite, and $A\in\mathbb R^{k\times n},$ then
$$
AX\sim\operatorname{\mathcal N}(A\mu,\, A\Sigma A^T)
$$
(and $A\mu\in\mathbb R^{k\times1}$ and $A\Sigma A^T\in\mathbb R^{k\times k}$).
So use that:
\begin{align}
& \langle X,\mu_X\rangle = \mu_X^T X \sim\operatorname{\mathcal N}(\|\mu_X\|^2,\, \sigma^2\mu_X^T \mu_X) = \operatorname{\mathcal N}(\|\mu_X\|^2, \sigma^2\|\mu_X\|^2). \\[6pt]
& \langle X,\mu_Y\rangle = \mu_Y^T X\sim\operatorname{\mathcal N}(\mu_Y^T\mu_X,\, \sigma^2\mu_Y^T\mu_Y) = \operatorname{\mathcal N(\mu_Y^T\mu_X,\, \sigma^2\|\mu_Y\|^2}).
\end{align}
You did not say what the joint distribution of $X$ and $Y$ is, but I will assume you meant they are independent.
Then
\begin{align}
& \langle X,\mu_X\rangle - \langle X,\mu_Y\rangle = \langle X,\,\mu_X-\mu_Y\rangle = (\mu_X-\mu_Y)^T X \\[8pt]
\sim {} & \operatorname{\mathcal N}((\mu_X-\mu_Y)^T\mu_X,\, \sigma^2\|\mu_X - \mu_Y\|^2), \\[8pt]
\text{and } & \langle X,\mu_X\rangle + \langle X,\mu_Y\rangle = \langle X,\,\mu_X+\mu_Y\rangle = (\mu_X+\mu_Y)^T X \\[8pt]
\sim {} & \operatorname{\mathcal N}((\mu_X+\mu_Y)^T\mu_X,\, \sigma^2\|\mu_X + \mu_Y\|^2).
\end{align}
And then you have to do some piecewise stuff about absolute values.
