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Let $X \in \mathbb{R}^N$ and $Z \sim \mathcal{N}(0, \sigma^2 I)$ be random vectors.

$Y = X + Z$

$X$ can be either $a_0 \in \mathbb{R}^N$ or $a_1\in \mathbb{R}^N$ with equal probability. So the decision rule is $$||y - a_0||^2 \overset{X = a_1}{\underset{X = a_0}{\gtrless}} ||y - a_1||^2$$

What is $P(\text{error} | X = a_0)$?

My attempt at solution:

\begin{align*} P(\text{error} | X = a_0) &= P(||Y - a_0||^2 > ||Y - a_1||^2 | X = a_0)\\ &= P(||Y - a_0||^2 > ||Y - a_1||^2) \, \, \text{ where $Y \sim \mathcal{N}(a_0, \sigma^2 I)$ }\\ &= \frac{1}{(2 \pi)^{N/2}} \frac{1}{\sigma^N} \int_{D} \exp \left( -\frac{1}{2 \sigma^2} ||y - a_0||^2 \right) dy \end{align*} where $D \subseteq \mathbb{R}^N$ is the region containing all points closer to $a_1$ than $a_0$.

Of course, this integral does not have an analytic formula, but can this be written in terms of single dimensional CDFs, exploiting the fact that $Y$'s components are independent random variables.

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\begin{align*}||Y - a_0||^2 &\overset{X = a_1}{\underset{X = a_0}{\gtrless}} ||Y - a_1||\\ \end{align*} can be written as: \begin{align*}(a_1 - a_0)^T Y &\overset{X = a_1}{\underset{X = a_0}{\gtrless}} \frac{||a_1||^2 - ||a_0||^2}{2}\\ \end{align*}

Note that $g(Y) = \left(a_1 - a_0 \right)^T Y$ is a sufficient statistic and a scalar quantity. You can prove that $\left ( g(Y) | X = a_0 \right) \sim \mathcal N \left( (a_1 - a_0)^T a_0, \sigma^2 ||a_1 - a_0||^2 \right)$

Hence, \begin{align*}P(\text{error} | X = a_0) &= P\left( g(Y) > \frac{||a_1||^2 - ||a_0||^2}{2} \Bigg | X = a_0\right) \end{align*}

can be easily evaluated using one-dimensional CDFs.

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