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.


\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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.