Let $X$ be an N-dimensional normal random vector with non-zero mean $\mu$ and diagonal covariance matrix $\sigma^2I$. I would like to understand if it is possible to derive the expected value of the random variable $\frac{X_i}{\|X\|^2_2}$; in other words \begin{equation} \mathbb{E}\left[ \frac{X_i}{\sum_{j=1}^N X^2_j} \right] \end{equation} with $i \in \{1, \dots, N\}$.

I found similar questions on StackExchange but the assumptions are usually that $\mu$ is the zero vector, or that the denominator is the norm $\|X\|_2$ (for which one talks about projected normal distribution) instead of the squared norm.

Here, I am only interested in the expectation, and not the distribution itself. A similar expression appears in, for example, the James Stein estimator, and to compute the bias one would need to compute the expectation in question.


1 Answer 1


In this question I posted, someone gave an answer to your question. For $X \sim \mathcal N(\mu,\sigma^2 I)$:

$$ \mathbb E\left( \frac{X}{||X||^k}\right) = \frac{\Gamma\left(\frac{n}{2}+1-\frac{k}{2}\right)}{(2\sigma^2)^{k/2}\Gamma\left(\frac{n}{2}+1\right)}{}_1F_1\left(\frac{k}{2};\frac{n+2}{2};-\frac{|\mu|^2}{2\sigma^2}\right)\,\mu,\tag{$\ast$} $$ where ${}_1F_1(a;b;z)$ is Kummer's confluent hypergeometric function.


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