$\DeclareMathOperator\E{\mathbb E} \DeclareMathOperator\Var{\mathrm{Var}} \newcommand\R{\mathbb R} \DeclareMathOperator\N{\mathcal N} \DeclareMathOperator\tr{\mathrm{tr}}$Suppose $X \sim \N(\mu, \Sigma)$, where $\mathrm{supp}(X) = \R^n$. We can assume $\Sigma$ to be nonsingular. What is $\E \lVert X \rVert$?

If $\Sigma = \sigma^2 I$, then $\lVert X \rVert / \sigma$ follows a noncentral chi distribution, which has known mean (in terms of a generalized Laguerre polynomial). That gives in this case $$ \E \lVert X \rVert = \sigma \sqrt{\frac{\pi}{2}} \; L_{{1/2}}^{{(n/2-1)}}\left({\frac {-\lVert \mu \rVert^{2}}{2 \sigma}}\right)\, ;$$ if we further assume $\mu = 0$ then it becomes a chi distribution and $$ \E \lVert X \rVert = \sigma \sqrt {2} \, \frac{\Gamma((n+1)/2)}{\Gamma(n/2)} .$$

In the general case, we have an easy upper bound via Jensen's inequality. Letting $C^T C = \Sigma$ and $Z \sim \N(0, I)$: \begin{align} \left( \E \lVert X \rVert \right)^2 &< \E \lVert X \rVert^2 \\&= \E \lVert \mu + C Z \rVert^2 \\&= \lVert \mu \rVert^2 + 2 \mu^T C \E Z + \E \tr( Z^T C^T C Z ) \\&= \lVert \mu \rVert^2 + \tr( C^T C \,\E Z Z^T ) \\&= \lVert \mu \rVert^2 + \tr( \Sigma ) ,\end{align} where the inequality is strict since we've assumed $\lVert X \rVert$ is not degenerate.

We also have a lower bound in the same way: since $\lVert \cdot \rVert$ is convex, $\E \lVert X \rVert \ge \lVert \E X \rVert = \lVert \mu \rVert$. (I previously had something more complicated here, based on bounding the variance of the function $z \mapsto \lVert \mu + C z \rVert$, but it gave a worse bound.)

Can we tighten these bounds or, preferably, find an exact expression for $\E \lVert X \rVert$ in the general case? If not, what about the diagonal but non-spherical case, or $\mu = 0$, or other interesting subcases?

There's been some discussion on this site of the distribution of $Y = \lVert X \rVert^2$, particularly here and here. If we have the full distribution of $Y = \lVert X \rVert^2$, we may be able to find $\E \sqrt{Y}$. That distribution is gross, though, and analytical answers seem difficult.

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    $\begingroup$ Also have alook at: math.stackexchange.com/questions/442472/… $\endgroup$ Commented Apr 5, 2015 at 16:21
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    $\begingroup$ For the case $\mu=0$, a good lower bound is $$\sqrt{\frac2\pi} \sqrt{\mathrm{tr}(\Sigma)}.$$ Roughly 80% of Jensen's bound, I'd say it gives a good idea of the value, perhaps better than the exact hypergeometric formula. And if you're interested in the proof: notice how $\mathbb E[\sqrt{Z^\top\Sigma Z}]$ is concave in $\Sigma$, then what is its minimum on the simplex of $\Sigma\succeq0$ diagonal of trace $1$? Conclude by homogeneity. $\endgroup$
    – Cryme
    Commented Oct 7, 2021 at 16:57

1 Answer 1


I will not write out a full answer, that will take to much time now. The complete answer to the question can be found in Mathai & Provost: "Quadratic forms in Random Variables", see for example my answer to this question on the Mathematics Stack Exchange site.

The answer is long, complicated and involves special functions such as "the Lauricella function". I don't know if there is an R package for that (I couldn't find one). The answer can be found in that book in the section 3.2b.7 (Theorem 3.2b.5): Positive Fractional moments of Quadratic forms, between pages 62--66.

The distribution of quadratic form in normal variables can be computed with the R package CompQuadForm, maybe you could try numerical integration with the help of that package.

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    $\begingroup$ Cool, that link is a nice trick to turn it into the (square root of) a linear combination of independent chi-squared variables. I don't actually have a need for this, was just made curious by it coming up in this problem, and so numerical integration isn't particularly appealing. I'll check out the book, and am still curious if either the upper or the lower bound can be tightened in a not-overly-complicated way.... $\endgroup$
    – Danica
    Commented Apr 5, 2015 at 22:28

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