Average value-at-risk of simple quadratic form

The average value-at-risk of a quadratic form y'y, where y\sim \mathcal{N}(0_n,I_n) is given by a particularly complex closed-loop formula which I’ll describe below.

Even the computation of the value-at-risk is given by a formula which is not very simple to work with either.

We know that if y\sim \mathcal{N}(0,I), then \|y\|^2\sim \chi^2(n). Then, the CDF of the quadratic form is given by

\mathrm{P}[\|y\|^2 \leq t] = P(\frac{n}{2},\frac{t}{2}),

where P is the regularised Gamma function. For convenience, let us define the function \Pi_n(t)=P(\frac{n}{2},\frac{t}{2}). The value-at-risk is then given by


For ease of notation, let us denote the value-at-risk by \theta_\alpha. Now the average value-at-risk is given by

\begin{aligned} \mathrm{AV@R}_{\alpha}[\|y\|^2] &= \mathbb{E}[\|y\|^2 \mid \|y\|^2 \geq \theta_\alpha]\\ &= \int_{\theta_\alpha}^{+\infty} z \frac{1}{2^{\frac{n}{2}}} z^{\frac{n}{2}-1}e^{-\frac{z}{2}}\mathrm{d}z\\ &= \frac{2}{\Gamma(\frac{n}{2})} \int_{\theta_\alpha}^{+\infty} t^{n/2}e^{-t}\mathrm{d}t \end{aligned}

Let us call the integral which appears in the last equation I_n. For even values of n this integral is given by the product of a polynomial by an exponential. For odd values, it involved the complementary error function.

For instance, for n=1

I_1 = \frac{\sqrt{\pi}}{2}\mathrm{erfc}(\sqrt{\frac{\theta_\alpha}{2}}) + \sqrt{\frac{\theta_\alpha}{2}}e^{-\frac{\theta_\alpha}{2}}

For n=2 we have

I_2 = \frac{1}{2}(\theta_\alpha+2)e^{-\frac{\theta_\alpha}{2}}

In case y\sim \mathcal{N}(0,\Sigma), then y'y follows a univariate Wishart distribution with scale matrix \Sigma and n degrees of freedom and similar manipulations can be done to derive the average value-at-risk formula.

When y is not centered at zero but its components are independently distributed then we need to use a non-central \chi^2 distribution or a non-central version of the Wishart distribution. In that case derivations become tough and even more so when y is not centered at zero and we do not have the assumption of component-wise independence.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Exploring and venting about quantitative issues

Look at the corners!

The math blog of Dmitry Ostrovsky

The Unapologetic Mathematician

Mathematics for the interested outsider

Almost Sure

A random mathematical blog


Mathematix is the mathematician of the village of Asterix

%d bloggers like this: