# 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

$\mathrm{V@R}_{\alpha}[\|y\|^2]=\Pi_n^{-1}(1-\alpha)$

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.

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