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Another integral (this time it looks like a lot of work but maybe it can be simplified).

I have the Student's t-distribution

$$\int_{-\infty}^\infty \frac{\Gamma(D/2+\nu/2)}{\Gamma(\nu/2)}\frac{|\Lambda|^{1/2}}{(\pi \nu)^{D/2}}\left(1+\frac{(x-\mu)^{T}\Lambda(x-\mu)}{\nu}\right)^{-D/2-\nu/2}\mathrm dx$$

and I want to calculate its covariance, so I started using the definition $\text{cov}[x] = E[(x-E[x])(x-E[x])^{T}]$ which simplifies to $E[xx^{T}]-\mu\mu^{T}$

Trying to solve the expectation $E[xx^{T}]$, we get something like this

$$E[xx^{T}]=C \int \left(1+\frac{(x-\mu)^{T}\Lambda (x-\mu)}{\nu}\right)^{-\alpha} xx^{T}dx$$

with $\alpha=D/2+\nu/2$ and $C$ as a normalization constant.

Changing variables via $y=\Lambda^{1/2}(x-\mu)$, we obtain

$$E[xx^{T}]=C \int \frac{1}{\left(1+\displaystyle \frac{y^{T}y}{\nu}\right)^\alpha} (\Lambda^{-1/2}y+\mu) (\Lambda^{-1/2}y+\mu)^{T}|\Lambda|^{-1/2}dy$$

At this point, I know the cross terms of the product $(\Lambda^{-1/2}y+\mu) (\Lambda^{-1/2}y+\mu)^{T}$ vanish and the last term is going to produce $\mu\mu^{T}$ because the distribution is normalized, but I'm having a hard time getting the first term

$$C \int_{\infty}^{\infty} \frac{\Lambda^{-1/2}y y^{T}\Lambda^{-1/2}}{\left(1+\displaystyle \frac{y^{T}y}{\nu}\right)^\alpha}|\Lambda|^{-1/2}dy$$

I tried to use spherical coordinates but since $yy^{T}$ is a $D×D$ matrix I would need to integrate every term. There should be a better procedure.

By the way, the result I should get is $\text{cov}[x]=\displaystyle \frac{\nu}{\nu-2}\Lambda^{-1}$.

Any help is appreciated.

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1 Answer 1

I haven't checked you algebra, but since you are interested in integration in polar coordinates, let us say that you want to compute that last integral $$ \int_{\mathbb{R}^n} \frac{ \Lambda^{-1/2} y y^T \Lambda^{-1/2}}{(1+\frac{y^T y}{\nu} )^{\alpha}}|\Lambda|^{-1/2} dy = |\Lambda|^{-1/2} \Lambda^{-1/2} \Big\{ \int_{\mathbb{R}^n} \frac{ y y^T }{(1+\frac{y^T y}{\nu} )^{\alpha}} dy \Big\} \Lambda^{-1/2} $$ by the linearity of the integral and that $\Lambda$ is constant. Now, in general $$ \int_{\mathbb{R}^n} f = n v_n \int_0^\infty \int_{S^{n-1}} f(r \theta) r^{n-1} d \sigma(\theta) dr $$ where $\theta$ is general point on the sphere $S^{n-1}$ and $\sigma$ is uniform measure on that sphere. Here $v_n$, is the volume of the unit ball in $\mathbb{R}^n$. (See An Elementary Introduction to Modern Convex Geometry, by K. Ball, p. 5.)

Letting $y = r\theta$ and noting that $y^Ty = r^{2}$ since $\theta^T \theta = 1$, we get $$ \int_{\mathbb{R}^n} \frac{ y y^T }{(1+\frac{y^T y}{\nu} )^{\alpha}} dy = n v_n \int_0^\infty \frac{r^2}{(1+r^2/\nu)^\alpha} r^{n-1} dr \int_{S^{n-1}} \theta \theta^T d\sigma(\theta). $$ Note that $\int_{S^{n-1}} \theta \theta^T d\sigma(\theta) $ is the covariance of a random variable uniformly distributed on the sphere $S^{n-1}$. By symmetry, it is equal to some constant, say $\alpha_n$, times the identity matrix $I_n$. To obtain that constant, you can verify that each coordinate of such a random variable is distributed as a Beta random variable. Thus, $$ \int_{\mathbb{R}^n} \frac{ y y^T }{(1+\frac{y^T y}{\nu} )^{\alpha}} dy = n v_n \alpha_n I_n \int_0^\infty \frac{r^2}{(1+r^2/\nu)^\alpha} r^{n-1} dr $$

PS. The distribution of $\theta$ is the same as the distribution of $\frac{X}{\|X\|}$ where $X$ is a standard normal random variable.

EDIT: If you need a rigorous proof of that integration formula in polar coordinate, one that I know of is based on the heavy machinery of geometric measure theory. In particular, if you have heard of co-area formula, then this is a special case. For example, in "Sobolev mappings, co-area formula and related topics" by Piotr Hajlasz, the following is derived (p. 232) $$ \int_{\mathbb{R}^n} g \,dx = \int_0^\infty \Big( \int_{\partial B(0,r)} g \,d \mathcal{H}^{n-1} \Big) dr $$ as an application of that general result. Here $\mathcal{H}^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure and $\partial B(0,r)$ is the boundary of the ball of radius $r$ centered at origin. Note that $H^{n-1}$ restricted to $\partial B(0,r)$ is in essence the distribution of a random variable uniformly distributed on the sphere of radius $r$. I will leave it to you to show that this version and the one I mentioned earlier are the same.

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Wow, that's quite something. Can you describe why $y=r\theta$ and $\theta^{T}\theta=1$? The representation of spherical coordinates I have seen always involves a lot of sines and cosines for the $n-1$ coordinates. –  Robert Smith Mar 10 '13 at 17:30
    
By the way, I read the description of K. Ball where he's using that change of variables but I still don't know the reason. –  Robert Smith Mar 10 '13 at 17:32
1  
This is a compact way of writing those sine-cosine relations, or rather a more abstract way of writing it. Every vector $y$ can be written as $ y = (y/\|y\|) \|y\|$. $\theta$ is the $y/\|y\|$ part and $r$ is $\|y\|$. (You can verify that this $\theta$ has unit norm $\theta^T \theta = \|\theta\|^2 = 1$.) This rather abstract and elegant way of writing the integration in polar coordinates is useful if you can use symmetry (in some sense) to effectively bypass detailed computations for that integral over the sphere. Otherwise, you have to go back to a lots of sines and cosines... –  passerby51 Mar 10 '13 at 18:52
    
... which might not be even worth the effort. –  passerby51 Mar 10 '13 at 18:53
    
+1. Interesting. I haven't found much information about that transformation. By the way, what is $v_{n}$? –  Robert Smith Mar 10 '13 at 19:14

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