Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to compute analytically the following expected value:

$E\left( \frac{X_i^2}{\sum_j \lambda_j^2 X_j^2}\right)$

where the $X_i \approx N(0,1)$ are iid.

It seems an elementary integral, but it is eluding me. Any pointer to a non-trivial solution technique, or the solution itself of course, is highly appreciated.

share|improve this question
    
Do you have some part answers, say estimates or upper or lower bounds, with only two variables but unequal $\lambda_j$? For instance $\lambda_1=1$ but $\lambda_2$ very large, what happens approximately? –  Will Jagy Sep 10 '10 at 0:37
    
Funny: for two random variables $E(X^2_1/(\lambda^2_1X^2_1+\lambda^2_2X_2^2))=1/(|\lambda_1|(|\lambda_1|+|\lambd‌​a_2|))$. I wonder why the $\ell^1$ norm enters the picture here... –  Did Sep 10 '10 at 16:12
    
Didier, if we demand all $ \lambda_j > 0$ and put $ \lambda_i^2$ in the numerator to make the sum to 1 obvious, do you think $$ E\left( \frac{ \lambda_i^2 X_i^2}{\sum_j \lambda_j^2 X_j^2}\right) = \frac{ \lambda_i}{\sum_j \lambda_j}? $$ –  Will Jagy Sep 10 '10 at 17:15
    
Will, the formula you wrote above is the natural generalization of the formula for $n=2$. Unfortunately the trick I used to prove the $n=2$ case fails for larger values of $n$ and since, at this moment, I see no "geometrical" reason (whatever that means) pointing to the $\ell^1$ norm, the answer to your question is: I do not know. Sorry. –  Did Sep 10 '10 at 19:02
    
Thanks, Didier. No sorrow necessary. –  Will Jagy Sep 10 '10 at 19:21
show 1 more comment

3 Answers

up vote 5 down vote accepted

Here are some preliminary computations.

One wants to compute $A_k^n=\lambda_k^2E\left(X_k^2S^{-1}\right)$, where $S=\sum\limits_{k=1}^n\lambda_k^2X_k^2$. Starting from the expression $$ S^{-1}=\int_0^{+\infty}\mathrm{e}^{-tS}\mathrm{d}t, $$ and using the independence property of the random variables $X_k$, one gets $$ A_1^n=\int_0^{+\infty}\lambda_1^2E(X_1^2\mathrm{e}^{-tS})\mathrm{d}t=\int_0^{+\infty}\lambda_1^2E(X^2\mathrm{e}^{-t\lambda_1^2X^2})E(\mathrm{e}^{-t\lambda_2^2X^2})\cdots E(\mathrm{e}^{-t\lambda_n^2X^2})\mathrm{d}t, $$ that is, $$ A_1^n=-\lambda_1^2\int_0^{+\infty}u'(t\lambda_1^2)u(t\lambda_2^2)\cdots u(t\lambda_n^2)\mathrm{d}t, \quad\text{where}\ u(t)=E(\mathrm{e}^{-tX^2}). $$ By some simple computations, $$ u(t)=(1+2t)^{-1/2},\quad u'(t)=-(1+2t)^{-3/2}, $$ hence $$ A_1^n=\int_0^{+\infty}\frac{\lambda_1^2\mathrm{d}t}{(1+2\lambda_1^2t)\sqrt{(1+2\lambda_1^2t)(1+2\lambda_2^2t)\cdots(1+2\lambda_n^2t)}}. $$ First example When $\lambda_k^2=\lambda^2$ for every $k$, the change of variable $s=\sqrt{1+2\lambda^2t}$, $s\mathrm{d}s=\lambda^2\mathrm{d}t$, yields $$ A_1^n=\int_1^{+\infty}\frac{\mathrm{d}s}{s^{n+1}}=\frac1n, $$ as was to be expected by symmetry.

Second example When $\lambda_1^2=\lambda^2$ and $\lambda_k^2=1$ for every $k\ge2$, the change of variable $s=\sqrt{1+2t}$, $s\mathrm{d}s=\mathrm{d}t$, with $1+2\lambda^2t=\lambda^2s^2+1-\lambda^2$ yields $$ A_1^n=\int_1^{+\infty}\frac{\lambda^2\mathrm{d}s}{s^{n-2}(\lambda^2s^2+1-\lambda^2)^{3/2}}=1-(n-1)\int_1^{+\infty}\frac{\mathrm{d}s}{s^{n}(\lambda^2s^2+1-\lambda^2)^{1/2}}. $$ When $n=2$ and $\lambda^2\ge1$, setting $\lambda^2=1/\cos^2 u$ yields $A_1^2=\displaystyle\frac1{1+\cos u}=\frac{\lambda}{1+\lambda}$. This last formula is also valid if $\lambda^2\le1$.

Further values for even integers $n$ are $$A_1^4=\dfrac{\lambda^2}{(1+\lambda)^2},\quad A_1^6=\dfrac{\lambda^2(1+3\lambda)}{3(1+\lambda)^3},\quad A_1^8=\dfrac{\lambda^2(1+4\lambda+5\lambda^2)}{5(1+\lambda)^4}. $$ When $n=3$ and $\lambda^2\ge1$, setting $\lambda^2=1/\cos^2 u$ and some further computations yield $$ A_1^3=\frac{1-u\cot u}{\sin^2u}. $$ Likewise, if $\lambda^2\le1$, setting $\lambda^2=1/\cosh^2 u$ yields $$ A_1^3=\frac{u\coth u-1}{\sinh^2 u}. $$

share|improve this answer
add comment

For the case $\lambda_1=\cdots=\lambda_n=1$, the answer is $1/n$ by symmetry.

Not using symmetry, $Y = \sum_{j\ne i} X_j^2$ has the distribution $\chi_{n-1}^2$. Now look up the $\chi^2$ density and fire up Maple. The integral over $Y$ gives something with an incomplete gamma function in it, then the integral over $X$ gives $1/n$.

I can find a few other specific values the same way. For example, if $\lambda_1=2$ and $\lambda_2=\cdots=\lambda_n=1$ then the answer is $1/6,1/9,7/81,29/405,523/8505,2483/45927$ for $n=2,4,6,8,10,12$ (what's the pattern?). This disproves Will's conjecture (if that's what it was).

share|improve this answer
    
And for $n=3$, $\lambda_1=1/\cos u$, $\lambda_2=\lambda_3=1$, the answer is $\text{cotan}^2(u)−u\text{cotan}^3(u)$. For $\lambda_1=2$, $u=\pi/3$ hence the answer is $(3\sqrt3−\pi)/(9\sqrt3)$. –  Did Aug 20 '11 at 16:47
add comment

I don't know about a fully analytical solution, but your problem seems tractable conceptually. The random variable $X\_i^2/\sum\lambda_j^2 X_j^2$ whose expected value you want depends only the ray from the origin through $X = (X_1,\ldots,X_n)$; i. e., it is a function of $T = X/||X||$, which is a random variable taking values on the unit sphere in $\mathbb{R}^n$. Since the components of $X$ are IID normal, $T$ is distributed uniformly on the unit sphere. Thus, your expected value is a simple (n-1)-dimensional surface integral over the unit sphere: $$\int \frac{T_i^2}{\sum_j \lambda_j^2 T_j^2}d\mu(T)$$ where $\mu$ is surface measure on the unit sphere. $d\mu(T)$ can be expressed in coordinates without too much difficulty, but that's all I'll attempt to say here. I don't know whether to expect an analytical solution.

share|improve this answer
4  
The easiest way to compute many integrals on the unit sphere is actually to reexpress them as expectations of functions of Gaussian random variables, so I'm not sure this buys anything. –  Mark Meckes Sep 10 '10 at 11:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.