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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 to be 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.

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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
up vote 7 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 $$ A_1^3=\frac{u\coth u-1}{\sinh^2 u}. $$

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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).

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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

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.

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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

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