PDF of combination of i.i.d. standard normal variables. I assume that $a_i$,$b_i$, and $c_i$ are independent standard normal variables by each other ($a_i$, $b_i$, $c_i$ ~ $N(0,1)$).
and equation which I want to get PDF or mean and variance is $$\frac{\sum_{k=0}^{10}a_k^2c_k^2}{\sum_{i=0}^{10}a_i^2+b_i^2}$$
I would like to get PDF or E[x] and V[x] of equation and If I get pdf or process of to get pdf, that will be more helpful for me. 
I have tried to get pdf and I think denominator of above equation follows Chi-square distribution.
but this equation seems that denominator and numerator are not independent so I have trouble solving this.
Using MATLAB, I inferred E[X] and V[X] in simulation which are 
E[$\sum_{i=0}^{10}\frac{a_i^2c_i^2}{a_i^2+b_i^2}$] = 0.5, V[$\sum_{i=0}^{10}\frac{a_i^2c_i^2}{a_i^2+b_i^2}$] = 0.1591.
But still I can't induce why this values are acquired.
If you give any clues for me I'll very appreciate.
 A: First of all, expected value is linear and variance of a sum of independent random variables is sum of the variances. So it all boils down to mean and variance of $a_1^2 c_1^2/(a_1^2 + b_1^2)$.  Now
$$ \eqalign{\mathbb E\left[ \frac{a_1^2  c_1^2}{a_1^2 + b_1^2}\right] &= \mathbb E\left[\frac{a_1^2}{a_1^2 + b_1^2}\right]\cr
}$$
Since by symmetry this is the same as $\mathbb E \left[ b_1^2/(a_1^2+b_1^2)\right]$, and the sum of these is $1$, we get
$$ \mathbb E\left[ \frac{a_1^2  c_1^2}{a_1^2 + b_1^2}\right]  = \frac{1}{2}$$
The variance turns out to be $7/8$.  This comes down to computing
$$ \mathbb E \left[ \frac{a_1^4}{(a_1^2 + b_1^2)^2} \right] $$
Express $(a_1,b_1)$ in polar coordinates as $[r \cos(\theta), r \sin(\theta)]$, where $\theta$ is uniformly distributed in $[0,2\pi]$.  Then $a_1^4/(a_1^2+b_1^2)^2 = \cos^4 (\theta)$
and $$ \mathbb E[\cos^4(\theta)] = 3/8$$
EDIT: The new problem changes things a lot.  Well, the expected value is still doable:
$$ \mathbb E\left[ \frac{a_i^2 c_i^2}{\sum_{k=0}^{10} a_k^2 + b_k^2} \right] = \mathbb E\left[ \frac{a_i^2}{\sum_{k=0}^{10} a_i^2 + b_i^2}\right] \mathbb E\left[ c_i^2\right] = \frac{1}{22}$$ 
so
$$ \mathbb E \left[ \frac{\sum_{k=0}^{10} a_k^2 c_k^2}{\sum_{k=0}^{10} a_k^2 + b_k^2}\right] = \frac{1}{2} $$
The variance is going to be more complicated.
