# Math Help - P.D.F of a norm of sheared Gaussian vector

1. ## P.D.F of a norm of sheared Gaussian vector

Question: Let $W_i \sim {\cal {N}}(0,\sigma ^2)$ for $i = 1,2,3,...,n$. For $(c_1,c_2,...,c_n) \in \mathbb{R}^n$, define $Y = \sum_{i = 1}^{i = n} c_i W_i ^2$. Compute $\text{Pr}\{Y \ge 0\}$
(Note that the constants $c_i$ can be negative,thus its not exactly non-central chi square)

My Attempt: I tried computing the pdf. I tried to use the M.G.F trick. We know that $M_{W_i ^2}(s) = \frac1{\sqrt{1 + 2s}}$. Clearly $M_Y(s) = \prod_{i=1}^{i=n} M_{W_i ^2}(sc_i) = \prod_{i=1}^{i=n}\frac1{\sqrt{1 + 2sc_i}}$

I am stuck here. I have forgotten Fourier Transforms, I think

Regards,
Srikanth

2. The problem with the MGF is that you cannot recognize the final distribution. With n=2 you can play with this to get an F.

$P(aX^2+bY^2>0)=P(X^2/Y^2>-b/a)=P(F_{1,1}>-b/a)$

3. Hello,

According to what you did, it looks like the $W_i$ are independent. I just want confirmation

$P(aX^2+bY^2>0)=P(X^2/Y^2>-b/a)$
But this is true only if a is positive (and not equal to 0)

4. Originally Posted by Moo
Hello,

According to what you did, it looks like the $W_i$ are independent. I just want confirmation
I apologise for not mentioning it. Actually $W_i$s are independent. A closed form for this probability would be ideal. However a bound on the probability involving the constants would be nice.

5. The Fourier transform is something like $\mathcal{F}(s)=\frac{1}{2^n} \prod_{k=1}^n \frac{1}{\sqrt{1-2i c_k s}}$

If we had a closed form for this, we would be able to use the inverse Fourier transform. But even then, it's not sure we'd get a closed form

I've thought of seperating the positive c_i's and the negative ones.
And work on the probability from this.. But I didn't really try.

6. Originally Posted by Moo
The Fourier transform is something like $\mathcal{F}(s)=\frac{1}{2^n} \prod_{k=1}^n \frac{1}{\sqrt{1-2i c_k s}}$

If we had a closed form for this, we would be able to use the inverse Fourier transform. But even then, it's not sure we'd get a closed form

I've thought of seperating the positive c_i's and the negative ones.
And work on the probability from this.. But I didn't really try.
Actually I tried the same thing and have now given up on the idea. As matheagle said "A linear combo of gamma's has no set distribution. X-Y can be negative hence its not another gamma."

Thanks for trying

7. I thought about dividing the sum by any one of them, say $W_n$ and then you have

$c_1F_1+c_2F_2+\cdots +c_{n-1}F_{n-1} +c_n$

where each of these F's are $F_{1,1}$ but they are dependent and I'm not sure what the sum of F's are either.

8. My next attempt is a non-trivial upper bound on the probability. I am far from it as of now

9. Where did this come from?
That may shed some light on this problem.
If you know something about the c's we can tranfrom this to a sum of gamma's. But having the c's negative messes that up.
It just looks like a nasty multivariate calculus problem where the you have the joint density which is your integrand
and you need to integrate over a region in $R^n$

10. Lol I was going to ask the same question : where does this come from ?

We can redefine Y :

Suppose $c_1,\dots,c_k ~| ~ 0\leqslant k\leqslant n$ (if =0, there is no sequence) are positive. The other ci's are negative.
There is no problem of permutation since the Wi's are iid.

Define $a_i=\frac{|c_i|}{\sigma}$

Then $Y=\sum_{i=1}^k a_i Z_i^2-\sum_{i=k+1}^n a_i Z_i^2$

Where $Z_i\sim \mathcal{N}(0,1)$

This may simplify a bit the pdf.

It just looks like a nasty multivariate calculus problem
It looks like it's worse than that.
If we want the pdf of Y, we'd have to make a substitution which, as far as I can see, wouldn't lead to a closed form...

11. Originally Posted by matheagle
Where did this come from?
That may shed some light on this problem.
Originally Posted by Moo
Lol I was going to ask the same question : where does this come from ?
It comes from a communication problem I am working on. That probability is the "Probability of error". I want to find a non trivial* condition(necessary or sufficient) on constants $c_i$s so that it minimize the probability of error OR minimizes the upper bound.

Actually I am no longer working on this problem. I changed the model and technique to make computations easier.

So thanks again. This discussion convinced me to change the technique

Thanks,
Srikanth

*choosing all c_i's negative is trivial

12. If the c's were positive and you could possibly keep this in the gamma family and then you would have a shot at it.
Having some positive and negative c's makes this nasty.