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Thread: P.D.F of a norm of sheared Gaussian vector

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

    Question: Let $\displaystyle W_i \sim {\cal {N}}(0,\sigma ^2)$ for $\displaystyle i = 1,2,3,...,n$. For $\displaystyle (c_1,c_2,...,c_n) \in \mathbb{R}^n$, define $\displaystyle Y = \sum_{i = 1}^{i = n} c_i W_i ^2$. Compute $\displaystyle \text{Pr}\{Y \ge 0\}$
    (Note that the constants $\displaystyle c_i$ 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 $\displaystyle M_{W_i ^2}(s) = \frac1{\sqrt{1 + 2s}}$. Clearly $\displaystyle M_Y(s) = \prod_{i=1}^{i=n} M_{W_i ^2}(sc_i) = \prod_{i=1}^{i=n}\frac1{\sqrt{1 + 2sc_i}}$

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

    Regards,
    Srikanth
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    MHF Contributor matheagle's Avatar
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    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.

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

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

    $\displaystyle P(aX^2+bY^2>0)=P(X^2/Y^2>-b/a)$
    But this is true only if a is positive (and not equal to 0)
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    Quote Originally Posted by Moo View Post
    Hello,

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

    I have been thinking about this problem for days
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    Moo
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    The Fourier transform is something like $\displaystyle \mathcal{F}(s)=\frac{1}{2^n} \prod_{k=1}^n \frac{1}{\sqrt{1-2i c_k s}}$

    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.
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    Lord of certain Rings
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    Quote Originally Posted by Moo View Post
    The Fourier transform is something like $\displaystyle \mathcal{F}(s)=\frac{1}{2^n} \prod_{k=1}^n \frac{1}{\sqrt{1-2i c_k s}}$

    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
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    MHF Contributor matheagle's Avatar
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    I thought about dividing the sum by any one of them, say $\displaystyle W_n$ and then you have

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

    where each of these F's are $\displaystyle F_{1,1}$ but they are dependent and I'm not sure what the sum of F's are either.
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    Lord of certain Rings
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    My next attempt is a non-trivial upper bound on the probability. I am far from it as of now
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    MHF Contributor matheagle's Avatar
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    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 $\displaystyle R^n$
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    Moo
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    Lol I was going to ask the same question : where does this come from ?

    We can redefine Y :

    Suppose $\displaystyle c_1,\dots,c_k ~| ~ 0\leqslant k\leqslant n$ (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 $\displaystyle a_i=\frac{|c_i|}{\sigma}$

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

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

    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...
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  11. #11
    Lord of certain Rings
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    Quote Originally Posted by matheagle View Post
    Where did this come from?
    That may shed some light on this problem.
    Quote Originally Posted by Moo View Post
    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 $\displaystyle c_i$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
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  12. #12
    MHF Contributor matheagle's Avatar
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    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.
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