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Science Forum Index » Statistics - Math Forum » Z = sqrt(X^2-Y^2), distribution?...
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| Pwyll... |
 Posted: Sun Jul 13, 2008 5:59 am |
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Guest
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Hi,
I have the following relation of normally distributed random variates X and Y:
Z = sqrt(X^2-Y^2)
What is the distribution of the non-zero real part of Z, especially in case E(X)=E(Y)?
Thought it would be easy, but I did not manage to find it...
Thanks for your input!
Dennis |
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| Jack Tomsky... |
| Posted: Mon Jul 14, 2008 4:40 pm |
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Guest
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Quote: Hi,
I have the following relation of normally distributed
random variates X and Y:
Z = sqrt(X^2-Y^2)
What is the distribution of the non-zero real part of
Z, especially in case E(X)=E(Y)?
Thought it would be easy, but I did not manage to
find it...
Thanks for your input!
Dennis
What you have is the square-root of the difference of two noncentral chi-squares, each with one degree of freedom. The noncentrality parameters are
dx^2 = (mux^2)/(sigx^2)
dy^2 = (muy^2)/(sigy^2)
If X^2-Y^2 < 0, then Z is purely imaginary. So what you want is the conditional distribution of the square-root of the difference of noncentral chi-squares, given that the difference is non-negative.
Unfortunately, that's as far as I can take it.
Jack |
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