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A point is picked randomly in space. Its three coordinates $X$, $Y$, and $Z$ are independent standard normal variables. Let $R = \sqrt{X^2+Y^2+Z^2}$ be the distance from the point from the origin. Find:

a) The density of $R^2$ (don't get how to set up the integral for this)
b) The density of $R$ (don't get part a)
c) $E(R)$
d) $\textrm{Var}(R)$

I don't get how to use the change of variables since we are dealing with $X$, $Y$ and a $Z$. Can you please explain how I can do this? Also, can it be done using spherical coordinates? I am lost on the coordinates available for us on this problem.

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2 Answers 2

Hint: Chi-square distribution.

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That is probably why such problem was given. So when introducing chi-square distribution you would already be familiar with it. –  mpiktas Apr 19 '11 at 9:20
    
can you please solve the problem? –  mary Apr 19 '11 at 9:21
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@user8917, did you bother to read the wikipedia page? Some of the proofs are given there. Do not expect people do your homework for you. –  mpiktas Apr 19 '11 at 9:22
    
This happens to involve the chi-square distribution, but the chi-square distribution is just a gamma distribution. –  Michael Lugo Apr 19 '11 at 14:13
    
@Lugo, not very specific in terms of what equations to use. –  mary Apr 27 '11 at 4:07

You can determine the density of the constituent variables simply by using a change of variables. Taking $X^{2}$ as the example:

$$ P(X^{2} \leq u) = P(-\sqrt{u} < X < \sqrt{u}) = \Phi(\sqrt{u}) - \Phi(-\sqrt{u}) = 2\Phi(\sqrt{u}) - 1 $$

You can obtain the density function of $X^{2}$ by differentiation, which will be the same as that of $Y^{2}$ and $Z^{2}$. Finally the density of the sum of two independent random variables is given by the convolution of the two density functions. Apply that formula twice to derive the density of $R^2$.

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