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Let $X_i, Y_i, Z_i$ be three iid family of standard normal random variables. Then the following random variable is distributed the same as the first coordinate of a uniform $2$-sphere:

$$ \frac{X_1}{\sqrt{X_1^2+X_2^2}} \frac{Y_1}{\sqrt{Y_1^2+Y_2^2}} + \frac{X_2}{\sqrt{X_1^2+X_2^2}} \frac{Y_2}{\sqrt{Y_1^2+Y_2^2}} \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}}$$.

Of course, $\frac{X_1}{\sqrt{X_1^2+X_2^2}}$ squared has $\beta(1/2,1/2)$ distribution.

I know this is true by looking at the matrix product $\gamma_{1,2}(\theta_1) \gamma_{2,3}(\theta_2) \gamma_{1,3}(\theta_3)$ where $\gamma_{i,j}(\theta)$ is the rotation matrix by $\theta$ in the $i \wedge j$ plane in $\mathbb{R}^3$. But is there a probabilistic way to prove it using beta-gamma algebra or otherwise?

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Since the support of $X$, $Y$ and $Z$ is $[-1,1]$ and these are independent, the support of $XZ(Y+1)$ is $[-2,2]$ hence it cannot be distributed as $Y$. –  Did Jan 28 '12 at 15:23
    
Not interested in a reformulation, or at least some explanations? –  Did Mar 1 '12 at 8:16
    
@Didier: you are absolutely right. I didn't see your comment from Jan 28. Please see my modified question above. Thanks! –  John Jiang Mar 2 '12 at 0:25

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