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A question in addition to: Cauchy and $\chi^{2}$ dist

I have to find the distribution of $Z = Y\sqrt X$

I know Y is given by a fraction of 2 standard normal random variables ($Y = \frac{V_{1}}{V_{2}}$) and $\sqrt X = V_{2}$ is a standard normal random variable and therefore I conclude that $Z = \frac{V_{1}}{V_{2}}\cdot V_{2} = V_{1}$ is standard normal distributed.

Is this correct??

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Is it truly $V_1/V_2$ and $V_2$? If so then yes. –  Apprentice Queue Jan 20 '12 at 21:15
    
My question is related to the fact that I use that Y is cauchy distributed (i.e it is a fraction between two standard normal distributed variables) and $X$ is $\chi$ distributed (i.e it's the square of a standard normal distributed variable) to reduce that $Z=Y\sqrt X$ is standard normal distributed. –  user11775 Jan 20 '12 at 21:23
    
If $Y = \frac{V_{1}}{V_{2}}$ and $X = V_{3}^2$, with the $V_i$ iid standard normals then $Y\sqrt X$ is not normally distributed. –  Henry Jan 20 '12 at 21:29

1 Answer 1

up vote 1 down vote accepted

$\sqrt X$ cannot be a standard normal random variable, since square-roots by convention non-negative. So the question is flawed.

There is also a minor issue if $V_2=0$, but this has zero probability.

If $X= V_{2}^2$, and if $V_1$ and $V_2$ are independent standard normal random variables then $\sqrt X = |V_2|$ and $Z = Y\sqrt X = \text{sign}(V_2) V_1$, which indeed has a standard normal distribution.

Indeed you only need the signs of $V_1$ and $V_2$ to be independent, $V_1$ to be symmetrically distributed about $0$, and $V_2$ to have zero probability of being $0$, for $Z$ and $V_1$ to have the same distribution.

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Thank you! I see what my problem is. How do you conclude that $Z = sign(V_{2}) V_{1}$ has a standard normal distribution? –  user11775 Jan 20 '12 at 21:42
    
@user11775: because $\Pr(Z \le k) = \Pr(V_2 \gt 0)\Pr(V_1 \le k) +\Pr(V_2 \lt 0)\Pr(V_1\ge -k) $ $= (\Pr(V_2 \gt 0) + \Pr(V_2 \lt 0))\Pr(V_1\le k) = \Pr(V_1\le k)$ –  Henry Jan 20 '12 at 21:50
    
Smart! :) Thank you again. –  user11775 Jan 20 '12 at 22:04

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