0
$\begingroup$

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??

  • 0
    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.2012-01-20

1 Answers 1

1

$\sqrt X$ 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.

  • 0
    Smart! :) Thank you again.2012-01-20