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If we have $U$, $V$, and $W$ as i.i.d normal RV with mean $0$ and variance $\sigma^2$, then what are the following expressed as a transformation of a known distribution (if known):

1) $\frac{U}{V + W}$ I don't think this can expressed as a distribution.

2) $\frac{U^2}{V^2}$ Both $U^2$ and $V^2$ are $\chi^2$ and the quotient of $\chi^2$ is an F-distribution. I am just not sure what the degrees of freedom are...

3) $\frac{U^2 + V^2}{V^2}$ $U^2 + V^2$ is simply a $\chi^2$ with the sum of the degrees of freedom of $U$ and $V$. A $\chi^2$ divided by another $\chi^2$ is F. But I am wondering is this can be simiplified to $\frac{U^2}{V^2} + 1$ (are you allowed to do that?) in which case it would be $1 + F$...

4) $\frac{U^2 + V^2}{W^2}$ I think it is the same as the one above... just an F distribution.

Please help me out with my reasoning! Thanks.

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1 Answer 1

up vote 5 down vote accepted
  1. $\frac{V+W}{\sigma^2} \sim N(0, 2)$ and the ratio of two iid standard normal is Cauchy(0,1). So $\frac{U}{(V+W)/\sqrt{2}} \sim \text{Cauchy}(0,1)$. You can then derive that $\frac{U}{V+W} \sim \text{Cauchy}(0, \sqrt{2})$.
  2. Yes, they are $\chi^2$. Specifically, $\frac{U^2}{\sigma^2}$ and $\frac{V^2}{\sigma^2}$ are $\chi^2(1)$. Then $\frac{U^2}{V^2} \sim F(1,1)$.
  3. $U^2 + V^2$ is not independent from $V^2$, so you can't say their ratio is a standard F-distribution. So yes you should write it as $\frac{U^2}{V^2} +1$. Then because $\frac{U^2}{V^2} \sim F(1,1)$, $\frac{U^2}{V^2} + 1$ looks like $F(1,1)$ but shifted to the right by 1.
  4. The sum of two independent $\chi^2(1)$ variables is $\chi^2(2)$, ie, $\frac{U^2 + V^2}{\sigma^2} \sim \chi^2(2)$. Then $\frac{U^2 + V^2}{W^2} \sim F(2,1)$.

All these relationships can be found in Wikipedia (normal, chi-square, Cauchy, F).

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