Given $Z_1, Z_2$ are i.i.d standard normal random variables. Let


Derive the pdf of $V$.

The numerator and denominator of $V$ are dependent, so the square of it is not F-distribution.

  • $\begingroup$ One has to suspect there are typographical errors in this question because "Z3" is never mentioned again: are you sure your formula for V is what you intend? $\endgroup$
    – whuber
    Commented Nov 17, 2018 at 19:41
  • $\begingroup$ Z3 is involved in another part of the question, and i derived the pdf by P(V = a) => P(F = f(a)) but i am not sure if it is correct $\endgroup$
    – Sean
    Commented Nov 18, 2018 at 15:47

1 Answer 1


A possible way to proceed:

Clearly $Z_1^2$ and $Z_2^2$ are independently distributed $\chi^2_1$ random variables.

Then noting that $Z_1=\text{sgn}(Z_1)|Z_1|$, we have $$V=\text{sgn}(Z_1)\sqrt{\frac{2Z_1^2}{Z_1^2+Z_2^2}}=\sqrt{2Z}\text{sgn}(Z_1)$$, where it is a standard exercise to show that $$Z=\frac{Z_1^2}{Z_1^2+Z_2^2}\sim \text{Beta}\left(\frac{1}{2},\frac{1}{2}\right)$$

Thanks to @whuber for mentioning a much easier to understand geometric approach:

Transform to polar coordinates: $$(Z_1,Z_2)\to (R,\Theta)$$ such that $$Z_1=R\cos\Theta\quad,\quad Z_2=R\sin\Theta$$

So our $V$ becomes


It is easy to verify that $R$ and $\Theta$ are independently distributed. And in particular, $$\Theta\sim U(0,2\pi)$$

This means $\cos\Theta$ has the so called Arcsine distribution with pdf $$f(t)=\frac{\mathbf1_{|t|<1}}{\pi\sqrt{1-t^2}}$$

So we directly get the density of $V$ as

\begin{align} f_V(v)&=\frac{1}{\sqrt 2}f\left(\frac{v}{\sqrt 2}\right) \\\\&=\frac{\mathbf1_{|v|<\sqrt 2}}{\pi\sqrt{2-v^2}} \end{align}

  • $\begingroup$ This is close, but not quite correct. The problem comes from the fact that $\sqrt{X^2} = |X| \neq X$. Hence, your answer gives a non-negative random variable, whereas the original form in the question is negative with probability one-half. Your answer should be correct if you multiply it by a random variable $\text{SGN} \sim 1-2 \cdot \text{Bern}(\tfrac{1}{2})$. $\endgroup$
    – Ben
    Commented Nov 18, 2018 at 22:45
  • $\begingroup$ @Ben Thank you for pointing out the error. Let me think how to proceed from here. $\endgroup$ Commented Nov 19, 2018 at 4:54
  • 1
    $\begingroup$ I like the geometric approach to this one: the ratio has a simple trigonometric interpretation. Drawing a picture should enable you to just write down the answer and know it's correct. $\endgroup$
    – whuber
    Commented Nov 19, 2018 at 15:32
  • 1
    $\begingroup$ @whuber Please see my edit. Is this what you were referring to? $\endgroup$ Commented Nov 19, 2018 at 16:30

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