4
$\begingroup$

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

$$V:=\frac{Z_1}{\sqrt{(Z_1^2+Z_2^2)/2}}$$

Derive the pdf of $V$.

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

$\endgroup$
2
  • $\begingroup$ 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
    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
    Nov 18, 2018 at 15:47

1 Answer 1

5
$\begingroup$

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

$$V=\sqrt{2}\cos\Theta$$

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}

$\endgroup$
4
  • $\begingroup$ 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
    Nov 18, 2018 at 22:45
  • $\begingroup$ @Ben Thank you for pointing out the error. Let me think how to proceed from here. $\endgroup$ 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
    Nov 19, 2018 at 15:32
  • 1
    $\begingroup$ @whuber Please see my edit. Is this what you were referring to? $\endgroup$ Nov 19, 2018 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.