3
$\begingroup$

Let $Z_1, Z_2 \sim \mathcal{N}(0,1)$ be i.i.d random variables. I wish to find the distribution of \begin{align} \frac{Z_1^2 + Z_2^2}{Z_1+Z_2} \,. \end{align} It is well known that $W = Z_1^2 + Z_2^2 \sim \chi^2_2$ and $Q = Z_1 + Z_2 \sim N(0,2)$. If $W$ and $Q$ were independent, then we might hope to invert this result relating to $t$-distributions, but of course they are not independent since \begin{align} Q^2 = W + 2Z_1 Z_2 \,. \end{align} Any thoughts on how to proceed are welcome!

$\endgroup$
3
  • $\begingroup$ It looks bimodal and heavy-tailed, vaguely reminding me of the reciprocal standard normal distribution though not as extreme $\endgroup$
    – Henry
    Dec 1, 2021 at 2:27
  • 3
    $\begingroup$ A random variable which functionally depends from another one is not necessarily stochastly dependent each other. $\endgroup$ Dec 1, 2021 at 3:35
  • $\begingroup$ @DaviAmérico Fair point. It looks like this post may be of use, though I still need to work through it to check for independence. $\endgroup$
    – lemmykc
    Dec 1, 2021 at 5:50

1 Answer 1

2
$\begingroup$

One way to proceed is to transform to polar coordinates $(Z_1,Z_2) \mapsto (R,\Theta)$ such that

$$Z_1=R\cos\Theta \quad, \quad Z_2=R\sin\Theta$$

Then,

$$\frac{Z_1^2+Z_2^2}{Z_1+Z_2}=\frac{R}{\cos\Theta+\sin\Theta}=\frac{R}{\sqrt 2\sin\left(\Theta+\frac{\pi}4\right)}$$

Now $R$ and $\Theta$ are independently distributed with $R$ having a Rayleigh distribution and $\Theta$ having a uniform distribution on $(0,2\pi)$. Independence of $R$ and $\Theta$ implies the independence of $R$ and $\sin\left(\Theta+\frac{\pi}4\right)$. So it is theoretically possible to derive this distribution.

$\endgroup$
5
  • 2
    $\begingroup$ Use $$\frac{Z_1^2+Z_2^2}{Z_1+Z_2} = \frac{1}{\sqrt 2}\left[\frac{X^2+Y^2}{X}\right]$$ where the i.i.d. assumption implies $X=(Z_1+Z_2)/\sqrt 2$ and $Y=(Z_1-Z_2)/\sqrt 2$ are independent standard Normal variables. This will (a) simplify the polar coordinate expression and (b) point out a missing factor of $\sqrt{2}$ in the analysis. Equivalently, let $(Z_1,Z_2) = R\sqrt{2}(\cos(\theta-\pi/4), \sin(\theta-\pi/4)).$ $\endgroup$
    – whuber
    Dec 1, 2021 at 15:45
  • $\begingroup$ @whuber I agree the expression looks better and simpler with that transformation. But does that make the problem easier to solve? $\endgroup$ Dec 1, 2021 at 16:28
  • $\begingroup$ Yes, because (1) no trigonometric manipulation is needed and (2) it automatically avoids a potential error in overlooking the factor of $1/\sqrt 2$ that would bite the casual reader. $\endgroup$
    – whuber
    Dec 1, 2021 at 16:33
  • $\begingroup$ @whuber Any advantage apart from that? The remaining job looks identical to what I presently have. $\endgroup$ Dec 1, 2021 at 16:51
  • $\begingroup$ Correct: that's why I offer this only as a comment, not as a separate answer. $\endgroup$
    – whuber
    Dec 1, 2021 at 17:05

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.