# Distribution of $\frac{Z_1^2 + Z_2^2}{Z_1+Z_2}$ where $Z_1, Z_2$ are standard normals

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!

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

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.

• 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 $(Z_1,Z_2) = R\sqrt{2}(\cos(\theta-\pi/4), \sin(\theta-\pi/4)).$
– whuber
Dec 1, 2021 at 15:45
• @whuber I agree the expression looks better and simpler with that transformation. But does that make the problem easier to solve? Dec 1, 2021 at 16:28
• 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.
– whuber
Dec 1, 2021 at 16:33
• @whuber Any advantage apart from that? The remaining job looks identical to what I presently have. Dec 1, 2021 at 16:51
• Correct: that's why I offer this only as a comment, not as a separate answer.
– whuber
Dec 1, 2021 at 17:05