# If $X$ and $Y$ are independent Normal variables each with mean zero, then $\frac{XY}{\sqrt{X^2+Y^2}}$ is also a Normal variable

I am trying to prove the statement:

If $$X\sim\mathcal{N}(0,\sigma_1^2)$$ and $$Y\sim\mathcal{N}(0,\sigma_2^2)$$ are independent random variables,

then $$\frac{XY}{\sqrt{X^2+Y^2}}$$ is also a Normal random variable.

For the special case $$\sigma_1=\sigma_2=\sigma$$ (say), we have the well-known result that $$\frac{XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$$ whenever $$X$$ and $$Y$$ are independent $$\mathcal{N}(0,\sigma^2)$$ variables. In fact, it is more generally known that $$\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$$ are independent $$\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$$ variables.

A proof of the last result follows by using the transformation $$(X,Y)\to(R,\Theta)\to(U,V)$$ where $$x=r\cos\theta,y=r\sin\theta$$ and $$u=\frac{r}{2}\sin(2\theta),v=\frac{r}{2}\cos(2\theta)$$. Indeed, here $$U=\frac{XY}{\sqrt{X^2+Y^2}}$$ and $$V=\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$$. I tried to imitate this proof for the problem at hand but it appears to get messy.

If I haven't made any error, then for $$(u,v)\in\mathbb{R}^2$$ I end up with the joint density of $$(U,V)$$ as

$$f_{U,V}(u,v)=\frac{2}{\sigma_1\sigma_2\pi}\exp\left[-\sqrt{u^2+v^2}\left(\frac{\sqrt{u^2+v^2}+v}{\sigma_1^2}+\frac{\sqrt{u^2+v^2}-v}{\sigma_2^2}\right)\right]$$

I have the multiplier $$2$$ above as the transformation is not one-to-one.

So density of $$U$$ would be given by $$\displaystyle \int_{\mathbb{R}}f_{U,V}(u,v)\,\mathrm{d}v$$, which isn't readily evaluated.

Now I am interested to know if there is a proof where I can only work with $$U$$ and don't have to consider some $$V$$ to show that $$U$$ is Normal. Finding the CDF of $$U$$ doesn't look so promising to me at the moment. I would also like to do the same for the case $$\sigma_1=\sigma_2=\sigma$$.

That is, if $$X$$ and $$Y$$ are independent $$\mathcal{N}(0,\sigma^2)$$ variables then I wish to show that $$Z=\frac{2XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}(0,\sigma^2)$$ without using a change of variables. If somehow I can argue that $$Z\stackrel{d}{=}X$$, then I'm done. So two questions here, the general case and then the particular case.

Related posts on Math.S.E:

Edit.

This problem is in fact due to L. Shepp as I found out in the exercises of An Introduction to Probability Theory and Its Applications (Vol. II) by Feller, alongwith a possible hint: Surely, $$U=\frac{XY}{\sqrt{X^2+Y^2}}=\frac{1}{\sqrt{\frac{1}{X^2}+\frac{1}{Y^2}}}$$ and I have the density of $$\frac{1}{X^2}$$ at hand.

Let's see what I could do now. Apart from this, a little help with the integral above is also welcome.

• While similar, the MGF approach for the joint $(U,V)$ is a bit easier. See the last answer of: math.stackexchange.com/a/2665178/22064 and: math.stackexchange.com/questions/2664469/… Mar 13, 2018 at 21:21
• @AlexR. Yes I had seen the joint mgf approach, which works quite well if I am to find the joint distribution for the equal variance case. But I already have the proof by change of variables in that case, which in my mind is easier. What I am trying to do is to work with $U$ alone, since that is the distribution I am after. Mar 13, 2018 at 21:45
• The trick is that the sum of $\frac{1}{X^2}$ and $\frac{1}{Y^2}$, which are scaled inverse chi-squared distributions, is also a scaled inverse chi-squared distribution (that is the property of stable distributions). So the magic happens in the third equation of the following: $$U = \frac{XY}{X^2+Y^2} = \frac{1}{\sqrt{\frac{1}{X^2}+\frac{1}{Y^2}}} = \frac{1}{\sqrt{\frac{1}{Z^2}}} = Z$$ Mar 15, 2018 at 14:54
• @MartijnWeterings Apparently that is the original proof given by Shepp. Mar 15, 2018 at 17:41
• I would not have come up with this myself if you hadn't mentioned the commentary by Shepp. But, I had the idea that you did not get this proof. Or at least this was unclear whether this was the case. Mar 15, 2018 at 17:47

The original solution of the problem by Shepp uses the concept of stable law property, which seems a bit advanced for me at the moment. So I could not comprehend the hint given in the exercise I cited in my post. I guess a proof involving only the single variable $U=\frac{XY}{\sqrt{X^2+Y^2}}$ and not using a change of variables is difficult to come up with. So I share three open access papers I found that provide an alternate solution to the problem:
The first one has convinced me not to go down the integration path I took with that choice of the variable $V$ to derive the density of $U$. It is the third paper that looks like something I can follow. I give a brief sketch of the proof here:
We assume without loss of generality $\sigma_1^2=1$, and set $\sigma_2^2=\sigma^2$. Now noting that $X^2\sim\chi^2_1$ and $\frac{Y^2}{\sigma^2}\sim\chi^2_1$ are independent, we have the joint density of $(X^2,Y^2)$. We denote it by $f_{X^2,Y^2}$.
Consider the transformation $(X^2,Y^2)\to(W,Z)$ such that $W=\frac{X^2Y^2}{X^2+Y^2}$ and $Z=\frac{X^2+Y^2}{Y^2}$. So we have the joint density of $(W,Z)$. Let us denote it by $f_{W,Z}$. Following the standard procedure, we integrate $f_{W,Z}$ wrt to $z$ to get the marginal density $f_W$ of $W$.
We find that $W=U^2$ is a Gamma variate with parameters $\frac{1}{2}$ and $2(1+\frac{1}{\sigma})^{-2}$, so that $(1+\frac{1}{\sigma})^2\,W\sim\chi^2_1$. We note that the density of $U$ is symmetric about $0$. This implies that $(1+\frac{1}{\sigma})U\sim\mathcal{N}(0,1)$, and hence $U\sim\mathcal{N}\left(0,\left(\frac{\sigma}{\sigma+1}\right)^2\right)$.