# 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/… – Alex R. Mar 13 '18 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. – StubbornAtom Mar 13 '18 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: $$U = \frac{XY}{X^2+Y^2} = \frac{1}{\sqrt{\frac{1}{X^2}+\frac{1}{Y^2}}} = \frac{1}{\sqrt{\frac{1}{Z^2}}} = Z$$ – Sextus Empiricus Mar 15 '18 at 14:54
• @MartijnWeterings Apparently that is the original proof given by Shepp. – StubbornAtom Mar 15 '18 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. – Sextus Empiricus Mar 15 '18 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 $U\sim\mathcal{N}\left(0,\left(\frac{\sigma}{\sigma+1}\right)^2\right)$.

according to this

Transforming two normal random variables

$$X=r\cos(\theta) \hspace{.5cm} Y=r\sin(\theta) \hspace{.5cm} X,Y \sim normal(0,1) \Leftrightarrow \theta \sim Uniform(0,2\pi) \hspace{.5cm} r^2\sim chi(2)$$.
$$X$$ and $$Y$$ are independent $$\Leftrightarrow$$ $$\theta$$ and $$r$$ are independent.

also $$\sin(\theta) \sim \cos(\theta) \sim \sin(2\theta) \sim 2\sin(\theta) \cos(\theta) \sim \cos(2\theta) \sim \cos(2\theta) \sim f$$ that $$f(z) =\frac{1}{\pi \sqrt(1-z^2)} I_{[-1,1]}(z)$$ since $$z=\sin(\theta) \Rightarrow f(z)=|\frac{d}{dz} \sin^{-1}(z)| f_{\theta}(\sin^{-1}(z)) + |\frac{d}{dz} (\pi-\sin^{-1}(z))| f_{\theta}(\pi -\sin^{-1}(z)) =\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} +\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} =\frac{1}{\pi\sqrt(1-z^2)}$$

similar for others.

$$\frac{2XY}{\sqrt(X^2+Y^2)}=\frac{2r^2 \cos(\theta) \sin(\theta)}{r}=2r \cos(\theta) \sin(\theta) =r \sin(2\theta) \sim r \sin(\theta) \sim N(0,1)$$

so we can show:

$$X= \sigma r \cos(\theta)$$ and $$Y=\sigma r \sin(\theta)$$

so

$$\frac{2XY}{\sqrt(X^2+Y^2)}=\frac{2r^2 \sigma \sigma \cos(\theta) \sin(\theta)}{r \sigma}=2\sigma r \cos(\theta) \sin(\theta) =\sigma r \sin(2\theta)\sim \sigma r \sin(\theta)\sim \sigma N(0,1)=N(0,\sigma^2)$$

to show independent

$$\frac{2XY}{\sqrt(X^2+Y^2)}= \sigma r \sin(\theta)$$

$$\frac{X^2-Y^2}{2\sqrt(X^2+Y^2)}=\frac{r^2 \sigma^2 (\cos^2(\theta)-\sin^2(\theta))}{2r\sigma} =\frac{1}{2} r \sigma (\cos^2(\theta)-\sin^2(\theta)) \sim \frac{1}{2} r \sigma \cos(2\theta)\sim \frac{1}{2} r \sigma \cos(\theta)$$ and easy to say they are independent .

• What if $\sigma_X \neq \sigma_Y$? – Sextus Empiricus Mar 10 '19 at 9:58
• i didn't think about it. but some calculation problems happen in $sqrt(X^2+Y^2)$ – masoud Mar 10 '19 at 11:49