Distribution of ratio $X/Y$ where $X$ is normal, $Y$ is half normal

Question

I know the ratio of 2 normally distributed random variables is a cauchy distribution...which, of course, has no variance.

But, if I have $X$ and $Y=|X|$ and take this ratio:

$Z=X/Y$, is it possible to find the variance of the ratio of the normal distribution to the half normal?

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Edit: (More details)

I am missing something a little in the answers so far, but I am extremly grateful. Let me provide a little more background.

Measuring data from sensors includes the process value and some noise. Xk = Vk + Nk, where Vk is the process measurement and Nk is the noise.

There are two issues, (1) Estimating Vk and (2) detecting a shift or change point.

I can estimate Vk using an exponential filter.

so F0k = a*(Xk) + (1-a)*F0k-1. The expected value of F0 -> E(F0) = E(X). This means exponential filtering will give me my expected value. The variance of F0 -> var(F0) = (a/(2-a))*var(X)

At this point I am interested in residuals: RES = Xk - F0k. This quantity should be 0 in the steady state. That is to say, the expected value is 0. I also assume that the value will be normally distributed (don't yell at me about this assumption). I can estimate the expected value online using a simple exponential filter (basically the same as an average):

F1k = b*(RES) + (1-b)*F1k-1.

The expected value of F1 -> E(F1) = E(RES) = 0. The variance of F1 -> var(F1) = (2/(2-a))(b/(2-b))*var(X)

Then I can take abs(RES) and use exponential filtering again:

F2k = b*(abs(RES)) + (1-b)*F2k-1

The expected value of F2 -> E(F2) = E(abs(RES)) = [sqrt(2/(2-a))*sqrt(2)/(sqrt(pi))]std(X). The variance of F2 -> var(F2) = (2/(2-a))(b/(2-b))*(1-(2/pi))*var(X)

All of these expected values and variances are just from either the exponential filtering properies or the expected value and variance of the normal or half-normal distribution.

My question was, what is the variance (and expected value) of Z = F1/F2

F1 will be some normally distributed value with expected value = 0. F2 will be some half-normally distributed value with expected value based on the standard deviation of x.

Can you find the var of Z?

I understand that X/abs(X) would just give you a bernoulli, but it seems it doesn't give me exactly what i'm looking for?

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Edit: (More details)

Here are some interesting simulation results:

F1 - exponential filter of Residual (noise added to original signal was Gaussian - result is normally distributed)

F2 - exponential filter of abs(Residual) (half-normal distribution)

Z - F1/F2 - simulated distribution of ratio of F1/F2 (notice the large variance around expected value of 0)

Answer 1

In general, the variance of $X/Y$, where $X$ is normal and $Y$ is half-normal, does not exist. Informally, this is because $Y$ has too much mass near 0. However, the situation you described is a special case, since $Y = |X|$.

In your description $X \sim N(\mu, \sigma^2)$ and $Y = |X|$. Therefore

$$ Z = \frac{X}{|X|} = {\rm sign}(X) = \begin{cases} 1 &\mbox{if } X > 0 \\ -1 & \mbox{if } X < 0. \end{cases} $$

So, $Z = 1$ with probability $p = P(X>0) = \Phi \left( \frac{\mu}{\sigma} \right) $, where $\Phi$ denotes the Normal CDF, which fully characterizes the distribution of $Z$, which is the question posed by the title.

To answer the question posed in the body of your post about the variance of $Z$, note $Z$ can be written as $2B - 1$ where $B \sim {\rm Bernoulli}(p)$ (see here for info on the Bernoulli distribution). Therefore,

$$ E(Z) = E(2B-1) = 2p-1 $$

$$ {\rm var}(Z) = {\rm var}(2B-1) = 4{\rm var}(B) = 4p(1-p)$$

Answer 2

Let $X,Y$~$N(0,1)$ and let $U=\frac{X}{Y}$ and $V=Y$

Here $f_{X,Y}(x,y)=\frac{1}{2\pi}e^{-\frac{1}{2}(x^2+y^2)},(x,y)\in\mathbb{R}^2$

Here $u=\frac{x}{y}$ and $v=y$

$=>~x=uv$ and $y=u$

Clearly $u=\frac{x}{y}$ and $v=y$ is a one-to-one transformation from $\mathbb{R}^2$ onto $\mathbb{R}^2$

Jacobian is $J=\begin{vmatrix} v & u \\ 0 & 1 \end{vmatrix}$

Now find the PDF of $(U,V)$ and find PDF of $U$ by integrating the PDF of $(U,V)$ over the range of $U$.

You get $U=\frac{X}{Y}$~$C(0,1)$.[Cauchy distribution]

Let $W=\frac{X}{|Y|}$

The distribution function of $W$ is $$F_W(w)=P[W\leq w]=P[W\leq w|Y>0].P[Y>0]+P[W\leq w|Y<0].P[Y<0]$$$=\frac{1}{2}{(P[U\leq w]+P[-U\leq w])}$$=\frac{1}{2}2(P[U\leq w]=F_U(w)$

[Since $U$ is symmetrically distributed about $0$]