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?
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?
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)