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

normal distribution

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)

enter image description here

F2 – exponential filter of abs(Residual) (half-normal distribution)
enter image description here

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

enter image description here

Best Answer

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,

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

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

Related Solutions

Solved – What Ratio of Independent Distributions gives a Normal Distribution

Let $Y_1 = Z \sqrt{E}$ where $E$ has an exponential distribution with mean $2 \sigma^2$ and $Z = \pm 1$ with equal probability. Let $Y_2 = 1 / \sqrt{B}$ where $B \sim \mbox{Beta}(0.5, 0.5)$. Assuming $(Z, E, B)$ are mutually independent, then $Y_1$ is independent of $Y_2$ and $Y_1 / Y_2 \sim \text{Normal}(0, \sigma^2)$. Hence we have

$Y_1$ independent of $Y_2$;
Both continuous; such that
$Y_1 / Y_2 \sim \text{Normal}(0, \sigma^2)$.

I haven't figured out how to get a $\text{Normal}(\mu, \sigma^2)$. It is harder to see how to do this since the problem reduces to finding $A$ and $B$ which are independent such that $$ \frac{A - B \mu}{B} \sim \text{Normal}(0, 1) $$ which is quite a bit harder than making $A/B \sim \text{Normal}(0,1)$ for independent $A$ and $B$.

Solved – Does confidence interval for odds ratio assume log-normal distribution

Given the comments, I have included the proof of the floating equation at the bottom of the response.

Given a two-by-two contingency table where the OR is $\frac{a/b}{c/d}$, if you take the log of both sides, the log odds ratio is $\log(a) - \log(b) - (\log(c) - \log(d))$. Hence the formula for the log odds ratio is additive. Because it is additive, we can assume that the log odds ratio converges to normality, much faster than the odds ratio, which has a multiplicative structure.

We can demonstrate the above using bootstrapping:

fake_dat <- data.frame(
  y = c(rep(1, 125), rep(0, 50), rep(1, 100), rep(0, 75)),
  g = c(rep("A", 175), rep("B", 175))
)
(mat <- table(fake_dat))
   g
y     A   B
  0  50  75
  1 125 100

Take A to be the treatment group, and B to be the control group.

# And the odds ratio is?
mat[2, 1] / mat[1, 1] / (mat[2, 2] / mat[1, 2])
[1] 1.875

We can resample the odds ratio and the log odds ratio from the data repeatedly and check their distributions:

or.s <- lor.s <- rep(NA, 3000)
for (i in 1:3000) {
  rand.samp <- sample(1:nrow(fake_dat), nrow(fake_dat), replace = TRUE)
  new_dat <- fake_dat[rand.samp, ]
  mat <- table(new_dat)
  or <- mat[2, 1] / mat[1, 1] / (mat[2, 2] / mat[1, 2])
  lor <- log(or)
  or.s[i] <- or
  lor.s[i] <- lor
}
par(mfrow = c(1, 2))
hist(or.s, main = "OR")
hist(lor.s, main = "LOR")
par(mfrow = c(1, 1))

We can see that at our sample size of 350, the estimate of the sampling distribution of the log odds ratio appears normally distributed. If you drop the sample size enough, you will arrive at a non-normally distributed estimate of the sampling distribution for the log odds ratio.

If we then apply the delta method to calculate the variance of the log odds ratio, we arrive at the equation you mentioned. And the delta method relies on the central limit theorem. So the lone requirement for normality is a large enough sample size, as the method is a large sample approximation.

One way to know if your sample size is large enough is to calculate a small sample odds ratio. One common small-sample formula for the odds ratio requires adding 0.5 to the cell counts before calculating the odds ratio and standard error in the same way. If the small sample odds ratio and confidence interval markedly differ from the standard odds ratio and CI, then the large sample approximation is probably inadequate. However, note that the small sample method is best used as a diagnostic rather than seen as a solution.

You can find coverage of these topics in chapter 2 of Agresti's Categorical Data Analysis and chapter 7 of Jewell's Estimation and Inference for Measures of Association. For the delta method, a good guide is Powell's Approximating variance of demographic parameters using the DELTA METHOD.

Origin of equation in question:

If we suppose that {$n_i, i = 1,2,3,4$} have a multinomial $(n, \pi_i)$ distribution, the quantity $\hat\pi_i$ has mean and variance:

\begin{equation} \mathrm{E}(\hat\pi_i)=\pi_i \quad\mathrm{and}\quad \mathrm{var}(\hat\pi_i)=\pi_i(1-\pi_i)/n=(\pi_i-\pi_i^2)/n \end{equation}

and for $i \neq j$:

\begin{equation} \mathrm{cov(\hat\pi_i,\hat\pi_j)}=-\pi_i\pi_j/n \end{equation}

We know that:

\begin{equation} OR = \frac{a / b}{c/d}=\frac{n\pi_a\times n\pi_d}{n\pi_b\times n\pi_c}=\frac{\pi_a\times \pi_d}{\pi_b\times \pi_c} \end{equation}

Then, $\log(OR) = \log(\pi_a) + \log(\pi_d) - \log(\pi_b) - \log(\pi_c)$.

For the delta method, we have the equation,

\begin{equation} \mathrm{var}(G)=\mathrm{var}[f(X_1,X_2,...,X_n)]\\ =\sum_{i=1}^n\mathrm{var}(X_i)\big[f'(X_i)\big]^2 + 2\sum_{i=1}^n\sum_{j=1}^n\mathrm{cov}(X_i,X_j)\big[f'(X_i)f'(X_j)\big] \end{equation}

Given this equation, the variance of $\big(\log{OR} = \log(\pi_a) + \log(\pi_d) - \log(\pi_b) - \log(\pi_c)\big)$ is:

\begin{equation} \frac{1}{\pi_a^2}\frac{\pi_a-\pi_a^2}{n} + \frac{1}{\pi_d^2}\frac{\pi_d-\pi_d^2}{n} + \frac{1}{\pi_b^2}\frac{\pi_b-\pi_b^2}{n} + \frac{1}{\pi_c^2}\frac{\pi_c-\pi_c^2}{n}\\ - \frac{2}{n}\frac{\pi_a\pi_d}{\pi_a\pi_d} + \frac{2}{n}\frac{\pi_a\pi_b}{\pi_a\pi_b} + \frac{2}{n}\frac{\pi_a\pi_c}{\pi_a\pi_c} + \frac{2}{n}\frac{\pi_d\pi_b}{\pi_d\pi_b} + \frac{2}{n}\frac{\pi_d\pi_c}{\pi_d\pi_c} - \frac{2}{n}\frac{\pi_b\pi_c}{\pi_b\pi_c}\\ = \frac{1}{n\pi_a}-\frac{1}{n} +\frac{1}{n\pi_b}-\frac{1}{n} +\frac{1}{n\pi_c}-\frac{1}{n} +\frac{1}{n\pi_d}-\frac{1}{n} +\frac{4}{n}\\ = \frac{1}{n\pi_a} +\frac{1}{n\pi_b} +\frac{1}{n\pi_c} +\frac{1}{n\pi_d}-\frac{4}{n}+\frac{4}{n}\\ =\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \end{equation}

Best Answer

Related Solutions

Solved – What Ratio of Independent Distributions gives a Normal Distribution

Solved – Does confidence interval for odds ratio assume log-normal distribution

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