# How do we call the distribution of a ratio consisting of a constant divided by a truncated normal random variable?

I have been thinking for a while about the following problem of Cohen's d effect size measure $$d={\frac {\mu _{1}-\mu _{2}}{\sigma}}$$ in random-effects meta-analyses for which I cannot find a more formal solution. Let us say we want to investigate the distribution of Cohen's d for a mean difference in the population for varying pooled standard deviations $$\sigma$$. Thus, the population parameter in terms of the unstandardized mean differences is constant, but population standard deviations may differ and $$\sigma$$, therefore, differs. Population standard deviations may differ just because different populations may differ with respect to how they react to, say, experimental treatments. So, just for added clarity, it's all about the distribution of the parameter, not about the estimator. Sampling variability does therefore not play a role here. The sample R code below generates such a distribution by assuming that $$\sigma$$ follows a truncated normal distribution of a variable in the positive real number range $${\displaystyle \mathbb {R} _{>0}=\left\{\sigma\in \mathbb {R} \mid \sigma>0\right\}}$$.

library(truncnorm)
set.seed(64378)
sigma <- rtruncnorm(10000, .01) # Distribution of pooled sigma
c <- 2 # Population mean difference
y <- c/x
plot(density(y))

mean_sd <- function(x){
avg <- round(mean(x),2)
std <- round(sd(x),2)
return(list(avg, std))

}

mean_sd(y)


While it is easy to generate the distribution, the real question is what kind of probability distribution this variable follows. This would probably enable one to calculate the distributional moments explicitly. The gamma distribution may be a candidate distribution, but I am far from being sure about that.

• You're not dividing a constant by a distribution (per your title), you're dividing it by a random variable, and then asking about the distribution of the ratio. It's going to be a particular of ratio distribution (which you might come up with several names for but it may be that there's no single term that''s particularly common). It would be unusual to assume a truncated normal for $\sigma$, however. Commented Jan 23, 2022 at 23:09
• Indeed, I must have been in a fit of absence of mind. Corrected accordingly. Thanks! Anyway, why is a truncated normal distribution "unusual" given that the sampling distribution of the std. deviation is approximatively normal when n --> infinity? Commented Jan 24, 2022 at 7:07
• Because $\sigma$ is naturally positive, truncated Normal models are not really appropriate. More suitable (and conventional) ones include inverse Gaussian, Gamma, and (sometimes) Lognormal.
– whuber
Commented Dec 9, 2022 at 18:17
• @whuber Totally valid point, I should have known better. Meanwhile, I found the answer on my own, at least one that provides a closed form here Commented Dec 10, 2022 at 10:45
• Unfortunately, that's not a solution. When you divide one random variable by another independent variable that follows a truncated Normal distribution, the expectation is either undefined or infinite and the variance is infinite.
– whuber
Commented Dec 10, 2022 at 16:57