# Distribution of the inverse square of a non-standard normal random variable multiplied by a constant

It's a somewhat complicated situation and sorry about my phrasing, but it's my first time here. Suppose I have random normal variable $$X$$ ~ $$N( \mu, \sigma^2)$$, which represents some true effect(s). Then I try to estimate the mean of this effects using the estimates from a few (sometimes underpowered) trials -> Meta-Analysis, and do the following transformation:

$$Y$$ = $$c/X^2$$, where $$c$$ is a constant $$> 0$$ and $$Y$$ must be $$> 0$$ and therefore x>0. As a side note, $$\sigma$$ is for example 90% of $$\mu$$ and divided by the 97.5%-quantile of the standard normal distribution. What distribution has $$Y$$? But most importantly how can I determine/estimate the central tendency of a sample from this resulting distribution?

What I found out til now, just squaring that $$X$$ would result in a non-central chi-square distribution with a non centrality parameter and then just do the inverse (maybe?) and still not knowing the central tendency. Also with that 90%-$$\sigma$$, I have some x<0 and with very small x I get a super long tail (which doesn't reflect any reality anymore, it's just a theoretical number). Tested with $$\mu=6$$ and $$\sigma=2.76$$, c=11128.6 and 3 trials 99%,80% and 10% powered depending on the trial size in a simulation with n=10000.

I worked with the Median, but it doesn't seem to catch the center well.

The description of your experiment is somewhat unclear, so I will limit to answer the question about distributions.

With your notation, $$X/\sigma \sim N(\mu/\sigma,1)$$ so the square of that, $$(X/\sigma)^2$$ has a noncentral chisquare distribution with 1 df (degree of freedom) and noncentrality parameter $$\lambda=(\mu/\sigma)^2$$.

Then we can write $$Y=c/X^2 = \frac{c/\sigma^2}{(X/\sigma)^2}$$ so it is distributed like a constant ($$c/\sigma^2$$) times an inverse chi-squared rv (random variable) with 1 df and the noncentrality parameter $$\lambda$$.

You will find implementations of the noncentral chi-squared many places, so it will be useful to express the density (cdf, ...) for the inverse noncentral chi-square using that. Let $$U$$ be a nonnegative rv and $$t>0$$, the density of $$U$$ is $$f$$. Then we find the density of $$1/U$$ as $$f_{1/U}(t)=f_U(1/t)/t^2$$ so you can apply that.

In R there is an implementation (on CRAN) in package invgamma. But it is very easy to write yourself:

dinvchisq  <-  function(x,df,ncp=0,log=FALSE) {
res <- dchisq(1/x,df,ncp,log=TRUE) - 2*log(x)
if(log) res else exp(res)
}

• Could you be more specific about which "inverse gamma distribution" this might be a special case of? Exactly how would one match the noncentrality parameter with the shape and scale parameters of the Inverse-Gamma? (It doesn't look possible.) – whuber Feb 2 at 18:51
• You are right, I remove that reference. – kjetil b halvorsen Feb 11 at 13:59