Bayesian Analysis – How to Parameterize the Ratio of Two Normally Distributed Variables

bayesiandistributionsmeta-analysisrandom variablevariance

Problem:
I am parameterizing distributions for use as a priors and data in a Bayesian meta-analysis. The data are provided in the literature as summary statistics, almost exclusively assumed to be normally distributed (although none of the variables can be < 0, some are ratios, some are mass, and etc.).

I have come across two cases for which I have no solution. Sometimes the parameter of interest is the inverse of the data or the ratio of two variables.

Examples:

  1. the ratio of two normally distributed variables:
    • data: mean and sd for percent nitrogen and percent carbon
    • parameter: ratio of carbon to nitrogen.
  2. the inverse of a normally distributed variable:
    • data: mass/area
    • parameter: area/mass

My current approach is to use simulation:

e.g. for a set of percent carbon and nitrogen data with means: xbar.n,c, variance: se.n,c, and sample size: n.n, n.c:

set.seed(1)
per.c <- rnorm(100000, xbar.c, se.c*n.c) # percent C
per.n <- rnorm(100000, xbar.n, se.n*n.n) # percent N

I want to parameterize ratio.cn = perc.c/perc.n

# parameter of interest
ratio.cn <- perc.c / perc.n

Then choose the best fit distributions with range $0 \rightarrow \infty$ for my prior 

library(MASS)
dist.fig <- list()
for(dist.i in c('gamma', 'lognormal', 'weibull')) {
    dist.fit[[dist.i]] <- fitdist(ratio.cn, dist.i)
}

Question:
Is this a valid approach? Are there other / better approaches?

Thanks in advance!

Update: the Cauchy distribution, which is defined as the ratio of two normals with $\mu=0$, has limited utility since I would like to estimate variance. Perhaps I could calculate the variance of a simulation of n draws from a Cauchy?

I did find the following closed-form approximations but I haven't tested to see if they give the same results… Hayya et al, 1975
$$\hat{\mu}_{y:x} = \mu_y/mu_x + \sigma^2_x * \mu_y / \mu_x^3 + cov(x,y) * \sigma^2_x * \sigma^2_y / \mu_x^2$$
$$\hat{\sigma}^2_{y:x} = \sigma^2_x\times\mu_y / mu_x^4 + \sigma^2_y / mu_x^2 – 2 * cov(x,y) * \sigma^2_x * \sigma^2_y / mu_x^3$$

Hayya, J. and Armstrong, D. and Gressis, N., 1975. A note on the ratio of two normally distributed variables. Management Science 21: 1338–1341

Best Answer

You might want to look at some of the references under the Wikipedia article on Ratio Distribution. It's possible you'll find better approximations or distributions to use. Otherwise, your approach seems sound.

Update I think a better reference might be:

See formulas 2-4 on page 195.

Update 2

On your updated question regarding variance from a Cauchy -- as John Cook pointed out in the comments, the variance doesn't exist. So, taking a sample variance simply won't work as an "estimator". In fact, you'll find that your sample variance does not converge at all and fluctuates wildly as you keep taking samples.

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