# Square root of inverse gamma distribution?

I asked this question on mathoverflow yesterday and got a suggestion to try it here. I apologize if this is an easy question, but I can't seem to find an answer anywhere.

I'm trying to duplicate a macroeconomic paper that uses MCMC analysis to derive time series and parameter values. For the variance of a white noise term the authors of the original paper choose an inverse gamma prior of mean 0.5 and having a 90% confidence interval between 0.21 and 0.79.

I'm using Dynare to run my estimation and Dynare only accepts these kinds of parameters in terms of standard error. Therefore, my question basically amounts to "how do I find the square root of an inverse gamma distribution"?

In case it's not clear what I'm asking, here's how the model equation appears in the paper:

$a_t=\rho_a\times{a_{t-1}}+\epsilon_a;\hspace{20 pt}\epsilon_a\sim{N}(0,\sigma_a)$

$a_t$ follows a simple AR(1) process and $\epsilon_a$ is meant to be white noise with mean 0 and variance $\sigma_a$. The prior for $\sigma_a$ is supposed to follow an inverse gamma distribution with the mean and confidence interval above. My question then is how can I find the "square root of the distribution" so I can write an expression for the standard error of $\epsilon_a$, instead of its variance?

If it's not possible to come up with a simple expression for this, would you have any suggestions for roughly equivalent specifications of the prior?

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 Thank you for your reply. In the original paper, one of the parameters that is estimated is $\sigma_{a}$, the variance of the technology shock. Unfortunately, in Dynare, while you are able to estimate transforms of parameters, you can't do so for the standard error of shocks, such as the technology shock in the model equation. Instead, you must assign a prior to the standard error of the shock directly. Given that the prior of the variance is inverse gamma I assumed that a suitable prior for the stderr would be the "square root" of this invgamma distribution. Does that make sense? – jefflovejapan Dec 4 '10 at 3:10 Thank you so much, that's exactly what I was looking for! – jefflovejapan Dec 4 '10 at 3:15