If $\sigma$ follows Inverse-Gamma, can I just square the result to get $\sigma^2$ if I do simulation? Suppose I have $\sigma|Y,Y*,\psi$~$Inverse-Gamma(A, B)$.  
How should I get $\sigma^2$.  Can I just square the result from Inverse-Gamma?  
To me, this seem to work, if you think of it this way: $\sigma$ is the standard deivation and $\sigma^2$ is the variance, and this relationship is one-to-one.  You are just squaring the number.
But, sometimes the distribution changes, say $X\tilde{\ } N(0,1)$, but $X^2\tilde{\ } \chi^2(k)$, so, I don't know, and I am very confused.  Thanks for your help!
 A: If you already have your $\sigma_i$, drawn from whatever distribution is appropriate, then you can, as you suspect, generate $\sigma_i^2$ by simply squaring the $\sigma_i$.  You can't, however, when you're trying to work through the mathematics of finding the probability distribution of $\sigma_i^2$.  In the former case, you're simply transforming variables; in the latter, you're finding the distribution of the transformed variables, which is a very different thing indeed.  
Note that if you did know the distribution of the transformed variables, you would then have two ways of generating random numbers - the first being to use the original distribution, then transform the output, and the second being to use the distribution of the transformed variable to generate the transformed variables directly.  In fact, random number generators (almost) all work by transforming a uniform variate, or several uniform variates, to another variate which has the desired distribution (to a reasonable degree of accuracy.)  
