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It is a question about finding the posterior parameters.

$X$ follows $\text{Gamma}(a,b)$ ----(Prior)

$y \vert x$ follows $\text{normal}(\mu,1/x)$ ----- (likelihood), that is variance $(\sigma^2) = 1/x$

The posterior distribution of $X$ is $\text{gamma}( a+n/2, b+(1/2)\sum(x_i-\mu)^2 )$ and of $1/x$ (variance) is $\text{invgamma}( a+n/2, b+(1/2)\sum(x_i-\mu)^2) $, it is the same except the change in the distributions.

I tried to solve for the posterior distribution of $1/(x^{1/2})$ but I am unable to find the value of the beta parameter, I got the alpha parameter to be $(2a)$ which I highly doubt to be right.

The Pointer
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  • Please format your equations using MathJax. – The Pointer Jan 27 '21 at 12:27
  • Please show how "tried to solve for" the posterior. This is, after all, a straightforward application of a change of variables formula, so the best resolution for you would be to detect the step at which you erred or got stuck. – whuber Jan 27 '21 at 14:33
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    When given the distribution of $W=1/X$, wirh $X>0$ with probability one, finding the distribution of $Z=1/√X$ is a mere change of variable, with Jacobian $2z$. – Xi'an Jan 27 '21 at 14:34
  • Some related Qs: https://stats.stackexchange.com/questions/54633/is-a-power-of-an-inverse-gamma-random-variable-itself-inverse-gamma, https://stats.stackexchange.com/questions/277567/variation-on-inverse-gamma-1-xr-inv-gamma, – kjetil b halvorsen Jan 27 '21 at 21:15

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