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Assume there is a data point $x$ sampled from a Normal distribution: $$\begin{align} x \sim \mathcal{N}(\mu,\frac{1}{yz}) \propto (yz)^{1/2} \exp [-\frac{1}{2} (x-\mu)^2yz] \end{align}$$ where $\mu$ is the known mean, $\frac{1}{yz}$ is the unknown variance, $y$ and $z$ are both unknown.

Further assume $y$ and $z$ are both sampled from Gamma distributions: $$\begin{align} y \sim Gamma(\alpha_1,\beta_1) \propto y^{\alpha_1 - 1} \exp[-\beta_1 y] \end{align}$$

$$\begin{align} z \sim Gamma(\alpha_2,\beta_2) \propto z^{\alpha_2 - 1} \exp[-\beta_2 z] \end{align}$$ whrere $\alpha_1,\alpha_2$ are the shape parameters and $\beta_1,\beta_2$ are the rate parameters.

The posterior distribution $p(y|x,z)$ is $$\begin{align} p(y|x,z) &\propto p(x|y,z) p(y) \\ & \propto z^{1/2} y^{\alpha_1 - 1 + 1/2} \exp[-\frac{1}{2}(x-\mu)^2 yz - \beta_1 y]\\ & = z^{1/2} y^{\alpha_1 -1 + 1/2} \exp\Big[ [-\frac{1}{2}(x-\mu)^2z - \beta_1] y \Big] \end{align}$$

My question is: by observing the above expression, is it correct to say the posterior distribution $p(y|x,z)$ is a Gamma distribution : $Gamma(\alpha_1 + 1/2, \frac{1}{2}(x-\mu)^2z + \beta_1)$? Is it correct to absorb the $z^{1/2}$ term in the above expression into the normalization constant of the Gamma distribution, because $z$ is considered given and fixed in this posterior distribution?

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    $\begingroup$ Yes completely correct. $\endgroup$ – Xi'an May 25 '20 at 10:24

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