# Normal distribution with known mean and unknown variance (product of two variables)

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?

• Yes completely correct. – Xi'an May 25 '20 at 10:24