 Solution

Let $\bar{X}=n^{-1}{\sum }_i{X}_i$ and $S={\sum }_i(X_i-\bar{X}{)}^2$. The Gaussian distribution likelihood is proportional to ${\tau }^{n/2}\exp [-\tau {\sum }_i(X_i-\mu {)}^2/2]$. Multiplying by the normal-gamma distribution prior gives the quadratic exponent ${\sum }_i(X_i-\mu {)}^2+K_0(\mu -{\mu }_0{)}^2+2{\beta }_0$. Completing the square,The power of $\tau$ is ${\tau }^{{\alpha }_0+n/2-1}\sqrt{\tau }$, so the posterior distribution has exactly the required family withandFor a proper prior, $K_0,{\alpha }_0,{\beta }_0>0$ and ${\mu }_0\in \mathbb{R}$. The separate $\sqrt{\tau }$ factor is retained in both densities; it does not add $1/2$ to the update of $\alpha$. This is a conjugate prior for the Gaussian mean and precision parameter.