# Specific step in the proof of conjugate prior for normal distribution with unknown mean and variance

I'm struggling to follow a specific step in the proof that

$$\tau \sim \text{Gamma}(\alpha, \beta), \quad \mu | \tau \sim \mathcal{N}(\nu, \frac{1}{k\tau})$$

is a conjugate prior distribution for a sample of iidrvs with distribution $$\mathcal{N}(\mu, \frac{1}{\tau})$$.

I'm able to derive that the posterior distribution is

$$\pi(\tau, \mu|\mathbf{x}) \propto \tau ^{\alpha + \frac{n}{2} - \frac{1}{2}}\text{exp}\Big(-\tau \Big[\beta + \frac{k}{2}(\mu - \nu)^2 + \frac{1}{2}\sum(x_i-\mu)^2\Big]\Big)$$

After that, my book (and all other references I could find) just says:

Complete the square to see that $$k(\mu-\nu)^2 + \sum(x_i - \mu)^2$$ $$=(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 + \frac{nk}{n+k}(\bar{x} - \nu)^2 + \sum(x_i - \bar{x})^2$$

I don't see how this follows without doing (what seems like) an unreasonable amount of algebra. I get the sense there must be some neat way of doing this that I'm not spotting.

Any help would be much appreciated. Thank you.

• Several answers already on site demonstrate "completing the square" in respect of similar Bayesian problems. It would be worthwhile taking a look at some of them Feb 19 '20 at 4:48

$$\pi(\tau, \mu|\mathbf{x}) \propto \tau ^{\alpha + \frac{n}{2} - \frac{1}{2}}\text{exp}\Big(-\tau \Big[\beta + \frac{k}{2}(\mu - \nu)^2 + \frac{1}{2}\sum(x_i-\mu)^2\Big]\Big)$$

$$\pi(\tau, \mu|\mathbf{x}) \propto \tau ^{\alpha + \frac{n}{2} - \frac{1}{2}}\text{exp}\Big(-\tau \Big[\beta + \frac{1}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 +\frac{1}{2} \frac{nk}{n+k}(\bar{x} - \nu)^2 + \frac{1}{2}\sum(x_i - \bar{x})^2\Big]\Big)$$

define

$$A=\frac{1}{2} \frac{nk}{n+k}(\bar{x} - \nu)^2 + \frac{1}{2}\sum(x_i - \bar{x})^2$$

$$\pi(\tau, \mu|\mathbf{x}) \propto \tau ^{\alpha + \frac{n}{2} - \frac{1}{2}}\text{exp}\Big(-\tau \Big[\beta+A + \frac{1}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 \Big]\Big)$$

for $$\tau$$

$$\pi(\tau|\mathbf{x}) \propto \int \tau ^{\alpha + \frac{n}{2} - \frac{1}{2}}\text{exp}\Big(-\tau \Big[\beta+A + \frac{1}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 \Big]\Big) d\mu$$

$$\pi(\tau|\mathbf{x}) \propto \tau^{\alpha + \frac{n}{2} - \frac{1}{2}}e^{-\tau (\beta+A) }\int e^{ -\frac{\tau}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 } d\mu$$

$$\pi(\tau|\mathbf{x}) \propto \tau^{\alpha + \frac{n}{2} - \frac{1}{2}}e^{-\tau (\beta+A) }\int e^{ -\frac{\tau}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 } d\mu \frac{\sqrt{\frac{1}{\tau(k + n)}}}{\sqrt{\frac{1}{\tau(k + n)}}}$$

$$\pi(\tau|\mathbf{x}) \propto \tau^{\alpha + \frac{n}{2} - \frac{1}{2}}e^{-\tau (\beta+A) } \sqrt{\frac{1}{\tau(k + n)}}\int \frac{1}{\sqrt{\frac{1}{\tau(k + n)}}} e^{ -\frac{\tau}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 } d\mu$$

do you see that integral $$\propto 1$$??(normal distribution).

$$\pi(\tau, \mu|\mathbf{x}) \propto \tau ^{\alpha + \frac{n}{2} - \frac{1}{2}}\text{exp}\Big(-\tau \Big[\beta + \frac{1}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 +\frac{1}{2} \frac{nk}{n+k}(\bar{x} - \nu)^2 + \frac{1}{2}\sum(x_i - \bar{x})^2\Big]\Big)$$

$$\pi( \mu|\tau,\mathbf{x}) \propto \text{exp}\Big(-\tau \Big[ \frac{1}{2}(k + n)\Big(\mu-\frac{k\nu + n\bar{x}}{k + n}\Big)^2 \Big]\Big)$$