# Posterior density for normal distribution

I have that $$Y_i$$ has distribution $$N(\beta x_i^2, 1)$$, $$i=1 \dotsc n$$ where $$\beta$$ has a prior distribution $$N(0, \sigma^2)$$. I need to calculate the posterior density of $$\beta$$, find its mean and variance.

My approach was to say that $$p(\beta, y) \propto f(y|\beta)p(\beta) \propto \exp \left\{ -\frac{(y_i - \beta x_i^2)^2}{2}\right\} \exp \left\{ -\frac{\beta^2}{2\sigma^2}\right\}$$.

To make the maths less messy, I omit the summation notation (but I do not forget about it!). I get that:

$$p(\beta|y) \propto \exp \left\{ - \frac{1}{2} (\beta^2 (x_i^4 - \sigma^{-2}) - 2\beta (x_i^2 y_i) + y_i^2)\right\}$$

Now, if I understand the question correctly I should be able to get a normal distribution-like expression inside the brackets. However, I fail to see how to make that happen.

The hint is that the variance of posterior density of $$\beta$$ is supposed to be $$(x_i^4 - \sigma^{-2})^{-1}$$ which tells me I should be quite close to getting the right answer.

The exponent term should be in the form (let $$s$$ be the deviation of the posterior): $$-(\beta-\mu)^2/2s^2=-\beta^2/2s^2+\beta\mu/s^2+C$$
If you match the terms, $$\beta^2/2s^2=\beta^2(x_i^4-\sigma^{-2})/2$$, so the variance, $$s^2$$, is as mentioned. Matching the other term yields $$\beta x_i^2y_i=\beta\mu/s^2\rightarrow\mu=s^2x_i^2y_i$$
• So it does not have to make a perfect fit when completing the square since the last bit can be a constant? When I was trying to complete the square, the last term was $y_i^2$ but yours would be $s^2 x_i^4 y_i^2$. Nov 8, 2020 at 13:53