Bayesian inference from normal distribution

Question

A friend asked me this question from Bertsekas's probability book. The question is about some algebraic manipulation at the bottom, but the entire passage is included for context

and $d$ is a constant that depends on $x_i$ but not $\theta$ (this last part got cut off in the image)
Our question is about why $$c_1c_2 \cdot \exp \big( -\sum_{i=0}^n \frac{(x_i- \theta)^2}{2 \sigma_i^2} \big)=d \cdot \exp \big( \frac{-( \theta -m)^2}{2v} \big)$$ As stated, this should just be by "some algebra, which involves completing the square...". Some confusions are that the square already looks complete on both sides, and whether $d$ may depend on the $\sigma_i^2$. We've tried to show this but the expressions just got uglier and uglier. Thanks in advance for any help.

Answer 1

In the step that you are referencing, the square is to be completed on $\theta$: the LHS is not complete because the exponent is a sum of squares, and it needs to be rewritten as a normal likelihood with respect to the posterior distribution of $\theta$.

How does the algebra actually proceed? Consider just the summation in the LHS expression:

$$\begin{align*} \sum_{i=1}^n \frac{(x_i - \theta)^2}{2\sigma_i^2} &= \frac{1}{2} \sum_{i=1}^n \left(\frac{\theta^2}{\sigma_i^2} - \frac{2x_i}{\sigma_i^2} \theta + \frac{x_i^2}{\sigma_i^2}\right) \\ &= \frac{1}{2} \left( \frac{\theta^2}{v} - 2 \frac{m}{v} \theta + \sum_{i=1}^n \frac{x_i^2}{\sigma_i^2} \right) \\ &= \frac{1}{2v} \left( (\theta^2 - 2m \theta + m^2) - m^2 + v \sum_{i=1}^n \frac{x_i^2}{\sigma_i^2} \right) \\ &= \frac{(\theta-m)^2}{2v} + C, \end{align*}$$ where $C$ is a constant that does not depend on $\theta$. Then, when taking the exponential, $\exp(-C)$ turns into the multiplicative constant $d$ used by the text.