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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.

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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 $$\beta x_i^2y_i=\beta\mu/s^2\rightarrow\mu=s^2x_i^2y_i$$

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  • $\begingroup$ 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 $s^2 x_i^4 y_i^2$. $\endgroup$
    – bajun65537
    Nov 8 '20 at 13:53
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    $\begingroup$ No, it doesn't, because constant terms can be part of the proportional operator. $\endgroup$
    – gunes
    Nov 8 '20 at 13:56

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