This is a screenshot from Coursera's class "Bayesian Statistics: Techniques and Models", Week 1, "Non-conjugate models" lecture (any one can audit the class and access the materials for free):

enter image description here

This is a repeat class, so there have been lots of eyeballs on it, and therefore I assume it is correct. However, the exponent in the last-but-one line is quadratic in $y$, while in the last one it is linear in $y$. How can this be correct? Did they just make an undetected mistake here or am I missing something really basic?

  • 1
    $\begingroup$ I guess the process here is dropping any element that is not a function of the model parameter $\mu$ $\endgroup$ – yoav_aaa Aug 12 '19 at 8:27

Note that he is calculating $P(\mu | y_1,y_2...)$. Thus the realisations $y_i$ are constant. Therefore, $C := -\frac{1}{2}\sum_{i=1}^n y_i^2$ is also constant and we get: $$P(\mu | y_1,y_2...) \propto\frac{1}{1+\mu^2} \text{exp}\left(C+n\left(\bar y\mu-\frac{\mu^2}{2} \right)\right)$$ $$=\frac{1}{1+\mu^2} \text{exp}(C)\cdot \text{exp}\left(n\left(\bar y\mu-\frac{\mu^2}{2} \right)\right)\propto \frac{1}{1+\mu^2} \text{exp}\left(n\left(\bar y\mu-\frac{\mu^2}{2} \right)\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.