0
$\begingroup$

I was reading through "Machine Learning: A Probabilistic Perspective" by Kevin Murphy and came across this example using priors but I don't understand how the posterior mean was calculated (page 168):

enter image description here

I tried computing the posterior mean myself, but I don't understand how it is 3.43. Since the likelihood and prior are conjugate Gaussian, we can use a closed form formula to get the posterior mean (Lemma 5 here):

$$ \theta = \frac{\sigma^2_0}{\sigma^2 + \sigma^2_0}x+\frac{\sigma^2}{\sigma^2 + \sigma^2_0}\mu_0 $$ where the prior is $\theta \sim N(\mu_0, \sigma^2_0)$, and $\sigma^2$ is the variance of the observed $x$'s distribution. So the posterior mean should be

$$ \theta = \frac{2.19^2}{1+2.19^2}5 + \frac{1}{1+2.19^2}0 = 4.14 $$

Am I doing something wrong or misinterpreting the example? Why is the posterior mean 3.43?

$\endgroup$
0
$\begingroup$

You are correct, but the book does not provide a correct variance of the prior distribution.

The correct variance of the prior distribution: $\sigma_0^2=1.4781^2$.

You can try to use the correct variance to verify that: $p(\theta\leq-1)=p(-1\leq\theta\leq0)=p(0\leq\theta\leq1)=p(1\leq\theta)=0.25$

Instead if you use the incorrect variance provided by the book, you will get : $p(\theta\leq-1)=0.324,\,\,p(-1\leq\theta\leq0)=p(0\leq\theta\leq1)=0.176,\,\, p(1\leq\theta)=0.324$

Now you can derive that:

$\theta = \frac{1.4781^2}{1+1.4781^2}5 + \frac{1}{1+1.4781^2}0 = 3.43$

$\endgroup$

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.