Computing the posterior mean using a Gaussian prior?

Question

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):

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?

Answer 1

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:

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