Originally Posted by

**I like Serena**
That's the formula yes, but if I'm not mistaken we should take $\mu_2=4$ instead of $\mu_2=5$.

To be honest, the problem statement is a bit confusing saying that the added Gaussian noise has $\mu=1$.

That just makes no sense - added Gaussian noise always has $\mu=0$.

I can only assume that they meant that the given Gaussian distribution was for $x_1$ with $\mu_1=1$, while the other Gaussian distributions would have the respective expectations for $x_i$ and $\sigma=2$. That is, $\mu_1=1,\mu_2=4,\mu_3=7,\mu_4=10$.

I asked about this, and this was the reply:

The noise has a shift of 1 (the mean), so if the real reading is 10, then the noisy reading (which will be the most probable) will be 11. Therefore, the difference between the expected reading at a certain x_i location (taking into account the noise effect ex: 11) and the real reading z_t should determine the probability p(z_t|x_i).