Conditional probability with normal distributions

Hi, I have two solutions to a problem one devised by me and the other by the instructor. They have the same result, but which is actually done correctly?

Problem:

Quote:

We have an initial location $\displaystyle x_{init}=1000$ with an uncertainty modelled with a Gaussian with $\displaystyle \sigma_{x}^2=900$. We take a GPS measurement $\displaystyle z=1100$ which has an error variance $\displaystyle \sigma_{z}^2=100$.

Write the probabibility density functions of the prior $\displaystyle p(x)$ and the measurement $\displaystyle p(z|x)$. And using Bayes rule, what is the posterior $\displaystyle p(x|z)$?

Solution 1:

Quote:

$\displaystyle p(x) = N(x; 1000, 900) $

$\displaystyle p(z|x) = N(z-x; 0, 100)$

$\displaystyle p(x|z) = \frac{p(z|x)p(x)}{p(z)}$

$\displaystyle p(z) = \int p(z|x)p(x) = N(z; 1000, 1000)$

$\displaystyle \vdots$

$\displaystyle p(x|z) = N(x; 1090, 90)$

Solution 2:

Quote:

$\displaystyle p(x) = N(x; 1000, 900) $

$\displaystyle p(z|x) = N(z; 1100, 100)$

$\displaystyle p(x|z) = \frac{p(z|x)p(x)}{p(z)}$

$\displaystyle p(z) = \int p(z|x)p(x) = \eta$ (a normalizing constant)

$\displaystyle \vdots$

$\displaystyle p(x|z) = N(x; 1090, 90)$

The specific calculations are not important, but I hope I've given enough information to say which $\displaystyle p(z|x)$ is actually correct. Or maybe they both are, or neither?