3
$\begingroup$

I am working though the exercises in the Computer Vision Models book.

Here I am at the Problem 2.7 which is described as:

The joint probability $\mathrm{Pr}(w, x, y, z)$ over four variables factorizes as

$\mathrm{Pr}(w, x, y, z) = \mathrm{Pr}(w)\mathrm{Pr}(z\mid y)\mathrm{Pr}(y\mid x,w)\mathrm{Pr}(x)$

Demonstrate that $x$ is independent of $w$ by showing that $\mathrm{Pr}(x,w) = \mathrm{Pr}(x)\mathrm{Pr}(w)$.

What I did in the given RHS was: $\mathrm{Pr}(w)\mathrm{Pr}(z,y\mid x,w)\mathrm{Pr}(x)$.

Is this right? I have a serious doubt about it, I have missed something big here. Just can't put my finger on it. And if this is wrong, any tips on how to proceed?

1 Answers 1

3

I am going to assume everything to be continuous (to simplify the steps. Otherwise replace integrals by sums etc...). Thus non-rigorously

\begin{eqnarray*} p \left( x, w \right) & = & \int p \left( x, w, y, z \right) d y d z\\ & = & \int p \left( z \left| y \right. \right) p \left( y|x, w \right) p \left( x \right) p \left( w \right) d y dz\\ & = & p \left( x \right) p \left( w \right) \int p \left( z \left| y \right. \right) p \left( y|x, w \right) d y d z\\ & = & p \left( x \right) p \left( w \right) \int p \left( y|x, w \right) \underbrace{\left( \int p \left( z|y \right) d z \right)}_{= 1} d y\\ & = & p \left( x \right) p \left( w \right) \underbrace{\int p \left( y|x, w \right) d y}_{= 1}\\ & = & p \left( x \right) p \left( w \right) \end{eqnarray*}

It would have been also possible to use Markov properties for graphs for something more elegant and a bit shorter.

  • 0
    That was very helpful. Tha$n$ks!!!2012-12-18