I'm reading a book on Computer Vision and it has this exercise:
The joint probability $Pr(w,x,y,z)$ over four variables factorizes as $$ Pr(w,x,y,z) = Pr(w)Pr(z|y)Pr(y|x,w)Pr(x) $$ Demonstrate that $x$ is independent of $w$ by showing that $Pr(x,w)=Pr(x)Pr(w)$.
The author is using $Pr(\bullet)$ to denote the probability density function. I have tried all sorts of manipulations using Baye's rule, but alas I can't find the proof. Could you please help?
So far the closest I've gotten was:
$$ \begin{array}{rcl} Pr(w,x,y,z) & = & Pr(w)Pr(z|y)Pr(y|x,w)Pr(x) \\ Pr(w,x,y,z) & = & Pr(w)Pr(z|y)Pr(y|x,w)Pr(x)\dfrac{Pr(w,x)}{Pr(w,x)} \\ Pr(w,x,y,z)Pr(w,x) & = & Pr(w)Pr(x)Pr(z|y)Pr(y,x,w) \end{array} $$
To complete the proof, one would need to show that $Pr(w,x,y,z)=Pr(z|y)Pr(y,x,w)$, but I don't see how this could be shown.
Hint: Use the Law of Total Probability.
$$\begin{align}\Pr(x,w) &= \iint_{\Bbb R^2} \Pr(x,y,z,w)\operatorname d(y,z) \\[1ex] &= \iint_{\Bbb R^2} \Pr(w)\Pr(z\mid y)\Pr(y\mid x,w)\Pr(x)\operatorname d(y,z)\end{align}$$