Bishop PRML Question 8.10: d-separation I have trouble with solving the second part of question 8.10 from Bishop's PRML (attached as image).
I tried several things. Here's my latest attempt:
\begin{align}
p(a, b, d) &= \int p(a)p(b)p(c|a,b)p(d|c) dc \\
&= \int p(a)p(b)p(c, d|a, b)dc \\
&= p(a)p(b)p(d)
\end{align}
\begin{align}
p(a, b|d) = \frac{p(a)p(b)p(d)}{p(d)} = p(a)p(b)=p(a|d)p(b|d)
\end{align}
Which results in the opposite of what has to be shown.
Help would be greatly appreciated.
Thanks

 A: You can show the desired result by showing that a and b are not d-seperated by the note d by using the traditional d-seperation criteria. 
a and b are d-seperated by d if every path from a to b is blocked by d. A path is blocked by d if it is not active wrt. d. Thus, if we can find an active path we are done. A path is active if 
a) the non-collider nodes are not in the conditioning set d
b) the collider nodes has descendants in the conditioning set d
As the path a -> c -> b has the collider node c with a descendant in the conditioning set d, the path is active. We have thus found an active path and we can conclude that a and b are not seperated by d.
A: Reading the conditional independences depicted on the graph, we have
$$
  p(a,b,c,d)=p(a)\,p(b)\,p(c\mid a,b)\,p(d\mid c).
$$
First, using Fubini, we get
\begin{align}
  p(a,b) &= \int \int p(a,b,c,d)\,dc\,d(d) \\ &= p(a)\,p(b) \int p(c\mid a,b)\left(\underbrace{\int p(d\mid c)\,d(d)}_{=1}\right)dc \\ &= p(a)\,p(b) \underbrace{\int p(c\mid a,b)\,dc}_{=1} \\ &= p(a)\,p(b).
\end{align}
Hence, $a\perp\!\!\!\perp b\mid \emptyset$.
[ Equivalently, you can also just notice that $$p(a,b,c,d)=p(a)\,p(b)\,p(c,d\mid a,b)$$ and marginalize over $(c,d)$. ]
Second, Bayes gives
$$
  p(a,b\mid d) = \frac{p(d\mid a,b)\,p(a)\,p(b)}{p(d)} \ne p(a\mid d)\,p(b\mid d).
$$
Therefore, $a\not\!\perp\!\!\!\perp b\mid d$.
Here, it's important to realize that $p(d\mid a,b)\neq p(d)$. The reason is that, if you receive information about $a$ and $b$, then you change your opinion about $c$; but then you change your opinion about $d$ (go back to the graph). What is true, for example, is that $p(d\mid a,b,c)=p(d\mid c)$. In words: once you have information about $c$, then, regarding your uncertainty about $d$, it doesn't matter what happened to $a$ and $b$.
The whole point of this exercise is to learn how to reason conditionally. By the way, tremendous author and great book.
A: There's a conceptual mistake here. 
The statement $a \not\perp b \mid d$ is equivalent to the statement $p(a,b \mid d) \ne f(a) g(b).$ That is, $a$ being indedpendent of $b$ given $d$ is another way of saying that the conditional probability of $a,b$ given $d$ is not a separable product of functions of $a$ and $b$. You still require the general form of the distribution $p(a,b \mid d)$, not assuming that $d$ is independent of everything. You assumed $d$'s independence in your derivation, when saying that $p(d \mid a,b) = p(d)$. You shouldn't do that. I understand how that can be confusing; it seems like that's what you should do when the question says "$d$ is observed".
The thing is that, $a$ and $b$ are dependent given $d$ because of the way $d$ depends on $a$ and $b$. It's the collider effect in play. If it didn't--if, for example, $d$ was a separate node detached from everything--then we would find, in fact, that $a \perp b \mid d$. So, by artificially imposing $d$'s independence of $a$ and $b$ in the derivation of $p(a,b \mid d)$, and leading to $a \perp b \mid d$, you are essentially making a circular argument.
