How to compute equation 1.68 of Bishop's book I am trying to obtain the right hand side of equation 1.68 in Pattern Recognition and Machine Learning by Bishop.
I was treating the problem as having four random variables $x,t,D,w$ where $D=(X,T)$ then I only obtain this:
$$P(t,x,D)=\int P(t,x,D,w)dw$$
$$P(t|x,D)P(x,D)=\int P(t|x,D,w)P(x,D,w)dw$$
$$P(t|x,D)=\int P(t|x,D,w)P(w|x,D)dw$$
Could anybody help me please? 
 A: The book sneakily invoked the concept of "conditional independence".
Suppose we have variables $A,$ $B,$ and $C,$ and that $A$ and $B$ are conditionally independent given $C.$ This means that $P(A \mid B, C) = P(A \mid C).$ That is, if $C$ is observed, then $A$ is independent of $B.$ However, that independence is conditional, so it's still true that $P(A \mid B) \ne P(A)$ in general.
In this case, $t$ is conditionally independent of $D$ given $w.$ The reason for this is that $t$ solely depends on $w$ and $x,$ but if you don't know $w$ then $D$ gives you a hint to the value of $w.$ However, if you do know $w$ then $D$ is no longer useful for determining the value of $t.$ This explains why $D$ was omitted from $P(t \mid x, w, D)$ but not from $P(t \mid x, D).$
Similarly, $w$ is entirely independent of $x$ so $P(w \mid x, D) = P(w \mid D).$
A: This is an old question but I thought I'd post this for anyone else googling this equation. The derivation, as Bishop says, proceeds by using the sum and product rules, and it's pretty straightforward that way. The following proof makes it clear just where the sum and product rules are being used.
\begin{align}
 p(t|x, \mathbf{x}, \mathbf{t}) &= \int p(t, \mathbf{w} | x, \mathbf{x}, \mathbf{t}) \rm{d}\mathbf{w}\tag{sum rule}\\
 &= \int p(t|x, \mathbf{w}, \mathbf{x}, \mathbf{t}) p(\mathbf{w}|x, \mathbf{x}, \mathbf{t}) \rm{d}\mathbf{w} \tag{product rule}\\
 &= \int p(t|x, \mathbf{w}) p(\mathbf{w}|\mathbf{x}, \mathbf{t}) \rm{d}\mathbf{w} \tag{drop variables}
\end{align}
We drop variables at the final step because, for example, the new prediction depends only on the new test point and the weights, not on the full training data.
