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For a grad school course I'm studying Bishop's Pattern Recognition and I can't follow how he derives the following formula.

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I've tried applying the sum rule first and then the product rule and vice versa but nothing lead me to the correct equation. Can anyone help out? Thanks.

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The trick here is that the prediction $t$ is conditionally independent of the data $\mathbf{x}, \mathbf{t}$ given the model parameters $w$: $$ t \perp\!\!\!\perp \mathbf{x}, \mathbf{t} \,\,\vert\,\, w $$ That means: If we know the model weights, knowing the training data will not alter the model prediction.

Another independence holds for the second factor—weights $w$ are independent on the current sample $x$, they only depend on the training data $\mathbf{x}, \mathbf{t}$.

The rest is simple:

$$ \begin{align} p(t|x,\mathbf{x},\mathbf{t})&=\int p(t,w|x,\mathbf{x},\mathbf{t})\,\mathrm{d}w \\ &=\int p(t|x,w,\mathbf{x},\mathbf{t})p(w|x,\mathbf{x},\mathbf{t})\,\mathrm{d}w \\ &=\int p(t|x,w)p(w|\mathbf{x},\mathbf{t}) \,\mathrm{d}w \end{align} $$

  1. Sum rule
  2. Chain rule
  3. Independence
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  • $\begingroup$ Thanks Jan! It makes sense to me, but would never have figured it out myself. $\endgroup$ – David Sep 6 '18 at 10:04
  • $\begingroup$ If you're stuck, it usually helps writing explicitly the formulas hidden behind $p$, e.g. $p(t|x,w,\mathbf{x},\mathbf{t}) = \mathcal{N}(t|y(x,w), \beta)$ (1.60), now you can immediately see that there is no $\mathbf{x},\mathbf{t}$ on the right side. $\endgroup$ – Jan Kukacka Sep 6 '18 at 10:13

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