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I have been trying to understand predictive posterior distribution.The following expression is given here

$p(x^*|x)=\int_\Theta c\times p(x^*,\theta|x)d\theta=\int_\Theta c\times p(x^*|\theta)p(\theta|x)d\theta$

I understand that the conditional probability is defined as

$f(A|B)=\frac{f(A,B)}{f(B)} = \frac{f(B|A)*f(A)}{f(B)}$

Can someone explain how the predictive posterior distribution is written using conditional probability?

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The first equality is the law of total probability. It is a special case of $f_X(x) = \int f_{X,Y}(x,y) \, dy$.

The second equality comes from computations with conditional densities similar to what you have written: $p(x^*, \theta \mid x) = p(x^* \mid \theta, x) p(\theta \mid x) = p(x^* \mid \theta) p(\theta \mid x)$, where the last step is due to $p(x^* \mid \theta, x) = p(x^* \mid \theta)$ by conditional independence of $x^*$ and $x$ given $\theta$.

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  • $\begingroup$ $p(x_k, x_{k-1}\mid y_{1:k-1}) = p(x_k \mid x_{k-1},y_{1:k-1}) p(x_{k-1} \mid y_{1:k-1})$ Could you please explain how this is written? $\endgroup$ – Natasha Apr 20 '18 at 17:34
  • $\begingroup$ @Natasha This is the same as $p(u,v) = p(u \mid v) p(v)$, except with an extra conditioning on another variable $z$: $p(u, v \mid z) = p(u \mid v,z) p(v \mid z)$. $\endgroup$ – angryavian Apr 20 '18 at 17:36
  • $\begingroup$ Thank you!That clarifies the second part.I am still not able to relate,"It is a special case of $f_X(x) = \int f_{X,Y}(x,y) \, dy$" to the first equality.$f_X(x) $" is the marginal density.Does the term on the left refer to conditioning on the marginal density?Could you please elaborate? $\endgroup$ – Natasha Apr 20 '18 at 17:48
  • $\begingroup$ @Natasha Again, it is an extra conditioning, e.g. $f(x \mid z) = \int f(x,y \mid z) \, dy$. $\endgroup$ – angryavian Apr 20 '18 at 19:29

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