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If we have a random variable $Y$ with pdf $P(Y|a,b)$, where $a$ and $b$ are parameters (with range $0$ to $\infty$).

As well as marginal posterior distributions for $a$ and $b$, these are $P(a\vert x)$ and $P(b\vert x)$, where $x$ is observed data.

Then would the predictive distribution of Y be

$$P(Y\vert x)=\int_0^\infty \left[\int_0^\infty P(Y|a,b)\cdot P(a|x) \,da\right] \cdot P(b\vert x) \,db$$

My confusion is because that we have two marginal distributions rather than a joint pdf.

Thanks in advance for any help it would be greatly appreciated.

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  • $\begingroup$ Do you know anything about the independence of the parameters $a,b$? If they are independent, then the joint pdf IS the product of the marginals, and your expression makes sense. $\endgroup$
    – rajb245
    Jan 27, 2014 at 18:14
  • $\begingroup$ I'm trying to do a general case, so they may or may not be independent. $\endgroup$ Jan 27, 2014 at 18:39

1 Answer 1

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Then would the predictive distribution of Y be

$$P(Y\vert x)=\int_0^\infty \left[\int_0^\infty P(Y|a,b)\cdot P(a|x)\,da\right] \cdot P(b\vert x) \,db$$

The answer to your question is, in general "no". The correct expression would be $$ P(Y\vert x)=\int_0^\infty \int_0^\infty P(Y|a,b)\cdot P(a,b|x)\,da \,db, $$ where $P(a,b|x)$ is the joint density of $a$ and $b$ for a given $x$. If, however, $a$ and $b$ are independent, then their joint density function is equal to the product of their marginal densities: $P(a,b|x)=P(a|x)P(b|x)$, and the expression you give is correct. If $a$ and $b$ are not independent, you must have an explicit relationship between them or some other way of getting the joint density to make this calculation in the general case.

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