Probability question: given $P(A|B)$ and $P(B)$ how do I find $P(A)$?

Question

I have a probability distribution for some quantity $A$ given a fixed $B$, i.e. $P(A|B)$. I also have a prior distribution $P(B)$ for $B$. I'm trying to find the distribution $P(A)$.

I had thought about using Bayes' theorem which implies that:

$P(A) = \frac{P(B)P(A|B)}{P(B|A)}$

But I don't have any information about $P(B|A)$, or at least I don't think I do...

I suspect I'm missing something obvious here, can anyone help out?

Thanks!

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EDIT

Based on the answers given I fear I might have tried to frame the question in too general a manner. The specific forms I'm working with are as follows:

$P(A|B) = A\exp{\left[-\frac{A^2 + B^2}{2}\right]} I_0\left(AB\right)$ (a slightly modified Rice distribution)

$P(B) \propto B^{-n}$ (generic power-law with $n>0$)

Does this change the situation at all? thanks!

Answer 1

I understand from your question that you have the distributions $P(B)$ and $P(A|B)$, which I assume are functions.

Then the joint distribution is $P(A, B)=P(A|B)P(B)$. Now to find the distribution $P(A)$ just sum $P(A, B)$ over $B$. Or if continuous case use integration over density functions. Refer http://en.wikipedia.org/wiki/Marginal_distribution for more information.

$$P(A=a)=\sum_{b}P(a,b)=\sum_{b}P(A=a|B=b)P(B=b).$$

I guess this is what you want.

Edit (to match the edit of the question): You have continuous case. So use integration as below. I guess $[0 +\infty)$ is sufficient as Rice is in non negative domain.

$$P(A=a)=\int_{-\infty}^{+\infty}P(a,b)db$$

Answer 2

You don't have enough information to find $P(A)$. That would be

$$ P(A) = P(B)\cdot P(A\mid B) + P(\neg B)\cdot P(A\mid \neg B) $$

but since $P(A\mid \neg B)$ can be anything, you can't do more than bound $P(A)$ between the limits given by assuming that $P(A\mid \neg B)$ is $0$ or $1$.

Answer 3

Unfortunately, you can't find $P(A)$ from the given information.

For instance: consider flipping a fair coin.

In the first scenario, let $A$ be the event "heads or tails" and $B$ the event "tails". Then $P(B)=\frac{1}{2}$, $$ P(A\mid B)=\frac{P((\text{heads or tails})\cap\text{tails})}{P(\text{tails})}=\frac{P(\text{tails})}{P(\text{tails})}=1, $$ and $P(A)=1$.

On the other hand, let $A$ and $B$ both the event "tails". Then $P(A\mid B)=1$ and $P(B)=\frac{1}{2}$, just like before, but now $P(A)=\frac{1}{2}$.