I am interested in finding the conditional probability $P(A|E_1,E_2,...,E_n)$ where the $E_i$ are mutually independent events. I know only $P(A)$ and $P(A|E_i)$.
Is this possible? If so, how? If not, what information is missing?
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1 Answer
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$\begin{align} \mathsf P(A\mid \bigcap_i E_i) & = \frac{\mathsf P(\bigcap_i E_i\mid A)\cdot\mathsf P(A)}{P(\bigcap_i E_i)} \\[1ex] & = \frac{\mathsf P(\bigcap_i E_i\mid A)\cdot \mathsf P(A)}{\prod_i \mathsf P(E_i)} \end{align}$
To proceed further you require a guarantee of conditional independence. That is: $\mathsf P(\bigcap_i E_i\mid A) = \prod_i\mathsf P(E_i\mid A)$
However, mutually independent events $\{E_1, E_2, \ldots E_n\}$ are not necessarily mutually, conditionally independent given event $A$.
answered May 19, 2015 at 0:33
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2$\begingroup$ Conditional independence took me awhile to grok. An example helps: let P(G|I,D) be the probability that a student gets a grade of G = {A,B,C,D,F} on an exam conditioned on their intellegence I and the exam difficulty D. Clearly I and D are independent until you have the evidence of their grade G; Once you know the grade, I and D are no longer indepedent (i.e., if the student got an F and they are really smart, then you know something about how difficult the exam was). $\endgroup$– wcochranAug 28, 2018 at 21:47
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$\begingroup$ Seems to be contradicting this answer : math.stackexchange.com/questions/568029/… $\endgroup$ Oct 30, 2018 at 9:36
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1$\begingroup$ @AkshayLAradhya Indeed, it is. Can you see why? Do you agree? $\endgroup$ Oct 30, 2018 at 12:09