8

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?

Anders
  • 183

1 Answers1

7

$\begin{align} \mathsf P(A\mid \bigcap_i E_i) & = \frac{\mathsf P(\bigcap_i E_i\mid A)\cdot\mathsf P(A)}{P(\bigcap_i E_i)} \\[1ex] & = \frac{\mathsf P(\bigcap_i E_i\mid A)\cdot \mathsf P(A)}{\prod_i \mathsf P(E_i)} \end{align}$

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 $A$.

Graham Kemp
  • 129,094
  • 2
    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). – wcochran Aug 28 '18 at 21:47
  • Seems to be contradicting this answer : https://math.stackexchange.com/questions/568029/express-multiple-conditional-probability-in-terms-of-single-conditional-probabil – DollarAkshay Oct 30 '18 at 09:36
  • 1
    @AkshayLAradhya Indeed, it is. Can you see why? Do you agree? – Graham Kemp Oct 30 '18 at 12:09