0

I need some help to calculate the following joint probability:

Let's consider a binary event $A=\{0,1\}$ with $P(A=1)=0.75$ and a set of finite events $B_i$, $i=1,\dots,n$, mutually independent. Let's also consider known the probabilities $P(A|B_i)=\alpha_i$ and $P(B_i)=\beta_i$, $i=1,\dots,n$. How can I derive the calculation of $P\big[A|\bigcap_{i=1}^n B_i\big]$?

Thank you in advance!

Luis Correia
  • 1
  • Except in special cases, I think you should not expect a unique answer. See if you can come up with two or more examples where $n=2$ and $P(A=1 \mid B_i)=0.75$ and $P(B_i)=0.25$ for $i=1$ and $2$. – Henry Mar 17 '22 at 23:43
  • Thank you Henry! I derived an additional comment from your suggestion (below). – Luis Correia Mar 18 '22 at 18:57

1 Answers1

0

Well, the problem is that while you are promised that the set $(B_i)_{i\in[[1..n]]}$ is of mutually independent events, that does not guarantee their conditional independence given $A$.

So Bayes' Theorem takes us to:

$$\begin{align}\textstyle\mathsf P(A=a\mid\bigcap_{i=1}^n B_i)~&=~\dfrac{\mathsf P(\bigcap_{i=1}^n B_i\mid A)~\mathsf P(A=a)}{\mathsf P(\bigcap_{i=1}^n B_i)}&&\text{Bayes' Theorem}\\[2ex]&=~\dfrac{\mathsf P(\bigcap_{i=1}^n B_i\mid A)~\mathsf P(A=a)}{\prod_{i=1}^n\mathsf P(B_i)}&&\text{Mutual Independence}\\[2ex]&=\dfrac{\mathsf P(\bigcap_{i=1}^n B_i\mid A)~0.75^a0.25^{1-a}}{\prod_{i=1}^n \beta_i}\,\mathbf 1_{a\in\{0,1\}}\end{align}$$

However, because you cannot automatically claim $\mathsf P(\bigcap_{i=1}^n B_i\mid A)$ equals $\prod_{i=1}^n\mathsf P(B_i\mid A)$, you may not go any further. So we maynot express this in terms of $\alpha_i$ .

Graham Kemp
  • 129,094
  • Thank you Graham, that really helped. I was wondering if we can reduce the $\bigcap_{i=1}^n B_i|A$ in some series of sets (i.e., unions, intersections, complements etc) where we could derive using the $\alpha_i$. For instance, if we take $n=2$ as suggested by \href{https://math.stackexchange.com/users/6460/henry}{Henry}, can we say $P(B_1\cap B_2|A)=P\big[(B_1\cap B_2)\cap A\big]/P(A)=P\big[(B_1\cap A)\cap(B_2\cap A)\big]/P(A)$ and consider $(B_i\cap A)$ mutually independent, $i=1,\dots,n$? – Luis Correia Mar 18 '22 at 18:57
  • @LuisCorreia "... and consider $(B_i\cap A)$ mutually independent." No; you cannot generally consider them to be so; rather you would have to construct a particular collection of $(B_i)$ that had that conditional independence property (for that $A$) and still retained unconditional mutual independence among themselves. – Graham Kemp Mar 20 '22 at 23:40
  • Hi Graham, apologies about my late response - I understood your point, I will have to lean over the problem and figure out how to design those B's sets. Thank you so much for your comments. – Luis Correia Mar 29 '22 at 06:57