Say the events $B$ and $C$ are conditionally independent in the presence of $A$, i. e.

$$B ⊥ C \vert A$$

and the task is to compute the following conditional probability:

$$P(B \vert C)$$

Would it be correct to assume that this equates to $P(B)$, since $P(B \vert C, A)$ should equal $P(B \vert A)$? If not, what would be the correct way to go about this, given $P(A)$ as well as the conditionals $P(B \vert A)$ and $P(C \vert A)$?


2 Answers 2


No, conditional independence does not imply independence. For example, if Alice and Bob are neighbors then the event that Alice wears a raincoat is conditionally independent of the event that Bob wears a raincoat given that it is raining outside. But, while these events are conditionally independent, they are not independent.

Without further information, I don't see a way to further simplify $P(B \mid C)$.

  • $\begingroup$ Thank you for your answer. We do have information on $P(A)$ and the conditionals of both $B$ and $C$, I made an edit to the original question. $\endgroup$
    – Biggzlar
    Apr 20, 2018 at 12:05

On your first question: no, because if $P(A)<1$ then conditional independence wrt $A$ does not imply independence. See the answer of littleO.

Further $P(B\mid C)$ cannot be expressed in $P(A)$, $P(B\mid A)$ and $P(C\mid A)$.

This is illustrated by as follows.

If $B=A$ and $C=\Omega$ then:$$P(B\cap C\mid A)=1=P(B\mid A)P(C\mid A)$$ and this with $P(B\mid A)=P(C\mid A)=1$

But if $B=\Omega$ and $C=A$ then this is also true.

In the first case $P(B\mid C)=P(A)<1$ but in the second case $P(B\mid C)=1$.

This in spite of the fact that in both cases $P(A)$, $P(B\mid A)$ and $P(C\mid A)$ have the same values.

edit for completeness.

There is an expression for $P(A)$, $P(B\mid A)$, $P(C\mid A)$ and $P(C)$.

We have: $$P(B\cap C)=P(B\cap C\mid A)P(A)=P(B\mid A)P(C\mid A)P(A)$$ so that: $$P(B\mid C)=\frac{P(B\cap C)}{P(C)}=\frac{P(B\mid A)P(C\mid A)P(A)}{P(C)}$$

Further note that: $$\frac{P(C\mid A)P(A)}{P(C)}=\frac{P(C\cap A)}{P(C)}=P(A\mid C)$$

So we can also write:$$P(B\mid C)=P(B\mid A)P(A\mid C)$$

  • $\begingroup$ Thank you for your answer. Would it not be possible to compute the joint probability of $P(B, C)$ and express $P(B \vert C)$ like so? $$P(B \vert C) = \frac{P(B, C)}{P(C)}$$ Reference of a similar question $\endgroup$
    – Biggzlar
    Apr 20, 2018 at 12:38
  • $\begingroup$ That is $P(B\mid C)$ by definition. Do you want an expression of $P(B\mid C)$ in $P(A)$, $P(B\mid A)$, $P(C\mid A)$ and $P(C)$ then? $\endgroup$
    – drhab
    Apr 20, 2018 at 12:47
  • $\begingroup$ Sorry, yes. I failed to include $P(B)$ and $P(C)$ as possible terms. $\endgroup$
    – Biggzlar
    Apr 20, 2018 at 13:00

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