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I understand $P(A\cap B)/P(B) = P(A|B)$. The conditional is the intersection of A and B divided by the whole area of B.

But why is $P(A\cap B|C)/P(B|C) = P(A|B \cap C)$?

Can you give some intuition?

Shouldn't it be: $P(A\cap B \cap C)/P(B,C) = P(A|B \cap C)$?

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    $\begingroup$ Maybe it's easier to comprehend in the multiplicative form: $P(A,B\mid C)=P(A\mid B,C)P(B\mid C)$? $\endgroup$ – Hagen von Eitzen Jan 28 '17 at 13:56
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Any probability result that is true for unconditional probability remains true if everything is conditioned on some event.

You know that by definition, $$P(A\mid B) = \frac{P(A\cap B)}{P(B)}\tag{1}$$ and so if we condition everything on $C$ having occurred, we get that $$P(A\mid (B \cap C)) = \frac{P((A\cap B)\mid C)}{P(B\mid C)}\tag{2}$$ exactly as your intuition tells you. But, you can set $D = B\cap C$ and start with the definition of $P(A\mid (B \cap C)) = P(A\mid D)$ as in $(1)$ $$P(A\mid (B \cap C)) = P(A\mid D) = \frac{P(A\cap D)}{P(D)} = \frac{P(A\cap (B \cap C))}{P(B\cap C)} = \frac{P(A\cap B \cap C)}{P(B\cap C)}\tag{3}$$ and then multiply and divide by $P(C))$ on the right of $(3)$ to write the final result as $(2)$ as in Taylor's answer.

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\begin{align*} \frac{P(A,B|C)}{P(B|C)} &= \frac{P(A,B,C)}{P(C)}\frac{P(C)}{P(B,C)} \\ &= \frac{P(A,B,C)}{P(B,C)} \\ &= P(A|B,C) \end{align*}

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    $\begingroup$ -1 Although quite correct, the question asked for some intuition, this does not contain any. $\endgroup$ – Jack Aidley Jan 27 '17 at 11:48
  • $\begingroup$ What is the meaning of $P(A,B)$? $\endgroup$ – Xi'an Jan 27 '17 at 12:48
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    $\begingroup$ it means P(A and B) :: the joint probability, $\endgroup$ – nyxee Jan 27 '17 at 13:18
  • $\begingroup$ @Xi'an I think it was the original notation $\endgroup$ – Taylor Jan 27 '17 at 15:03
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Just draw the Venn diagram. We then have $$\Pr[A \cap B \mid C] = \frac{\text{"1"}}{\text{"C"}}, \quad \Pr[B \mid C] = \frac{\text{"1"} + \text{"2"}}{\text{"C"}}, \quad \Pr[A \mid B \cap C] = \frac{\text{"1"}}{\text{"1"} + \text{"2"}},$$ and the relationship follows by dividing the first expression by the second.

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My intuition is the following ...

Conditioning on $C$ means that we are considering only the cases when $C$ is given. Now, suppose that I live in a world where $C$ is always given.

My pepole know nothing about and cannot imagine a world without $C$. For some reason, our mathematicians denote probability of $X$ by $\hat{P}(X)$. They have also already discovered the rule

$$\hat P(A|B) = \frac{\hat P(A\cap B)}{\hat P(B)}\text{.}$$

Now, you as an Earthling, know a world where $C$ is not part of the assumptions in everyday life. So, when you come to our planet you can immediately notice, that every our probability $\hat P(X)$ actually correspond to your $P(X|C)$.

You are immediately able to rewrite the RHS, following the upper discovery:

$$\frac{P(A\cap B\mid C)}{P(B \mid C)}\text{.}$$

But ... What is the LHS? Well, what is the probability of $A$ when $B$ is given when $C$ is (also) given? Precisely $$P(A\mid B\cap C)\text{,}$$ hence the formula.

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A random variable conditioned on some event is another random variable (conditioning is like truncating your universe).

For example, let $A,B$ be the height and weight of people (drawn randomly from the current world population). Let $C$ be their nationality. Then the variables $A,B$ conditioned on some value of $C$ (say, $C=$French) correspond simply to new random variables (say $A',B'$ : height and weight of French people).

Then, because you know the definition of conditional probability you can write $P(A'\cap B')/P(B') = P(A'|B')$ which is the same as $P(A\cap B | C)/P(B |C) = P(A|B C)$

This is what the accepted answer aptly summarizes:

Any probability result that is true for unconditional probability remains true if everything is conditioned on some event.

Of course, the conditioning must be "global", that is, it must be applied to all the variables.

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  • $\begingroup$ -1 @leonboy: If you redefine $A$ and $B$ as random variable instead of events (sets) it makes no sense to talk about the intersection $A\cap B$. $\endgroup$ – Jarle Tufto Jan 28 '17 at 21:49

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