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(My question is prompted by this post.)

I'm not understanding why $\frac{P(A\cap B\cap C)}{P(B\cap C)}$is not equal to $\frac{P(A)P(B)P(C)}{P(B)P(C)} = P(A)$. I know it's wrong but I cannot understand why. Can you explain it please?

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    $\begingroup$ Not sure why this was downvoted so hard. Seems to be a reasonable, albeit elementary, question regarding dependence/independence. $\endgroup$
    – Josh
    Feb 21 '18 at 16:28
  • $\begingroup$ The first step you give is only true if A, B and C are mutually independent; otherwise it is not true. $\endgroup$
    – R18
    Feb 22 '18 at 11:39
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    $\begingroup$ Are we sure about the tags here? There's nothing inherently Bayesian about intersecting probabilities. $\endgroup$
    – Firebug
    Feb 22 '18 at 12:31
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If $A$, $B$, and $C$ are mutually independent you can say $P(A\cap B\cap C)= P(A)P(B)P(C)$. But, in general case, they are not independent, and you can't always say that.

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  • $\begingroup$ can you be specific about the "vice versa" that doesn't hold? $\endgroup$
    – AdamO
    Feb 21 '18 at 15:53
  • $\begingroup$ @AdamO I mean if $P(A\cap B) = P(A)P(B)$, we can't conclude that $A$ and $B$ are independent. $\endgroup$
    – OmG
    Feb 21 '18 at 15:55
  • $\begingroup$ @OmG what you said is incorrect; that is the definition of independence of two events $A$ and $B$. If that equation is satisfied, $A$ and $B$ are independent. However, in order for three events $A,B,C$ to be independent, four criteria must be satisfied: $$P(A\cap B) = P(A)P(B), \quad P(A\cap C) = P(A)P(C), \\ P(B\cap C) = P(B)P(C), \quad P(A\cap B \cap C) = P(A)P(B)P(C).$$ $\endgroup$
    – Martin L
    Feb 21 '18 at 16:16
  • $\begingroup$ @MartinL I've updated my answer. $\endgroup$
    – OmG
    Feb 21 '18 at 16:19
  • $\begingroup$ It's still not correct. Here is an example when $A, B, C$ are not independent, but $P\left(A \cap B \cap C \right) = P\left(A\right) P\left(B\right) P\left(C\right)$: (link in next comment): $\endgroup$
    – oszkar
    Feb 22 '18 at 11:49
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Suppose the three different domains A, B, C as shown below. A∩B∩C is the red colored surface. What you ask is similar to the Bayes' rule:

Given that A and B occur, what is the probability that C also occurs?

enter image description here

This is basically the probability $P(C|A,B)$ or if you prefer $P(C|A∩B)$, and this is given by Bayes' rule:

$$P(C|A,B) = \frac{P(A,B,C)}{P(A,B)}$$

If I use the colors from the picture above $\frac{P(A,B,C)}{P(A,B)} = \frac{\text{red}}{\text{red+brown}}$.

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    $\begingroup$ Isn't $\frac{P\left(A, B, C\right)}{P\left(A, B\right)} = \frac{\text{red}}{\text{red} + \text{brown}}$? $\endgroup$
    – oszkar
    Feb 21 '18 at 21:57
  • $\begingroup$ Yes you are right. I will change it! $\endgroup$
    – ekoulier
    Feb 21 '18 at 22:24
  • $\begingroup$ So that's OK and correct. But $\frac{P\left(A\right) P\left(B\right) P\left(C\right)}{P\left(B\right) P\left(C\right)} = P\left(A\right) = \text{green} + \text{light brown} + \text{red} + \text{brown}$. Can you say if it's equal to $\frac{\text{red}}{\text{red} + \text{brown}}$ or not? $\endgroup$
    – oszkar
    Feb 21 '18 at 23:01
  • $\begingroup$ Thanks for the answer, but at this point my question is: isn't it red=P(A)P(B)P(C)? I got it is like this if they are mutually exclusive, but if not how can i make it? $\endgroup$
    – Lorenzo B.
    Feb 23 '18 at 11:48
  • $\begingroup$ @LorenzoB. Indeed $P(A,B,C) = \text{red}$ but you can't say that $P(A)P(B)P(C) = \text{red}$. $\endgroup$
    – ekoulier
    Feb 26 '18 at 9:55
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It could be wrong, and it could be right, based on the independency/dependency of the events. Of course it's not generally true.

An example when it's not true:

Lets say \begin{align*} \Omega &:= \left\{ x \in \mathbb{Z} \; : \; 1 \le x \le 31 \right\}, \\ A &:= \left\{ x \in \Omega \; : \; 2|x \right\}, \\ B &:= \left\{ x \in \Omega \; : \; 3|x \right\}, \\ C &:= \left\{ x \in \Omega \; : \; 5|x \right\} \end{align*}

In this case: \begin{align*} A \cap B \cap C &:= \left\{ x \in \Omega \; : \; 30|x \right\}, \\ B \cap C &:= \left\{ x \in \Omega \; : \; 15|x \right\} \end{align*}

And the corresponding probabilities: \begin{align*} P\left(A\right) &= \frac{15}{31} \\ P\left(B\right) &= \frac{10}{31} \\ P\left(C\right) &= \frac{6}{31} \\ P\left(A \cap B \cap C\right) &= \frac{1}{31} \\ P\left(B \cap C\right) &= \frac{2}{31} \end{align*}

Hence: \begin{align*} \frac{P\left(A \cap B \cap C\right)}{P\left(B \cap C\right)} &= \frac{\frac{1}{31}}{\frac{2}{31}} = \frac{1}{2}\\ \frac{P\left(A\right) P\left(B\right) P\left(C\right)}{P\left(B\right) P\left(C\right)} &= P\left(A\right) = \frac{15}{31} \end{align*}

And they are not equal. (And also $P\left(A \cap B \cap C\right) \ne P\left(A\right) P\left(B\right) P\left(C\right)$, so $A, B, C$ are not independent.)

(Some homework (as an example when it's true, remark what a small difference is enough): check for $\Omega := \left\{ x \in \mathbb{Z} \; : \; 1 \le x \le 30 \right\}$. It will be true, and $A, B, C$ will be independent. So it can be true, but not always, just in the case of independence.)

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  • $\begingroup$ I always wandered what's wrong with this answer (as it was downvoted). Can anyone help? $\endgroup$
    – oszkar
    Apr 17 '20 at 8:14
  • $\begingroup$ -1 you spelled wandered as wondered $\endgroup$
    – John D
    Nov 3 '20 at 0:00

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