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Intuitively, if events A and B are independent, then any subset of A should also be independent of B, that is, let $C \subset A$, then is it true that $P(C \cap B) = P(C) P(B)$?

Here is my attempt to prove this statement:

We have $P(A \cap B) = P(C \cap B) + P((A \setminus C) \cap B)$ and

$P(A)P(B) = P(A \setminus C)P(B) + P(C)P(B)$.

The independence of A and B implies that $P(C \cap B) = P(C) P(B)$ if and only if $P((A \setminus C) \cap B) = P(A \setminus C)P(B)$. In words, if A and B are independent, then any subset of A is independent of B if and only if its complement in A is independent of B.

So if my original statement is wrong, can anybody give a counter example?

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2 Answers 2

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You can try with $C:=A\cap B$; this is a subset of $A$ and $$\mathbb P(C\cap B)=\mathbb P( A\cap B) =\mathbb P(A)\mathbb P(B) $$ while $$\mathbb P(B)\mathbb P(C)=(\mathbb P(B))^2\mathbb P(A).$$ Unless we are in a degenerated case (that is, $\mathbb P(A)=0$ or $\mathbb P(B)\in\{0,1\}$), the set $A\cap B$ is not independent of $B$.

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Davide's example is correct, though I thought I'd try to clarify it using a simple example (since I like simple examples).

Suppose we roll a six-sided die. Let $A$ be the event "rolling a 1 or a 2" and let $B$ be the event "rolling an odd number". That is $A=\{1,2\}$ and $B=\{1,3,5\}$. Since we are rolling a six-sided die, it follows that $\mathbb{P}(A)=\frac{2}{6}=\frac{1}{3}$ and $\mathbb{P}(B)=\frac{3}{6}=\frac{1}{2}$.

$A\cap B$ is the event "rolling a number that's both an odd number and either a 1 or a 2", i.e. the event "rolling a 1". The probability of this is $\frac{1}{6}$. Noting that $\mathbb{P}(A)\cdot \mathbb{P}(B)=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$, we can conclude that $A$ and $B$ are indeed independent events.

Just as in Davide's example, let $C=A\cap B$. As just discussed, this is the event "rolling a 1" and $\mathbb{P}(C)=\frac{1}{6}$.

If $B$ would be independent of $C$ it would follow that $\mathbb{P}(B\cap C)=\mathbb{P}(B)\cdot\mathbb{P}(C)=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$. However, we see that $B\cap C = \{1,3,5\}\cap \{1\}=\{1\}$ and thus that $\mathbb{P}(B\cap C)=\frac{1}{6}$. We can thus conclude that the events are not independent.

In less technical terms, while knowing that somebody rolled a 1 or a 2 on the die doesn't give you any information about whether or not he or she rolled an odd number, giving you the information that he or she rolled a 1 (a subset of the set $\{1,2\}$) gives you all the information you need.

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