How to think about multiple independent events?

Question

Suppose $\mathcal E$ and $\mathcal F$ are independent events in a probability space, and also that $\mathcal E$ and $\mathcal G$ are independent. Is $\mathcal F \cap \mathcal G$ independent of $\mathcal E$? If so, how can I demonstrate that?

Answer 1

Let $A$ and $B$ be independent events, and let $A$ and $C$ be independent events. How do I show that $A$ and $B\cup C$ are independent events as well?

You cannot show this result because it does not hold for all $A, B, C$ enjoying these properties. Consider the following counter-example.

Consider two independent tosses of a fair coin. Let $B=\{HT,HH\}$ and $C=\{HT,TT\}$ be the events that the first and second tosses resulted in Heads and Tails respectively. Let $A=\{HT,TH\}$ be the event that exactly one toss resulted in Heads.

Then, $P(A)=P(B)=P(C) = \frac 12$ while $P(A\cap B) = P(A\cap C) = \frac 14$ and so $A$ and $B$ are independent events as are $A$ and $C$ independent events. Indeed, $B$ and $C$ are also independent events (that is, $A$, $B$, and $C$ are pairwise independent events). However, $$P(A) = \frac 12 ~ \text{and}~ P(B\cup C)=\frac 34 ~ \text{while}~ P(A\cap(B\cup C)) =\frac 14 \neq P(A)P(B\cup C)$$ and so $A$ and $B\cup C$ are dependent events.

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Putting away our counter-example, let us consider what conditions are needed to make $A$ and $B\cup C$ independent events. The other answers have already done the work for us. We have that \begin{align} P(A\cap (B\cup C)) &= P((A\cap B) \cup (A\cap C))\\ &= P(A\cap B) + P(A\cap C) - P(((A\cap B) \cap (A\cap C))\\ &= P(A)P(B) + P(A)P(C) - P(A\cap B \cap C)\\ &= P(A)\left(P(B) + P(C) - P(B\cap C)\right) + \left(P(A)P(B\cap C) - P(A\cap B \cap C)\right)\\ &= P(A)P(B\cup C) + \left[P(A)P(B\cap C) - P(A\cap B \cap C)\right] \end{align} and so $P(A\cap (B\cup C))$ equals $P(A)P(B \cup C)$ (as is needed to prove that $A$ and $B\cup C$ are independent events) exactly when $P(A)P(B\cap C)$ equals $P(A\cap B \cap C) = P(A\cap (B\cap C))$, that is when $A$ and $B\cap C$ are independent events.

$A$ and $B\cup C$ are independent events whenever $A$ and $B\cap C$ are independent events.

Notice that whether $B$ and $C$ are independent or not is not relevant to the issue at hand: in the counter-example above, $B$ and $C$ were independent events and yet $A = \{HT, TH\}$ and $B\cap C = \{HT\}$ were not independent events. Of course, as noted by Deep North, if $A$, $B$, and $C$ are mutually independent events (which requires not just independence of $B$ and $C$ but also for $P(A\cap B \cap C) = P(A)P(B)P(C)$ to hold), then $A$ and $B\cap C$ are indeed independent events. Mutual independence of $A$, $B$ and $C$ is a sufficient condition.

Indeed, if $A$ and $B\cap C$ are independent events, then, together with the hypothesis that $A$ and $B$ are independent, as are $A$ and $C$ independent events, we can show that $A$ is independent of all $4$ of the events $B\cap C, B\cap C^c, B^c\cap C, B^c\cap C^c$, that is, of all $16$ events in the $\sigma$-algebra generated by $B$ and $C$; one of these events is $B\cup C$.

Answer 2

Two things.

1) Is there some way you know to rewrite the event $A \cap (B\cup C)$. Intuitively, we know how A,B and A,C interact, but we don't know how B,C interact. So $(B\cup C)$ is getting in our way.

2) Is there some way you know of rewriting $P(X\cup Y)$?

Even if you don't immediately get the answer, please edit your answer with the answers to these questions and we'll go from there.

edit

Please check me on this. I believe I have a counterexample.

Rolling a die to get X.

A: X < 4

B: X in {1, 4}

C: X in {1, 5}

Answer 3

As per Dilip Sarwate's comment, these events are demonstrably not independent.

The typical way I would try to prove independence proceeds like this:

\begin{align*} P(A, B \cup C) & = P(\{A, B\} \cup \{A, C\}) & \text{distributive property} \\ & = P(A, B) + P(A, C) - P(A,B,C) & \text{sum rule} \end{align*}

and here you'd like to factor $P(A)$ out of the expression in order to establish the property $P(A, B \cup C) = P(A)P(B \cup C)$, which would be sufficient to prove independence. However if you try to do that here, you get stuck:

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

Note that the braced expression is almost $P(B) + P(C) - P(B,C)$, which would get you to your goal. But you have no information that allows you to reduce $P(B,C \, | \, A)$ any further.

Note that in my original answer I had sloppily asserted that $P(B, C \, | \, A) = P(A)P(B, C)$ and thus erroneously claimed that the result asked to be proved was true; it's easy to mess up!

But given that it proves to be difficult to demonstrate independence in this way, a good next step is to look for a counterexample, i.e. something that falsifies the claim of independence. Dilip Sarwate's comment on the OP includes exactly such an example.

Answer 4

$P[A \cap(B \cup C)]=P[(A \cap B) \cup (A \cap C)]=P(A \cap B)+P(A \cap C)-P[( A \cap B)\cap (A \cap C)]=P(A)*P(B)+P(A)*P(C)-P(A \cap B \cap C)$

$P(A)*P(B \cup C)=P(A)[P(B)+P(C)-P(B \cap C)]=P(A)*P(B)+P(A)*P(C)-P(A)*P( B \cap C)$

Now, we need to show $P(A \cap B \cap C)=P(A)*P( B \cap C)$

If $A, B,C$ are mutually independent,the results are obvious.

While the condition is $A$ and $B$ are independent and $A$ and $C$ are independent do not guarantee independent of $B$ and $C$

Therefore, the OP may need to reexamine the condition of the question.