# How to think about multiple independent events?

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?

• Please add the [self-study] tag & read its wiki. Commented Sep 19, 2015 at 10:31
• I find it a bit disappointing that people just did the problem here. Regardless of whether the "self-study" tag is there, we all know what it's like to me told an answer and what's it's like to be lead to one. The latter is almost always more meaningful. Commented Sep 19, 2015 at 11:19
• I upvoted you, now I am even wondering there are something missing for both my solution and jtobin's solution. Since both of us assume that A , B and C are mutually independent which might not be correct. Commented Sep 19, 2015 at 11:25
• Hmmm. That's a good point. I'm gonna actually work this out myself. Commented Sep 19, 2015 at 13:16
• What is especially disappointing is that this question has received three incorrect answers, though two may yet be modified. Consider two independent tosses of a fair coin, and let $B= \{HT,HH\}$ and $C=\{HT,TT\}$ be the events that the first and second tosses resulted in Heads and Tails respectively, and $A=\{HT,TH\}$ the event that exactly one toss resulted in Heads. Thus, $P(A)=P(B)=P(C)=\frac 12$, $P(A\cap B)=P(A\cap C)=\frac 14$, so that $A,B$ are independent as are $A,C$. But $P(B\cup C)=\frac 34,P(A\cap(B\cup C)=\frac 14 \neq P(A)P(B\cup C)$, that is, $A$ and $B\cup C$ are dependent. Commented Sep 19, 2015 at 13:59

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.

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 $B\cup C$.

• I would add that a trivial way to make the framed condition hold is $B$ and $C$ disjoint, since then $P(B\cap C)=0$. Commented May 16, 2018 at 8:19
• @Miguel Yes, that is another sufficient condition for $A$ and $B\cup C$ to be independent events, just like mutual independence of $A,B,C$ is a sufficient condition as my answer says. My answer is about what is the necessary condition for $A$ and $B\cup C$ to be independent events. Commented May 16, 2018 at 13:20

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 $P(X\cup Y)$?

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}

• I would go by this answer! Try to work it out yourself! you do not gain too much by just seeing the answer! Commented Sep 19, 2015 at 13:10

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.

• Why is $P(A,B,C)$ on the second line equal to $P(A)P(B,C)$ on the third line? It is not given that $A$ is independent of $B\cap C$, just of $B$, and of $C$ _separately. Commented Sep 19, 2015 at 14:04
• So, after your edit, is it just the derivation that is sloppy but the result claimed is itself correct, that is, $A$ is indeed independent of $B\cup C$ as the OP is tasked with proving? Or is it that the derivation does not prove the claim that $A$ is independent of $B\cup C$? Commented Sep 19, 2015 at 21:08
• @DilipSarwate My derivation does not prove the claim; my edit also changed the erroneous $=$ assertion to $\neq$ in an attempt to make this clear. I'll edit the answer again to be more explicit. Commented Sep 20, 2015 at 0:41
• OK, +1 for fixing your answer. Commented Sep 20, 2015 at 2:55

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

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

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.

• In your second long equation, you got a $-P(A)P(B\cap C)$ term when you multiplied out that middle expression. But you wrote $-P(A\cap B \cap C)$ instead, that is, you equated $P(A)P(B\cap C)$ and $P(A\cap B \cap C)$, in effect assuming that $A$ and $B\cap C$ are independent. Why is that? Commented Sep 19, 2015 at 13:47
• Thanks, it is an assumed independent which may not be correct. Commented Sep 20, 2015 at 0:29