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?

According to the definition of independent events, $A$ and $B\cup C$ are independent if and only if $$P(A\cap (B\cup C)) = P(A)P(B\cup C).$$

Since $A$ and $B$ and $A$ and $C$ are independent, I know that $$P(A\cap B) = P(A)P(B) \quad\text{and}\quad P(A\cap C)=P(A)P(C).$$

However, I have no idea how to solve this. I attempted to apply the probability rules I know but got nowhere.

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ Sep 19 '15 at 10:31
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    $\begingroup$ 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. $\endgroup$ Sep 19 '15 at 11:19
  • $\begingroup$ 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. $\endgroup$
    – Deep North
    Sep 19 '15 at 11:25
  • $\begingroup$ Hmmm. That's a good point. I'm gonna actually work this out myself. $\endgroup$ Sep 19 '15 at 13:16
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    $\begingroup$ 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. $\endgroup$ Sep 19 '15 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 $B\cup C$.

  • $\begingroup$ I would add that a trivial way to make the framed condition hold is $B$ and $C$ disjoint, since then $P(B\cap C)=0$. $\endgroup$
    – Miguel
    May 16 '18 at 8:19
  • $\begingroup$ @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. $\endgroup$ May 16 '18 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 $P(X\cup Y)$?

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.


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}

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    $\begingroup$ I would go by this answer! Try to work it out yourself! you do not gain too much by just seeing the answer! $\endgroup$
    – Gumeo
    Sep 19 '15 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.

  • $\begingroup$ 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. $\endgroup$ Sep 19 '15 at 14:04
  • $\begingroup$ 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 $B\cup C$? $\endgroup$ Sep 19 '15 at 21:08
  • $\begingroup$ @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. $\endgroup$
    – jtobin
    Sep 20 '15 at 0:41
  • $\begingroup$ OK, +1 for fixing your answer. $\endgroup$ Sep 20 '15 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.

  • $\begingroup$ 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? $\endgroup$ Sep 19 '15 at 13:47
  • $\begingroup$ Thanks, it is an assumed independent which may not be correct. $\endgroup$
    – Deep North
    Sep 20 '15 at 0:29

P{A(B+C)}=P(AB+BC)=P(AB)+P(AC)-P(ABC) =P(A)P(B)+P(A)P(C)-P(A)P(BC) [A,B,C are mutually independent] =P(A)[P(B)+P(C)-P(BC)] =P(A)P(B+C) Hence A and B+C are independent.


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