10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Sep 19 '15 at 10:31
  • 3
    $\begingroup$ 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. $\endgroup$ – jlimahaverford 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. $\endgroup$ – jlimahaverford Sep 19 '15 at 13:16
  • 3
    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Sep 19 '15 at 13:59
11
$\begingroup$

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 $B\cup C$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate May 16 '18 at 13:20
6
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate 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 $B\cup C$? $\endgroup$ – Dilip Sarwate 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. $\endgroup$ – Dilip Sarwate Sep 20 '15 at 2:55
1
$\begingroup$

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

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

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Dilip Sarwate 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
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.