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Let W be a sample space, and let A,B $\subseteq$ W be two independent events.

  1. Prove or disprove: $\bar{A}$ and $\bar{B}$ are necessarily independent.
  2. Prove or disprove: A and $\bar{B}$ are necessarily independent.
  3. Prove or disprove: $\bar{A}$ and $\bar{A}$ are necessarily independent.
  4. Prove or disprove: It is possible that A = B
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Your #1 and #2 are the same. – Mike Wierzbicki Oct 18 '11 at 2:16
Please don't order us to do things (this is an objection to the way you have phrased your "question", if it is a question, as an order). – Gerry Myerson Oct 18 '11 at 2:16
There is more than one definition of independence. Are you using the definition $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$? – André Nicolas Oct 18 '11 at 2:16
So, what happens when you take that definition and try to do the first problem? – GEdgar Oct 18 '11 at 2:36
Hint for question 4 (the easiest): Take $A=B$. Independence would say that $P(A\cap B)=P(A)P(B)$, meaning that $P(A\cap A)=P(A)P(A)$. Under what circumstances can this happen? – André Nicolas Oct 18 '11 at 3:08
up vote 1 down vote accepted

I will assume that you are at a certain level in your probability course, and that in particular you know the following:

(i) $P(\overline{E})=1-P(E)$

(ii) $P(E \cup F)=P(E)+P(F)-P(E \cap F)$. If $E$ and $F$ are disjoint, then $P(E \cup F)=P(E)+P(F)$.

Problem 1: If $A$ and $B$ are independent, then $\overline{A}$ and $\overline{B}$ are always independent. To show this, we need to show that $P(\overline{A} \cap \overline{B})=P(\overline{A})P(\overline{B})$.

But the event $\overline{A} \cap \overline{B}$ is the same as the event $\overline{A \cup B}$. This can be most easily seen by drawing a Venn diagram.

It follows that $$P(\overline{A}\cap \overline{B})=1-P(A \cup B). \qquad\text{(Equation 1)}$$ But $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B) \qquad\text{(Equation 2)}$$ (the last equality follows from the fact that $A$ and $B$ are independent). Putting Equations 1 and 2 together, we find that $$P(\overline{A}\cap \overline{B})=1-(P(A)+P(B)-P(A)P(B))=1-P(A)-P(B)+P(A)P(B).$$ But $$1-P(A)-P(B)+P(A)P(B)=(1-P(A))(1-P(B))=P(\overline{A})P(\overline{B}),$$ and the independence of $A$ and $B$ follows.

Note: The above argument is particularly mechanical and formula-bound. It would be better to do Problem 2 first. Two applications of the idea used there would give a nicer solution of Problem 1.

Problem 2: We show that if $A$ and $B$ are independent, then $A$ and $\overline{B}$ are always independent. Note that $A= (A\cap B) \cup (A \cap \overline{B})$. Moreover, the events $A\cap B$ and $A \cap \overline{B}$ are disjoint (we cannot have have both happen).

It follows that $$P(A)=P(A\cap B) +P(A\cap \overline{B}).$$ Using the independence of $A$ and $B$, we conclude that $$P(A)=P(A)P(B) +P(A\cap \overline{B}).$$ Rewrite the above equation as $$P(A\cap \overline{B})=P(A)-P(A)P(B)=P(A)(1-P(B))=P(A)P(\overline{B}),$$ and it follows that $A$ and $\overline{B}$ are independent.

Problem 3: The question asks whether $P(\overline{A}\cap \overline{A})$ is always equal to $P(\overline{A})P(\overline{A})$. Note that $\overline{A}\cap \overline{A}=\overline{A}$, so the question asks whether we always have $P(\overline{A})=P(\overline{A})P(\overline{A})$. Certainly not! The equation $x=x^2$ has only two solutions, $x=0$ and $x=1$. So if we can find an $A$ and $B$, independent, such that $P(\overline{A})$ is neither $0$ nor $1$, we will have an example that shows that $P(\overline{A}\cap \overline{A})$ is not always equal to $P(\overline{A})P(\overline{A})$.

Almost anything you can think of will do. For example, let $A$ be the event we get a head when tossing a fair coin, and $B$ the event we get a $6$ when tossing a fair die. Then $A$ and $B$ are independent. (Don't pay attention to the $B$ stuff, $B$ is basically irrelevant.)

We have $P(\overline{A})=1/2$, and therefore $P(\overline{A})$ is not equal to $P(\overline{A})P(\overline{A})$. This single example all by itself shows that $\overline{A}$ and $\overline{A}$ are not necessarily independent. Intuitively, that sounds very reasonable, but a formal definition sometimes has surprising consequences, so we did need to check.

Problem 4: For some of the groundwork, please see Problem 3. Can we have $A$ and $B$ independent, where $A=B$? So we want to check whether we can have $P(A\cap A)=P(A)P(A)$, or equivalently whether we can have $P(A)=P(A)P(A)$. Consider the equation $x=x^2$. This has the solutions $x=0$ and $x=1$. So if $P(A)=0$ or $P(A)=1$, the equation $P(A\cap A)=P(A)P(A)$ will be satisfied, and we will have independence.

An easy example of independence is this. Toss a two-headed coin, and let $A$ (and $B$) be the event we get a head. Clearly $P(A)=1$, so $A$ and $A$ are independent. Or else, same coin, let $A$ be the event we get a tail. Then $P(A)=0$, so $A$ and $A$ are independent.

So the answer is yes, even if $A=B$, it is possible that $A$ and $B$ are independent. One can produce many such examples, all of them boring. Deal a $5$ card hand. Let $A$ be the event that the cards are all of the same kind. (They are all of the same kind if all are Aces, or all are Kings, and so on.) Of course $P(A)=0$, unless one of the players is a particularly stupid cheater. So if $A=B$, then $A$ and $B$ are independent.

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