Finding example of independent event in coin toss

Question

This is from the book All of Statistics - Larry Wasserman, Chapter 1, excercise 23

Here we will get some experience simulating conditional probabilities. Consider tossing a fair die. Let $A = [2, 4, 6]$ and $B = [1,2,3,4]$. Then, $P(A) = 1/2, P(B) = 2/3$ and $P(A \cap B) = 1/3$. Since $P(A > \cap B) = P(A)P(B)$, the events A and B are independent.

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Now find two events $A$ and $B$ that are not independent.

Suppose I choose A = {2, 4}, then $P(A) = 1/3$, and $P(A \cap B)$ is still $1/3. P(A \cap B) \neq P(A)P(B)$. However, I think A and B are still independent because I could not see how choosing B and A influences another. How can one find an define $A$ and $B$ so that they are dependent?

Answer 1

A useful way to think about dependency is this: If the chance of $B$ happening is effected by the happening or not happening of $A$, then we can say that they are dependent.

As an extreme case, consider the situation where $A=B$:

Well, when $A$ happens, $B$ is bound to happen as well, while if $A$ does not happen, $B$ cannot happen either. So, the chance of $B$ happening is very much effected by $A$ happening or not. Indeed, one could say they are 'perfectly' or 'maximally' dependent.

The math shows this as well. With $A=B=[2,4,6]$ we get $P(A \cap B)=\frac{1}{2}$ but $P(A)\cdot P(B)=\frac{1}{4}$

But using the above way of thinking, you can also test for dependency as follows: If $P(A|B)=P(A|B')$, then $A$ and $B$ are independent, otherwise they are dependent. In this case: $P(A|B)=1$, but $P(A|B')=0$, so they are dependent.

Also note that $P(A|B)=P(A|B')$ if and only if $P(A|B)=P(A)$, so that is another tèst you can do: if $P(A|B) \not = P(A)$ then they are dependent: once again, the chance of $A$ happening is effected by the happening of $B$.

Now, your example works too! As you found out, with $A=[2,4]$ and $B=[1,2,3,4]$, we have $P(A \cap B)=P(A)=\frac{1}{3}$, which does not equal $P(A)\cdot P(B)=\frac{2}{9}$, so they are dependent.

But more intuitively, why is this so?

Well, note that we have $A \subseteq B$, so whenever $A$ happens, we know $B$ definitely happens. However, if $A$ does not happen, $B$ may or may not happen. So, the happening of $A$ will effect the chances of $B$ happening, making them dependent.

Seen the other way around: if $B$ happens, then $A$ may or may not happen, but if $B$ does not happen, than $A$ definitely cannot happen. So, the happening of $A$ will effect the chances of $B$ happening. So again, they will influence each other, and are thus dependent.

Indeed, to use that other test:

$P(B|A)=1$, but $P(B|A')=\frac{1}{2}$. Not the same, so dependent.

$P(A|B)=\frac{1}{2}$, but $P(A|B')=0$. Again, not the same, so dependent

Answer 2

Hint: $A = \{ 1,2\}, B = \{ 3,4\}$

Answer 3

Here is an example where the events are neither mutually exclusive nor disjoint nor one a subset of the other:

$A$: get an even number, $\{2,4,6\}$

$B$: get a number greater than $3$, $\{4,5,6\}$.

$P(A)=P(B)=\frac12 $, but $P(A\cap B)=P(\{4,6\})=\frac13$. So $P(A\cap B)\ne P(A)\cdot P(B)$.