Can A and B be independent if A is a subset of B? (except P(B)=1 and P(A)=0)

Question

I have a question about the independence of events.

Let's say I have positive integers from 1 to 10. Event A is integers below 2 and event B is integers below 3, then P(A)=0.2, P(B)=0.3 and A is a subset of B.

I am wondering whether A and B are independent. The probability that one event (P(A)) occurs in no way affects the probability of the other event occurring (P(B)) here. So from the definition of the independent events, A and B are independent. But it looks like there is some dependency between A and B because A is a subset of B. A always occurs where B occurs.

Any suggestions are appreciated.

Answer 1

Assuming independence and $A\subset B$ yields $$p(A)p(B)=p(A\cap B)=p(A)$$ This implies $p(B)=1$ or $p(A)=0$.

Answer 2

In your example, $A \subset B$. Think about this:

I choose a number at random. $P(A)$ is the probability of choosing $1$ or $2$ which has probability $0.2$. But what if I then told you that the number I chose is definitely in $B = \{1,2,3\}$? Would this change the probability that the number is in $A = \{1,2\}$? Your answer should be yes! It becomes much more likely ($\frac{2}{3}$, in fact). Assuming $B$ has happened affects that probability that $A$ has happened, hence they are not independent.

In general, if $A\subset B$ we have $P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(A)}{P(B)}$. In your example, this would be $\frac{0.2}{0.3} = \frac{2}{3},$ as expected.

To be clear, if $X$ and $Y$ are random variables uniformly distributed over $\{1,2,\ldots,10\}$ then the event $X \in A$ is not independent of $X \in B$. This conditions the value of the single choice $X$ on knowledge of the subset in which $X$ lies. However, if you are talking about making two selections $X$ and $Y$ then $Y \in A$ is independent of $X \in B$.

Answer 3

If we have events $A, B$ s.t. $A \subseteq B$ they are independent iff

$$P(A)P(B) = P(A \cap B)$$

$$\iff P(A)P(B) = P(A)$$

$$\iff P(A) = 0 \ \color{red}{\text{or}} \ P(B) = 1$$

Hence, we need not have

$$P(A) = 0 \ \color{red}{\text{and}} \ P(B) = 1$$