To show that $A$ and $B$ are independent if and only if the Bernoulli random variables $X$ and $Y$ corresponding to these events are independent, we need to approach the problem in two parts.

1. Showing that $A$ and $B$ are independent implies $X$ and $Y$ are independent.

Assumptions:

• $A$ and $B$ are independent events, which means: $P(A \cap B) = P(A) \cdot P(B)$

Corresponding Bernoulli random variables:

• Let $X$ and $Y$ be the Bernoulli random variables for events $A$ and $B$ respectively: \begin{cases} 1, & \text{if the outcome is in } A \\ 0, & \text{if the outcome is in } A^c \end{cases}$$ $$Y = \begin{cases} 1, & \text{if the outcome is in } B \\ 0, & \text{if the outcome is in } B^c \end{cases}$$

Objective:

We need to show that $E(X^n Y^m) = E(X^n) \cdot E(Y^m)$ for all non-negative integers $n$ and $m$ when $A$ and $B$ are independent.

Proof:

Since $X$ and $Y$ are Bernoulli random variables, their powers are simple:

• $X^n = X$ for any $n \geq 1$ (since $X \in \{0, 1\}$),
• $Y^m = Y$ for any $m \geq 1$.

Thus, for any $n$ and $m$, we are really proving: $E(XY) = E(X) \cdot E(Y)$

Now, compute each term:

• $E(X) = P(A)$ (since $X = 1$ when the outcome is in $A$, and 0 otherwise),
• $E(Y) = P(B)$,
• $E(XY) = P(A \cap B)$ (since $XY = 1$ only if both $X = 1$ and $Y = 1$, which happens if and only if the outcome is in both $A$ and $B$).

Since $A$ and $B$ are independent, we have: $P(A \cap B) = P(A) \cdot P(B)$ Thus: $E(XY) = P(A \cap B) = P(A) \cdot P(B) = E(X) \cdot E(Y)$ Therefore, $E(XY) = E(X) \cdot E(Y)$, which shows that $X$ and $Y$ are independent.

---

2. Showing that $X$ and $Y$ are independent implies $A$ and $B$ are independent.

Assumptions:

• $X$ and $Y$ are independent random variables, which means: $E(XY) = E(X) \cdot E(Y)$

Objective:

We need to show that this implies: $P(A \cap B) = P(A) \cdot P(B)$

Proof:

As before, we know that:

• $E(X) = P(A)$,
• $E(Y) = P(B)$,
• $E(XY) = P(A \cap B)$.

Since $X$ and $Y$ are independent, we are given: $E(XY) = E(X) \cdot E(Y)$ Substituting the expressions for $E(X)$ and $E(Y)$, we get: $P(A \cap B) = P(A) \cdot P(B)$ This proves that $A$ and $B$ are independent.

---

Conclusion:

We have shown that $A$ and $B$ are independent if and only if the corresponding Bernoulli random variables $X$ and $Y$ are independent, which completes the proof.

Follow-up questions:

1. How do Bernoulli random variables differ from general random variables?
2. What happens if $A$ and $B$ are dependent events? How does that affect the proof?
3. Can this result be generalized to non-Bernoulli random variables?
4. How can we extend this proof to more than two events?
5. What is the significance of the expectation operator in proving independence?

Tip:

When dealing with Bernoulli random variables, remember that their powers are either 0 or 1, which simplifies many calculations involving products and expectations.