Let's solve this step by step for each function of the Bernoulli random variables $X_1$ and $X_2$.

Given Information:

• $X_1 \sim \text{Bern}(p)$, $X_2 \sim \text{Bern}(p)$.
• $X_1$ and $X_2$ are independent.
• We need to derive the PMFs or CDFs for two functions of $X_1$ and $X_2$:
  • $g(X_1, X_2) = X_1 + X_2$
  • $f(X_1, X_2) = X_1 \cdot X_2$

Solution:

1. Distribution of $g(X_1, X_2) = X_1 + X_2$

Since $X_1$ and $X_2$ are Bernoulli random variables, each can take values 0 or 1 with probabilities:

• $P(X_i = 1) = p$
• $P(X_i = 0) = 1 - p$

The possible values of $g(X_1, X_2) = X_1 + X_2$ are:

1. $g(X_1, X_2) = 0$, if $X_1 = 0$ and $X_2 = 0$.
2. $g(X_1, X_2) = 1$, if either $X_1 = 1$ and $X_2 = 0$, or $X_1 = 0$ and $X_2 = 1$.
3. $g(X_1, X_2) = 2$, if $X_1 = 1$ and $X_2 = 1$.

Let's calculate the probabilities for each case.

• $P(g(X_1, X_2) = 0)$: $P(X_1 = 0 \text{ and } X_2 = 0) = P(X_1 = 0) \cdot P(X_2 = 0) = (1 - p)(1 - p) = (1 - p)^2$

• $P(g(X_1, X_2) = 1)$: There are two ways this can happen: $(X_1 = 1, X_2 = 0)$ or $(X_1 = 0, X_2 = 1)$. $P(g(X_1, X_2) = 1) = P(X_1 = 1, X_2 = 0) + P(X_1 = 0, X_2 = 1) = p(1 - p) + (1 - p)p = 2p(1 - p)$

• $P(g(X_1, X_2) = 2)$: $P(X_1 = 1 \text{ and } X_2 = 1) = P(X_1 = 1) \cdot P(X_2 = 1) = p \cdot p = p^2$

Thus, the PMF of $g(X_1, X_2)$ is:

(1 - p)^2, & k = 0, \\ 2p(1 - p), & k = 1, \\ p^2, & k = 2. \end{cases}$$ #### 2. Distribution of $$f(X_1, X_2) = X_1 \cdot X_2$$ The possible values of $$f(X_1, X_2) = X_1 \cdot X_2$$ are: 1. $$f(X_1, X_2) = 0$$, if $$X_1 = 0$$ or $$X_2 = 0$$. 2. $$f(X_1, X_2) = 1$$, if $$X_1 = 1$$ and $$X_2 = 1$$. Let's calculate the probabilities for each case. - **$$P(f(X_1, X_2) = 0)$$:** This happens if at least one of $$X_1$$ or $$X_2$$ is 0. The complementary event is that both are 1, so: $$P(f(X_1, X_2) = 0) = 1 - P(X_1 = 1 \text{ and } X_2 = 1) = 1 - p^2$$ - **$$P(f(X_1, X_2) = 1)$$:** This happens if both $$X_1 = 1$$ and $$X_2 = 1$$. $$P(f(X_1, X_2) = 1) = P(X_1 = 1) \cdot P(X_2 = 1) = p^2$$ Thus, the PMF of $$f(X_1, X_2)$$ is: $$P(f(X_1, X_2) = k) = \begin{cases} 1 - p^2, & k = 0, \\ p^2, & k = 1. \end{cases}$$ ### Summary 1. For $$g(X_1, X_2) = X_1 + X_2$$, the PMF is: $$P(g(X_1, X_2) = k) = \begin{cases} (1 - p)^2, & k = 0, \\ 2p(1 - p), & k = 1, \\ p^2, & k = 2. \end{cases}$$ 2. For $$f(X_1, X_2) = X_1 \cdot X_2$$, the PMF is: $$P(f(X_1, X_2) = k) = \begin{cases} 1 - p^2, & k = 0, \\ p^2, & k = 1. \end{cases}$$ Would you like more details on the derivations, or do you have any