Let $Y_1 = aX_1 \sim \text{Bernoulli}(p)$ and $Y_2 = bX_2 \sim \text{Bernoulli}(p)$, what is the PMF of $Z = Y_1 + Y_2$ for $a > 0$, $b > 0$ and $a \neq b$?

Can somebody check my result?

$$p_{Y_1}(x) = \begin{cases}p & \text{if } x = a\\1 - p & \text{if } x = 0\end{cases}$$

$$p_{Y_2}(x) = \begin{cases}p & \text{if } x = b\\1 - p & \text{if } x = 0\end{cases}$$

Then the convolution is

$$p_Z(z) = \sum_{k=-\infty }^{\infty} p_{Y_1}(k) p_{Y_2}(z-k)$$

Case 1: $z = 0$, $k = 0$

$$p_Z(0) = p_{Y_1}(0)p_{Y_2}(0) = (1-p)^2$$

Case 2: $z = a$, $k = 0$ or $z = a$, $k = a$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(a) + p_{Y_1}(a)p_{Y_2}(0) = (1-p)\cdot 0 + p(1-p)$$

Case 3: $z = b$, $k = 0$ or $z = b$, $k = b$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(0) = (1-p)\cdot p + 0(1-p)$$

Case 4: $z = a+b$, $k = a$ or $z = a+b$, $k = b$

$$p_Z(a) = p_{Y_1}(a)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(a) = p^2$$

Then the PMF is

$$p_{Z}(z) = \begin{cases}(1-p)^2 & \text{if } z = 0\\p(1-p) & \text{if } z = a\\p(1-p) & \text{if } z = b\\p^2 & \text{if } z = a+b\end{cases}$$

I test my result by comparing it to the expected value

$$E[Z] = aE[X_1] + bE[X_2] = (a + b)p$$

\begin{align*}E[Z] &= \sum_{x} p_Z(x)x\\ &= p(1-p)a + p(1-p)b + p^2(a+b)\\ &\neq p(a+b)\\ &= (a+b)p \end{align*}

Let $Y_1 = aX_1 \sim \text{Bernoulli}(p)$ and $Y_2 = bX_2 \sim \text{Bernoulli}(p)$, what is the PMF of $Z = Y_1 + Y_2$ for $a > 0$, $b > 0$ and $a \neq b$?

Can somebody check my result?

$$p_{Y_1}(x) = \begin{cases}p & \text{if } x = a\\1 - p & \text{if } x = 0\end{cases}$$

$$p_{Y_2}(x) = \begin{cases}p & \text{if } x = b\\1 - p & \text{if } x = 0\end{cases}$$

Then the convolution is

$$p_Z(z) = \sum_{k=-\infty }^{\infty} p_{Y_1}(k) p_{Y_2}(z-k)$$

Case 1: $z = 0$, $k = 0$

$$p_Z(0) = p_{Y_1}(0)p_{Y_2}(0) = (1-p)^2$$

Case 2: $z = a$, $k = 0$ or $z = a$, $k = a$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(a) + p_{Y_1}(a)p_{Y_2}(0) = (1-p)\cdot 0 + p(1-p)$$

Case 3: $z = b$, $k = 0$ or $z = b$, $k = b$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(0) = (1-p)\cdot p + 0(1-p)$$

Case 4: $z = a+b$, $k = a$ or $z = a+b$, $k = b$

$$p_Z(a) = p_{Y_1}(a)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(a) = p^2$$

Then the PMF is

$$p_{Z}(z) = \begin{cases}(1-p)^2 & \text{if } z = 0\\p(1-p) & \text{if } z = a\\p(1-p) & \text{if } z = b\\p^2 & \text{if } z = a+b\end{cases}$$

I test my result by comparing it to the expected value

$$E[Z] = aE[X_1] + bE[X_2] = (a + b)p$$

\begin{align*}E[Z] &= \sum_{x} p_Z(x)x\\ &= p(1-p)a + p(1-p)b + p^2(a+b)\\ &= p(a+b)\\ &= (a+b)p \end{align*}

Let $Y_1 = aX_1 \sim \text{Bernoulli}(p)$ and $Y_2 = bX_2 \sim \text{Bernoulli}(p)$, what is the PMF of $Z = Y_1 + Y_2$ for $a > 0$, $b > 0$ and $a \neq b$?

My attempt:

$$p_{Y_1}(x) = \begin{cases}p & \text{if } x = a\\1 - p & \text{if } x = 0\end{cases}$$

$$p_{Y_2}(x) = \begin{cases}p & \text{if } x = b\\1 - p & \text{if } x = 0\end{cases}$$

Then the convolution is

$$p_Z(z) = \sum_{k=-\infty }^{\infty} p_{Y_1}(k) p_{Y_2}(z-k)$$

Case 1: $z = 0$, $k = 0$

$$p_Z(0) = p_{Y_1}(0)p_{Y_2}(0) = (1-p)^2$$

Case 2: $z = a$, $k = 0$ or $z = a$, $k = a$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(a) + p_{Y_1}(a)p_{Y_2}(0) = 2p(1-p)$$

Case 3: $z = b$, $k = 0$ or $z = b$, $k = b$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(0) = 2p(1-p)$$

Case 4: $z = a+b$, $k = a$ or $z = a+b$, $k = b$

$$p_Z(a) = p_{Y_1}(a)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(a) = 2p^2$$

Then the PMF is

$$p_{Z}(z) = \begin{cases}(1-p)^2 & \text{if } z = 0\\2p(1-p) & \text{if } z = a\\2p(1-p) & \text{if } z = b\\2p^2 & \text{if } z = a+b\end{cases}$$

I test my result by comparing it to the expected value

$$E[Z] = aE[X_1] + bE[X_2] = (a + b)p$$

\begin{align*}E[Z] &= \sum_{x} p_Z(x)x\\ &= 2p(1-p)a + 2p(1-p)b + 2p^2(a+b)\\ &= 2p(a+b)\\ &\neq (a+b)p \end{align*}

Since a direct computation of the expected value gives a different result, my PMF is wrong.

Let $Y_1 = aX_1 \sim \text{Bernoulli}(p)$ and $Y_2 = bX_2 \sim \text{Bernoulli}(p)$, what is the PMF of $Z = Y_1 + Y_2$ for $a > 0$, $b > 0$ and $a \neq b$?

Can somebody check my result?

$$p_{Y_1}(x) = \begin{cases}p & \text{if } x = a\\1 - p & \text{if } x = 0\end{cases}$$

$$p_{Y_2}(x) = \begin{cases}p & \text{if } x = b\\1 - p & \text{if } x = 0\end{cases}$$

Then the convolution is

$$p_Z(z) = \sum_{k=-\infty }^{\infty} p_{Y_1}(k) p_{Y_2}(z-k)$$

Case 1: $z = 0$, $k = 0$

$$p_Z(0) = p_{Y_1}(0)p_{Y_2}(0) = (1-p)^2$$

Case 2: $z = a$, $k = 0$ or $z = a$, $k = a$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(a) + p_{Y_1}(a)p_{Y_2}(0) = (1-p)\cdot 0 + p(1-p)$$

Case 3: $z = b$, $k = 0$ or $z = b$, $k = b$

$$p_Z(a) = p_{Y_1}(0)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(0) = (1-p)\cdot p + 0(1-p)$$

Case 4: $z = a+b$, $k = a$ or $z = a+b$, $k = b$

$$p_Z(a) = p_{Y_1}(a)p_{Y_2}(b) + p_{Y_1}(b)p_{Y_2}(a) = p^2$$

Then the PMF is

$$p_{Z}(z) = \begin{cases}(1-p)^2 & \text{if } z = 0\\p(1-p) & \text{if } z = a\\p(1-p) & \text{if } z = b\\p^2 & \text{if } z = a+b\end{cases}$$

I test my result by comparing it to the expected value

$$E[Z] = aE[X_1] + bE[X_2] = (a + b)p$$

\begin{align*}E[Z] &= \sum_{x} p_Z(x)x\\ &= p(1-p)a + p(1-p)b + p^2(a+b)\\ &\neq p(a+b)\\ &= (a+b)p \end{align*}