Question
$X_1 \sim \mathrm{Bi}(n,p_1)$ and $X_2 \sim \mathrm{Bi}(n,p_2)$ and $X_1$ and $X_2$ are independent, and $p_1 <\frac{1}{2}$ and $p_2 < \frac{1}{2}$ then $X_1 + X_2 \sim \mathrm{Bi(n, p_1 + p_2)}.$ True or false. Provide mathematical reasoning.
First attempt
I have looked at the uniqueness property of moment generating functions to show that this statement is false.
Consider the moment generating functions for $X_1$ and $X_2$. The properties of MGF's stipulate for two random variables $X_1$ and $X_2$, and if they areindependent, $X_1 + X_2$ is calculated through the product of $X_1$ and $X_2$ given they are distributed the same way. This will be demonstrated below.
\begin{align} \mathrm{M}_{X_1}(t) &= \left(1-p_1-pe^{t}\right)^{n_1} \\ \mathrm{M}_{X_2}(t) &= \left(1-p_2-pe^{t}\right)^{n_2} \end{align}
Provided $X_1$ and $X_2$ are independent random variables, the joint moment function $\mathrm{M}_{X_1, X_2}(X_1 + X_2)$ is given as the product of $\mathrm{M}_{X_1}(t)$ and $\mathrm{M}_{X_2}(t)$
\begin{align} \mathrm{M}_{X_1, X_2}(X_1 + X_2) &= \mathrm{M}_{X_1}(t) \cdot \mathrm{M}_{X_2}(t) \\ &= \left(1-p_1-p_1e^{t}\right)^{n} \cdot \left(1-p_2-p_2e^{t}\right)^{n_2} \\ &= \left(q_1-p_1e^{t}\right)^{n} \cdot \left(q_2-p_2e^{t}\right)^{n_2} \end{align}
As observed from the top result, the product cannot be expressed as a single MGF in the form of $(q + pe^{t})^{n}$. The uniqueness property of MGFs state the random variables $X_1$ and $X_2$ cannot follow the same distribution. As such, the statement $X_1 + X_2 \sim \mathrm{Bi(n, p_1 + p_2)}$ is false.
Would be approach be correct and are there areas for myself to re-evaluate my approach.
Thank you!