Let $(X_1,X_2,X_3,X_4)\sim Mult(n,4,p_1,p_2,p_3,p_4)$. What is the distribution of $X_1+X_2$?
So, we have $X_1\sim Bin(n,p_1),X_2\sim Bin(n,p_2)$. I guess the sum should also be off the binomial distribution, but I don't know how to get the formal proof.
$$P(X_1+X_2=k)=\sum_{k_1}P(X_1=k_1 \text{ and } X_2=k-k_1)=\sum_{k_1}P(X_1=k_1)P(X_2=k-k_1),$$where the last step follows by independence. I'm stuck here. Any help?
Best Answer
So, you take $n$ i.i.d. samples from the distribution $p$ on $\{1,2,3,4\}$ with probability mass function $p(i)=p_i$. $X_1$ is the number of samples equal to $1$, $X_2$ is the number equal to $2$.
You are correct, $X_i$ is distributed as $\operatorname{Bin}(n,p_i)$ (but the $X_i$'s are not independent).
Now, $X_1+X_2$ is the number of samples taking value either $1$ or $2$. If you think of it this way, it is immediate that each of the $n$ samples has probability $p_1+p_2$ to be in $\{1,2\}$, and that, since these $n$ samples are independent, $$ X_1+X_2 \sim \operatorname{Bin}(n,p_1+p_2)\,. $$