Show equality in a probability problem. Help understanding steps in a solution.

Question

Let $X$ and $Y$ be independent Possion variables with parameters $\lambda$ and $\mu$. Show that

i) $X+Y$ is Possion with parameter $\lambda +\mu.$

ii) The conditional distribution of $X$, given that $X+Y=n$, is binomial and find the parameters.

The book solution: Let us notice that for $X$ and $Y$ non-negative independent random variables, then

$$P(X+Y=n)=\sum_{k=0}^{n}P(X=n-k)P(Y=k), \tag{1}$$

so in our specific case of Possion random variables we have

$$P(X+Y=n)=\sum_{k=0}^{n}\frac{e^{\lambda}\lambda^{n-k}}{(n-k)!}\cdot\frac{e^{-\mu}\mu^k}{k!}=\frac{e^{-(\lambda+\mu)}}{n!}\sum_{k=0}^{n}{n\choose k}\lambda^{n-k}\mu^k=\frac{e^{-(\lambda+\mu)}(\lambda+\mu)^n}{n!}.\tag{2}$$

For the second part of the question:

$$P(X=k|X+Y=n)=\frac{P(X=k,X+Y=n)}{P(X+Y=n)}=\\ \frac{P(X=k)P(Y=n-k)}{P(X+Y=n)}= {n\choose k}\frac{\lambda^k\mu^{n-k}}{(\mu+\lambda)^n},$$

Hence the conditional distribution is binomial with parameters $n$ and $\frac{\lambda}{\lambda + \mu}.$

---

Questions:

1. What is (1) telling me and why does it hold?
2. How can I compute the second sum in (2)? (Need to know the arithmetic behind it
3. In i), we set $Y=k$ and in ii) we set $X=k,$ why is it okay to do this? Could we have set $X=k$ in the first as well?

Answer 1

1. In order to have $X+Y=n$, you need to have $Y$ take some value between $0$ and $n$ (let's call that value $k$), and then you need to have $X=n-k$ in order to satisfy $X+Y=n$. Since these are disjoint events, their probabilities need to be added up, leading to $$\sum_{k=0}^n P(X=n-k\cap Y=k)$$Finally, using the fact that $X$ and $Y$ are independent, $$P(X=n-k\cap Y=k)=P(X=n-k)P(Y=k).$$

2. This is the binomial theorem, which says that for any $a,b$ and natural number $n$, $$ (a+b)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k} $$ Just let $a=\mu$ and $b=\lambda$.

3. You could also have used $X=k$ in the first part. Then it would look like $\sum_{k=0}^n P(X=k)P(Y=n-k)$. This corresponds to adding up the numbers in the reverse order, which gives you the same result.
   
   The $k$ is unrelated between the two problems: in the first, it is just a "dummy variable" which we introduce in order to write out the summation, and in the second, it represents a value which $X$ can take conditional on $X+Y=n$.