How to derive a marginal PMF from this cumbersome joint PMF

Question

I've been learning probability theory recently and got stuck with this problem: knowing that the joint PMF of a 2 dimensional random variable $(X,Y)$ :
\begin{equation}P(X=n,Y=m)=\frac{\lambda^np^m(1-p)^{n-m}}{m!(n-m)!}e^{-\lambda}, m=0,1,…,n,\ n=0,1,2,…\end{equation}
notice that $\lambda>0,\ 0<p<1$ in the formula above. The only related knowledge I know now is that you can derive a marginal PMF from a joint one by the formula:
\begin{equation}p_X(x)=\sum_yp_{X,Y}(x,y)\end{equation}
and I tried to solve the problem with this formula and found:
\begin{equation}\frac{\lambda^np^m(1-p)^{n-m}}{m!(n-m)!}e^{-\lambda}\\
=\frac{n!\lambda^np^m(1-p)^{n-m}}{n!m!(n-m)!}e^{-\lambda}\\
=\binom{n}{m}p^m(1-p)^{n-m} \cdot \frac{\lambda^n}{n!}e^{-\lambda}\end{equation}
the item on the left hand side of the dot is the PMF of a binomial random variable with parameters n and p, while the item on the right hand side is PMF of Poisson random variable. But the problem is: m and n is somewhat "intertwined" here, for example, if I want to derive the PMF of $Y$, taking sum:
\begin{equation}\sum_{n=0}^{\infty}\binom{n}{m}p^m(1-p)^{n-m} \cdot \frac{\lambda^n}{n!}e^{-\lambda}\end{equation}
does not make sense to me: for different n, the range of m is always different, when $n=0$, the item here becomes\begin{equation}\binom{0}{m}p^m(1-p)^{-m} \cdot e^{-\lambda}\end{equation}
how can I understand $\binom{0}{m}$ here? Or where am I wrong? I wish someone could help me to explain this "intertwining problem" and solve this problem step by step if it's convenient for you.

Answer 1

One way of computing $ p_Y $ directly with the original PMF: Observe that $ P(X = n, Y = m) = 0 $ whenever $ m > n $. Then, \begin{align*} p_Y(m) &= \sum_{n \geq 0} P(X = n, Y = m) = \sum_{n \geq m} P(X = n, Y = m) \\ &= \frac{p^m}{m!}e^{-\lambda} \sum_{n \geq m} \frac{\lambda^n (1-p)^{n-m}}{(n-m)!}. \end{align*} Adjusting the summation index and taking out a factor of $ \lambda^m $, \begin{align*} \phantom{p_Y(m)} &= \frac{(\lambda p)^m}{m!}e^{-\lambda} \sum_{n \geq 0} \frac{\lambda^n (1-p)^{n}}{n!} \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &= \frac{(\lambda p)^m}{m!} e^{-\lambda} e^{\lambda(1-p)} = \frac{(\lambda p)^m}{m!} e^{-\lambda p}, \end{align*} i.e. the marginal distribution of $ Y $ is $ \mathrm{Poi}(\lambda p) $.

---

Regarding your question on the binomial coefficients: ${{n}\choose{k}} = 0 $ (by definition) whenever $k > n$. This makes sense when thinking of the combinatorial interpretation of ${{n}\choose{k}}$ as the number of $k$-element subsets of the set $\{1,\ldots n\} $. For this example, we have $ {n \choose m} = 0$ for $n < m $, in line with what we observed above about the joint PMF.