# Summation Bounds When Finding Transformation of 2 Poisson Random Variables

I am reviewing some material on functions of several random variables from Section 7.4 of John E. Freund's Mathematical Statistics, 6th Edition, and I'm stumped on how the author gets the upper bound of a summation during a transformation, so I'm hoping someone could help me understand this. The particular worked example is on page 249 and 250. Here is the question:

If $$X_1$$ and $$X_2$$ are independent random variables having Poisson distributions with the parameters $$\lambda_1$$ and $$\lambda_2$$, find the probability distribution of the random variable $$Y=X_1+X_2$$.

The question is straight forward, and I fully understand every step until the author shows the upper bound for $$X_2$$ (and in fact, I can derive the answer through other methods, but I'm interested in working through this method). Here is the solution the author provides and then I'll explain exactly what my issue is:

Since $$X_1$$ and $$X_2$$ are independent, their joint distribution is given by:

$$\begin{eqnarray*} f(x_{1},x_{2}) & = & \frac{e^{-\lambda_{1}}\left(\lambda_{1}\right)^{x_{1}}}{x_{1}!}\cdotp\frac{e^{-\lambda_{2}}\left(\lambda_{2}\right)^{x_{2}}}{x_{2}!}\\ & = & \frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{1}\right)^{x_{1}}\left(\lambda_{2}\right)^{x_{2}}}{x_{1}!x_{2}!} \end{eqnarray*}$$

for $$x_{1}=0,1,2,\ldots$$ and $$x_{2}=0,1,2,\ldots$$ . Since $$y=x_{1}+x_{2}$$ and hence $$x_{1}=y-x_{1}$$, we can substitute $$y-x_{2}$$ for $$x_{1}$$, getting

$$\begin{eqnarray*} g(y,x_{2}) & = & \frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{2}\right)^{x_{2}}\left(\lambda_{1}\right)^{y-x_{2}}}{x_{2}!\left(y-x_{2}\right)!} \end{eqnarray*}$$

for $$y=0,1,2,\ldots$$ and $$x_{2}=0,1,2,\ldots,y$$, for the joint distribution of $$Y$$ and $$X_{2}$$. Then summing on $$x_{2}$$ from $$0$$ to $$y$$, we get

$$\begin{eqnarray*} h(y) & = & \sum_{x_{2}=0}^{y}\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{2}\right)^{x_{2}}\left(\lambda_{1}\right)^{y-x_{2}}}{x_{2}!\left(y-x_{2}\right)!}\,\,\,\,\,\,\,(*)\\ & = & \frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}}{y!}\cdotp\sum_{x_{2}=0}^{y}\frac{y!}{x_{2}!\left(y-x_{2}\right)!}\left(\lambda_{2}\right)^{x_{2}}\left(\lambda_{1}\right)^{y-x_{2}} \end{eqnarray*}$$

Now, what I don't understand is how we obtain an upper bound of $$y$$ on the summation for $$x_2$$ on the last two lines, denoted with $$(*)$$? Since $$x_2$$ is poisson, shouldn't the upper bound be infinity? Can someone help me understand where "$$y$$" comes from? Thanks.

You have to recall that $$X_1$$ and $$X_2$$ are Poisson random variables and this means the support for $$X_1$$ and $$X_2$$ is 0, 1, 2, etc. for each. In other words, $$x_1$$ and $$x_2$$ are integers such that $$x_1\ge 0$$ and $$x_2\ge 0$$. Then, combine these facts with the fact that $$x_1=y-x_2$$ and substituting this value for $$x_1$$ into $$x_1\ge 0$$ reveals $$y-x_{2} \ge 0 \implies y\ge x_{2}$$. Obviously, this means $$x_2$$ the upper bound cannot exceed $$y$$.