Suppose we have a random variable that follows a Poisson Point Process: $ X \sim poisson(\lambda t) $ and a function $f(x) = ax + b $ where $a,b \in \mathbb{R}$.
What is the pdf of $Y = aX + b$?
I know for continuous random variables if $f(x)$ is strictly increasing:
$$f_y(Y = y) = \frac{1}{|a|} f_x\left(\frac{y-b}{a}\right)$$
However I am having trouble doing this for the discrete case. My attempt is as follows suppose for simplicity $t = 1$, thus $X \sim poisson(\lambda)$, Now we have:
\begin{align*}
P(Y \leq y) &= P(aX + b \leq y)\\
&= P\left(X \leq \frac{y – b}{a}\right)\\
&= F_x\left(\frac{y-b}{a}\right)\\ \\
P(Y = k) &= P(Y \leq k) – P(Y \leq k – 1)\\
&=F_y(Y = k) – F_y(Y = k – 1)\\
&= F_x(\frac{k-b}{a}) – F_x(\frac{k-1-b}{a}) \\
&= \sum_{n = 0}^{\frac{k-b}{a}} e^{\lambda} \frac{\lambda^n}{n!} – \sum_{n = 0}^{\frac{k-1 -b}{a}} e^{\lambda} \frac{\lambda^n}{n!}
\end{align*}
At this point I am stuck because clearly the factorials could take noninteger values, so this difference seems to be meaningless. Can anyone give me a hint as to how to continue?
Best Answer
continue from your last line that $\sum_{n=0}^{\frac{k-b}{a}} e^{\lambda} \frac{\lambda^n}{n!} - \sum_{n=0}^{\frac{k-b-1}{a}} e^{\lambda} \frac{\lambda^n}{n!}$ = $e^{\lambda}\frac{\lambda^{\frac{k-b}{a}}}{\frac{k-b}{a}!}$
Since k is an integer calculated from k = aq + b (for some q satisfying X-distribution)