Let’s denote the interval between consecutive Poisson events with random variable Y, during the interval that extends from a to a + y, the number of Poisson events k has the probability \(P(k)=e^{-\lambda y} \frac{(\lambda y)^k}{k!}\), if \(k=0\), \(e^{-\lambda y}\frac{(\lambda y)^0}{0!}=e^{-\lambda y}\) means there is no event during the (a,a+y) time period.
Because there will be no occurrences in the interval (a, a + y) only if Y > y, so \(P(Y > y)=e^{-\lambda y}\), then the cdf is \(F_Y(y)=P(Y \le y)=1-P(Y > y)=1-e^{-\lambda y}\). Then the pdf is \(\frac{d}{dy}F_Y(y)=\lambda e^{-\lambda y}, \quad y>0\), which is a Exponential random variable.
The moment-generating function for a Exponential random variable Y is: \[ \begin{align} M_Y(t)=E(e^{tY})&=\int_{0}^{+\infty}e^{ty}\lambda e^{-\lambda y}dy\\ &=\lambda\int_{0}^{+\infty}e^{-(\lambda-t)y}dy\\ &=-\frac{\lambda}{\lambda-t}e^{-(\lambda-t)y}\Big|_{y=0}^{y\to+\infty}\\ &=\frac{\lambda}{\lambda-t} \end{align} \]
The expected value is: \(M_Y^{(1)}(t)=\frac{d}{dt}\frac{\lambda}{\lambda-t}=\frac{\lambda}{(\lambda-t)^2}|_{t=0}=1/\lambda=E(Y)\)
\(M_Y^{(2)}(t)=\frac{2\lambda}{(\lambda-t)^3}|_{t=0}=2/\lambda^2=E(Y^2)\) Then, the Variance is:\(Var(Y)=E(Y^2)-(E(Y))^2=2/\lambda^2-1/\lambda^2=1/\lambda^2\)