Question

Suppose events are occurring randomly in time. The number of events is a Poisson random variable with parameter λ. Prove the amount of time one has to wait until a total of n events has occurred will be the gamma random variable with parameters (n,1/λ).

Answer 1

The number of events at time is a random variable $X_t$ can be modelled by Poisson distribution. The Poisson PMF is

$P\left ( X_t=n \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^n}{n!};n=0,1,2,3,.....$ Here $\lambda$ is the Poisson parameter - the average number of events in unit time.

1) The event that it takes more than time for the is the union of disjoint events namely 0, 1,2,3,..., in time . That is

$Y_n>t=\left (X_t=0 \right )\cup \left (X_t=1 \right )\cup \left (X_t=2 \right )\cup ....\cup \left (X_t=n-1 \right )$
So the probability,

$P\left (Y_n>t \right )=P\left (X_t=0 \right )+P \left (X_t=1 \right )+P \left (X_t=2 \right )=..+P \left (X_t=n-1 \right )\\$

Since the probability of union of disjoint events are the sum of the probabilities of individual events.

Using the Poisson PMF formula, we have

$P\left ( X_t=0 \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^0}{0!}\\ P\left ( X_t=1 \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^1}{1!}\\ P\left ( X_t=2 \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^2}{2!}\\ .....................\\ P\left ( X_t=n-1 \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^{n-1}}{\left ( n-1 \right )!}\\$

2) Using the complement of event and the definition of CDF

$P\left (Y_n\leqslant t \right)=1-P\left (Y_n>t \right)\\ F_{Y_n} \left ( t \right)=1-e^{-\lambda t}-e^{-\lambda t}\left ( \lambda t \right )-\frac{1}{2}e^{-\lambda t}\left ( \lambda t \right )^2-...-e^{-\lambda t}\frac{\left (\lambda t \right ) ^{n-1}}{\left ( n-1 \right )!}$

3) The PDF is obtained from CDF by differentiating (Note the cancellation of successive terms after diffrentiation)

$f_{Y_n} \left ( t \right)=\frac{\mathrm{d} }{\mathrm{d} t}F_{Y_n} \left ( t \right)\\ f_{Y_n} \left ( t \right)= \left [1-e^{-\lambda t}-e^{-\lambda t}\left ( \lambda t \right )-\frac{1}{2}e^{-\lambda t}\left ( \lambda t \right )^2-...-e^{-\lambda t}\frac{\left (\lambda t \right ) ^{n-1}}{\left ( n-1 \right )!} \right ]\\ f_{Y_n} \left ( t \right)=- \sum_{i=0}^{n-1}\frac{\mathrm{d} }{\mathrm{d} t}\left [e^{-\lambda t}\frac{\left (\lambda t \right ) ^{i}}{i!} \right ]\\ f_{Y_n} \left ( t \right)=\lambda e^{-\lambda t}+\lambda \sum_{i=1}^{n-1}\left [ e^{-\lambda t}\frac{\left (\lambda t \right ) ^{i}}{i!} -e^{-\lambda t}\frac{\left (\lambda t \right ) ^{i-1}}{\left ( i-1 \right )!} \right ]$

$f_{Y_n} \left ( t \right)=\lambda e^{-\lambda t}+\lambda e^{-\lambda t} \sum_{i=1}^{n-1}\left [ \frac{\left (\lambda t \right ) ^{i}}{i!} -\frac{\left (\lambda t \right ) ^{i-1}}{\left ( i-1 \right )!} \right ]\\ f_{Y_n} \left ( t \right)=\lambda e^{-\lambda t}+\lambda e^{-\lambda t} \left [ \frac{\left (\lambda t \right ) ^{n-1}}{\left ( n-1 \right )!} -1 \right ]\\ f_{Y_n} \left ( t \right)=\lambda e^{-\lambda t} \frac{\left (\lambda t \right ) ^{n-1}}{\left ( n-1 \right )!} \\ f_{Y_n} \left ( t \right)=\lambda ^ne^{-\lambda t} \frac{t ^{n-1}}{\left ( n-1 \right )!} ;t>0$

Which the Gamma PDF with parameters $Y_n\sim \Gamma \left ( n,1/\lambda \right )$ .