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I found the following result on Wikipedia relating to the CDF of the Gamma Distribution when the shape parameter is an integer. (Note: there is a slight difference on how I have defined the scale parameter and how it is given on the Wikipedia page) $$ X \sim Gamma(n, \lambda)\\ f_x(x; n, \lambda) = \frac{e^{-\lambda}x^{n-1}}{\lambda^n \Gamma(n)}\\ F(x;n,\lambda) = 1 - \sum_{i=0}^{n-1}{\frac {(\lambda x)^i} {i!}e^{-\lambda x}} $$

Can we prove the above result by considering a Poisson Process with parameter $\lambda$?

I reasoned this out as follows: (I believe this is not a very rigorous proof)

Let $X_t$ denote the Poisson Process with parameter $\lambda$. We know that: $$ X_t \sim Poi(\lambda t) $$

Also, denote by $T_i$ the inter-arrival time between the ${(i-1)}^{th}$ and $i^{th}$ occurrence of the event. Now, since we are dealing with a Poisson Process with parameter $\lambda$ so, $T_i$'s are $iid$ $Exp(\lambda)$ distributed.

As a result, $$ \sum_{i=1}^{n}{T_i} \sim Gamma(n,\lambda) $$ Now, $$ \begin{align} \mathbb{P}[\sum_{i=1}^{n}{T_i} > t] &= Pr[by\ time\ t\ less\ than\ n\ events\ have\ occured]\\ &= \mathbb{P}[X_t < n]\\ &= \sum_{x=0}^{n-1}{\frac {(\lambda t)^x} {x!}e^{-\lambda t}} \end{align} $$

Finally, subtracting the above value from $1$ gives us the CDF in the required form.

Is this reasoning technically correct? I would highly appreciate if anyone could suggest a different method.

Kaustav Sen
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  • The Wikipedia article does not say $f_x(x; n, \lambda).$ That notation is nonsense. $f_X(x; n, \lambda),$ on the other hand, might make sense. Failure to attend to this distinction between $X$ and $x$ will render you unable to understand some routine exercises for undergraduates and other things. – Michael Hardy Dec 27 '21 at 05:51

1 Answers1

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Yes, you are correct, except that you are using $X_i$ in the place of $T_i$.

If $\{X_t,t \geq 0\}$, is a Poisson process with intensity parameter $\lambda$, then the inter-arrival times $T_{1},T_{2},\cdots$ are are iid exponential random variables with mean $1/\lambda$.

If $W_n$ denotes the waiting time for the $n^{th}$ event to occur, then $$W_{n}=T_{1}+T_{2}+\cdots +T_{n}$$.

Then, we have, $W_{n}>t \ \ \Leftrightarrow \ \ X_t<n$.

$$P\{W_{n}>t\}=P\{X_t<n\} = \sum_{x=0}^{n-1}\dfrac{(\lambda t)^{x}}{x!}e^{\lambda t}$$

Jean Marie
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Lella
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