Let $$T \sim Gamma(\alpha, \lambda)$$ $$f(t) = \frac1{ \Gamma(\alpha)}{\lambda^\alpha}t^{\alpha-1}{e^{-\lambda t}} \qquad t,\alpha,\lambda > 0$$

The CDF result : $$F(t) = 1 - \sum_{i=0}^{\alpha-1}{\frac {(\lambda t)^i} {i!}e^{-\lambda t}}, \qquad t,\alpha,\lambda > 0$$


$$F(t) =e^{-\lambda t}\sum_{i=0}^{\alpha-1}{\frac {(\lambda t)^i} {i!}}, \qquad t,\alpha,\lambda > 0$$


This discrete summation works only for integer-valued $\alpha$, and there's a reason to that. That reason is itself the perspective of what the identity expresses (it's an expression for real), rather than merely an algebraic manipulation.

$F(t) = P(T_{\alpha} < t)$ is the probability that $\alpha$ occurrence of the underlying Poisson process takes place before time $t$.

Here I attached the subscript $\alpha$ for $T$ as an index to emphasize the duality that will be clear soon.

In other words, up to time $t$, there are at least $\alpha$ occurrences. Therefore, consider the random variable $X_t$ that follows the discrete Poisson distribution (counts of occurrences).

$$P( X_t = i) = e^{-\lambda t} \frac{ (\lambda t)^i }{ i !}$$

It is subscripted (indexed) by $t$ to indicate that it is over time interval $[0, t]$, with $E[X_t] = \lambda t$. Namely, $(\lambda t)$ as a whole is the "parameter" of the Poisson distribution.

$$F_T(t) \equiv P(T_{\alpha} < t) = P( X_t \geq \alpha) \\ \implies F_T(t) = 1 - P( X_t < \alpha)$$

which is the desired expression.

  • $\begingroup$ Thanks a lot , very helpful now to understand first time i thought they used a definite integration for exponential to proof it. $\endgroup$ – Andray Jamil Almakhadmeh Oct 24 '18 at 18:49

It is the series expansion of the CDF. For $T \sim \text{Gamma}(a,λ)$, the standard CDF is the regularized Gamma $Γ$ function :

$$F(x;a,λ) = \int_0^x f(u;a,λ)\mathrm{d}u= \int_0^x \frac1{ \Gamma(a)}{\lambda^a}t^{a-1}{e^{-\lambda u}}\mathrm{d}u = \frac{γ(a,λx)}{Γ(α)}$$

where $γ$ is the lower incomplete gamma function.

If α is a positive integer (i.e., the distribution is an Erlang distribution), the cumulative distribution function has the following series expansion:

$$F(x;a,λ) = 1 - \sum_{i=0}^{a-1}{\frac {(\lambda x)^i} {i!}e^{-\lambda x}} = e^{-\lambda t}\sum_{i=0}^{a-1}{\frac {(\lambda x)^i} {i!}}$$

  • $\begingroup$ Thanks a lot, actually i read all about of this before , and more suitable to me that be the CDF is incomplete Gamma function divided by gamma function . My ask about how did they got this series (mathematically proof ) . $\endgroup$ – Andray Jamil Almakhadmeh Oct 24 '18 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.