# How did they get this proof for CDF of gamma distribution?

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

or

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

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.

• Thanks a lot , very helpful now to understand first time i thought they used a definite integration for exponential to proof it. – 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:

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

• 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 ) . – Andray Jamil Almakhadmeh Oct 24 '18 at 18:29