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

Question

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

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

Answer 1

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.

Answer 2

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:

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