How to Differentiate CDF of Gamma Distribution to Obtain PDF

Question

CDF of a gamma distribution ($X \sim \mathcal{G}(n, \lambda)$) looks like

$$F(x) = \frac{\Gamma_x(n)}{\Gamma(n)}$$

Where $\Gamma_x(n) = \int_0^x t^{n-1} e^{-t} \, dt$ the incomplete gamma function. Ok so far?

But how do I differentiate such an expression?

$$\frac{d}{dx} \frac{\int_0^x t^{n-1} e^{-t} \, dt}{\int_0^\infty t^{n-1} e^{-t} \, dt}$$

UPDATE

With help from @MhenniBenghorbal, I have gotten:

$$\frac{d}{dx} \frac{\int_0^x t^{n-1} e^{-t} \, dt}{\int_0^\infty t^{n-1} e^{-t} \, dt} = \frac{1}{\Gamma(n)} x^{n-1}e^{-\lambda t}$$

but then its missing the $\lambda^n$ term of the original PDF of a gamma distribution. How do I get that back? I must have overlooked something? But I can see where …

Answer 1

Here is how you apply it

$$ \frac{d}{dx}\int_{0}^{x}t^{n-1}e^{-t}dt= x^{n-1}e^{-x}. $$

Note: In your case you the fundamental theorem of calculus is enough

Theorem: Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by $$F(x) = \int_a^x\!f(t)\, dt.$$ Then, $F$ is continuous on $[a, b]$, differentiable on the open interval (a, b), and $$F'(x) = f(x)\,$$ for all $x$ in $(a, b)$.