I am trying to understand the sum of equal Gamma distributions
Let $X_1,...,X_n$ be iid where all $X_i$ each follow a Gamma distribution with $shape=\lambda,scale=\beta$
Is the following then true?
$Y=\sum_{i=1}^n X_i$ follows a Gamma distribution with $shape=n\cdot \lambda,scale=\beta$
Yes, it is true.
For two independent r.v. $X_1 \sim Gamma(\lambda_1, \beta)$ and $X_2 \sim Gamma(\lambda_2, \beta)$,
$$ X_1 + X_2 \sim Gamma(\lambda_1 + \lambda_2, \beta). $$
Proof. Let $X_i\sim Gamma(\lambda_i, \beta)$, $i = 1, 2$, with p.d.f. $$ f_{X_i} = \frac{1}{\Gamma(\lambda_i)\beta^{\lambda_i}}x^{\lambda_i-1}e^{-\frac{x}{\beta}}. $$
$$ \begin{aligned} f_{X_1+X_2}(x) &= \int_{-\infty}^{+\infty}f_{X_1}(y)f_{X_2}(x-y)dy = \\ &= |f_{X_1}(y) \ne 0, \text{ for }y > 0, \text{ and }f_{X_2}(x-y) \ne 0, \text{ for }y < x| = \\ &= \int_{0}^{x}f_{X_1}(y)f_{X_2}(x-y)dy = \\ &= \int_{0}^{x}\frac{1}{\Gamma(\lambda_1)\beta^{\lambda_1}}y^{\lambda_1-1}e^{-\frac{y}{\beta}}\frac{1}{\Gamma(\lambda_2)\beta^{\lambda_2}}(x-y)^{\lambda_2-1}e^{-\frac{x-y}{\beta}}dy = \\ &= \frac{1}{\Gamma(\lambda_1)\Gamma(\lambda_2)\beta^{\lambda_1+\lambda_2}}e^{-\frac{x}{\beta}}\int_{0}^{x}y^{\lambda_1-1}(x-y)^{\lambda_2-1}dy = \\ &= |y = xt \Leftrightarrow dy = xdt| = \\ &= \frac{1}{\Gamma(\lambda_1)\Gamma(\lambda_2)\beta^{\lambda_1+\lambda_2}}x^{\lambda_1 + \lambda_2 - 1}e^{-\frac{x}{\beta}}\int_{0}^{1}t^{\lambda_1-1}(1-t)^{\lambda_2-1}dt = \\ &= \left|\frac{1}{\Gamma(\lambda_1)\Gamma(\lambda_2)}\int_{0}^{1}t^{\lambda_1-1}(1-t)^{\lambda_2-1}dy = \frac{1}{\Gamma(\lambda_1)\Gamma(\lambda_2)} Beta(\lambda_1, \lambda_2) = \frac{1}{\Gamma(\lambda_1+\lambda_2)}\right| = \\ &= \frac{1}{\Gamma(\lambda_1+\lambda_2)\beta^{\lambda_1+\lambda_2}}x^{\lambda_1 + \lambda_2 - 1}e^{-\frac{x}{\beta}}, \end{aligned} $$
which is a p.d.f. of $Gamma(\lambda_1 + \lambda_2, \beta)$.
If $X_1, X_2$ are i.i.d. ($\lambda_1 = \lambda_2 = \lambda$), then
$$ X_1 + X_2 \sim Gamma(2\lambda, \beta). $$
You can extend this for $n$ i.i.d. $Gamma(\lambda, \beta)$ random variables.
