My question is coming in the end (that part about convolution).
I have independent $X_1,X_2,...,X_n$ that is assumed to be gamma($a,b$)-distributed, where a (shape) and b (rate). I want to find the distribution (cdf) of $-(c_1X_1+...+c_nX_n)$, where all c's is positive.
I know that $c_iX_i \sim \text{gamma}(a,\frac{b}{c_i})$ for $i\in 1,...,n$. If all the c's are equal then we will have the same rateparamter, and hence we can find the distribution $(c_1X_1+...+c_nX_n)$ by summing the shape paramteres to obtain
$(c_1X_1+...+c_nX_n) \sim$ gamma($\sum_{i=1}^{n}a,\frac{b}{c_i}$) = gamma($na,n\cdot\frac{b}{c_1}$) , where the last equality follows by all the c's are equal.
I know that it is a special case that all the c's are equal to each other, so I also want to handle it in the case where they are not equal.
I found this from a paper:
If $X_1,X_2,...,X_n$ belong to a class that is closed under convolution, the distribution of $-(c_1X_1+...+c_nX_n)$ is similar to the individual factor $X_i$, where Gamma-distribution is an example of that.
My question: Why is the gamma-distribution closed under convolution and why is the distribution of $-(c_1X_1+...+c_nX_n)$ then similar to the distribution of one the X's (for instance $X_1$)?