TITLE: Density of $X_1+X_2$ for $X_1, X_2$ iid Pareto(1,1)
QUESTION [0 upvotes]: Let $X_1,X_2$ be iid $\text{Pareto}(1,1)$ distributed. I want to determine the density of $X_1+X_2$
We defined the following density for $X\sim \text{Par}(\alpha,x_0)$:
$$f(x)=\frac{\alpha x_0^\alpha}{x^{\alpha+1}},\ \ \ x>x_0$$
$x_0>0$ and $\alpha>0$.

Is the density given by the solution of this integral:

$$h(z)=\int f_2(z-x)f_1(x)\,dx=\int_{x_0}^{z}\frac{1}{(z-x)^2}\frac{1}{x^2}\,dx\,?$$

REPLY [0 votes]: $X_1,X_2\sim f(x)=\frac1{x^2},x>1$
Let $Z=X_1+X_2,X=X_2\implies X_1=Z-X,X_2=X$.

$X_1>1\implies Z-X>1$
$X_2>1\implies X>1$

Thus you have $1<X<Z-1$.
The absolute value of $J\left(\frac{X_1,X_2}{Z,X}\right)$ is $1$.
Thus $f_{ZX}(z,x)=f_{X_2}(z-x)f_{X_1}(x)$.$$\implies f_Z(z)=\int_1^{z-1}f_2(z-x)f_1(x)dx$$
Because for a particular $z,x$ ranges from $1\to z-1$.
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