understanding the convolution of random variable formula

Question

Consider summing two iid exponential r.v. We know for a fact that this is Gamma distribution with $\alpha = 2, \beta = \theta$. However, when using the convolution formula $Z=X+Y$, we have $$ f(z)= \int_{-\infty}^\infty f(x)f(z-x)dx = \int_{0}^\infty f(x)f(z-x)dx $$ because $x\in(0,\infty)$ not $x \in (0,z)$, It leads to divergence and sums to infinity, what is the intuition behind integrating to $z$ and not infinity?

Answer 1

$f(x)$ has support on $x > 0$ as you correctly note.

Therefore $f(z-x) = 0$ whenever $z - x < 0 \iff z < x$

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UPDATE

From the definition of the exponential random variable, $f(y) = 0$ if $y<0$. Thus, letting $y = z-x$, we have $f(z-x) = 0$ if $z-x<0 \iff z < x \iff x > z$. Thus, the product $f(x)f(z-x)$ is nonzero only when $x > 0$ and $x < z$, which means we should only be integrating over $x \in (0,z)$, since the product vanishes elsewhere.