$\Bbb P(E_0+E_1>x)$ where $E_0,E_1$ are exponential random variables

Question

Let $E_0\sim$ Exp($\lambda$) and $E_1\sim$ Exp($1$) be independetn r.v. and let $\lambda>1$ and $x>0$.

I am not sure if the bounds of the following integral should be $(0,1)$ or $(0,\infty)$ $$\Bbb P(E_0+E_1>x)=\int\limits_0^1 \Bbb P(E_0>x-s)f_{E_1}(s)ds$$ where $f_{E_1}$ is the density function of $E_1$

Answer 1

It is $\int_x^{\infty} e^{-s}ds+\int_0^{x}\int_{x-s}^{\infty} \lambda e^{-\lambda x}dx e^{-s} ds$. [$s>x$ cannot be ignored in the computation].

Answer 2

The integral in theory should be over $(-\infty,\infty)$

In practice this reduces to an integral over $[0,x]$ plus another over $(x, \infty)$ since

• $f_{E_1}(s)=0$ for $s \lt 0$ and
• $\Bbb P(E_0>x-s) = 1$ for $s \gt x$