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

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 $$E_1$$

• Why do you think the bounds should be $0$ and $1$? – angryavian Jun 12 at 7:47
• @Henry But $s$ is the dummy variable – angryavian Jun 12 at 7:48
• Oh the right bound should be $x$ right? I realize now that $1$ doesn't make sense – John Cataldo Jun 12 at 7:51
• $P(E_0 >x-s)$ is not $0$ when $x-s$ is negative. It is $1$. So don't ignore $s >x$. – Kabo Murphy Jun 12 at 7:54
• – Math-fun Jun 12 at 8:01

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].

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 $$s \gt x$$
• Wrong answer. The probability in the last line is $1$, not $0$. – Kabo Murphy Jun 12 at 7:58
• @KaviRamaMurthy - Corrected - thank you – Henry Jun 12 at 8:01