Re: Question about "adding" pdf's

From: Markus Triska (triska_at_gmx.at)
Date: 06/28/04 

Date: Mon, 28 Jun 2004 00:02:30 +0000

>
> So, Integrate[f(x)f(t-x)dx]. Sure, the the first pdf is 0 from
> -Infinity to 0, but what about the second one? How do you enforce the
> fact that for (t-x) < 1, it equals 0? Where and how is this done?
>

Since from -\infty < x < 1, f(x) is 0, it suffices to integrate from 1
(changing countably and therfore also finitely many function values does
not change the value of the definite integral, you can therefore really
integrate from 1, and not from 1 - 0). Conversely, f(t - x) is zero for
t - x < 1, and therefore for t < 1+x. It therefore suffices to integrate
up to 1+x (same argument as above). You end up integrating from 1 to
1+x. That's all.