# CDF for f(x) = 0.5e^-|x|

This is the full question: "If a random variable has density f(x)= 0.5e^-|x|, for x∈R, find the cumulative distribution function".

I know that to find cdf from the pdf you would have to integrate from the lower bound to x, but what would be the lower bound in this case? negative infinity? so the integration would just be from negative infinity to x?

• Yes. When the support is not mentioned, I'd assume that it is all the space. Other than that - this is actually a known and useful distribution. Note that questions that arise from homework and coursework should be marked as self-study – tmrlvi Jul 31 '19 at 22:12
• ($x \in R)$ = ($-\infty < x < \infty)$ – user158565 Jul 31 '19 at 22:25
• The definition of the CDF of $X$ is that its value at any number $x$ is the chance $X\le x.$ Therefore the region of integration of the density must be the set of numbers $\{y\mid y \le x\}.$ If the lower bound were a finite number $a,$ say, you wouldn't get the correct answer because $\Pr(X\le a)$ is never zero. – whuber Jul 31 '19 at 22:31

One way to verify the support (bounds) is to show that $$f(x)$$ integrates to 1 on them. As an example:
$$\int_0^\infty 0.5e^{-|x|}dx=0.5\int_0^\infty e^{-x}dx=0.5(-e^{-\infty}+e^0)=0.5\neq 1.$$
So clearly $$[0,\infty)$$ is not the support. You can verify that $$(-\infty,\infty)$$ is.
• There's a subtle lapse in the logic here, which is revealed by considering the function $f(x)=e^{|x|}(1/2 + x + |x| + \sin(x))/3.$ It, too, integrates to unity over $(-\infty,\infty),$ but--as you can readily check--$f$ is not a valid density, even though it could be used as a density over $[0,\infty).$ – whuber Jul 31 '19 at 22:39
• @whuber: obviously the region on which $f$ is defined has to require that $f$ is non negative. Yet, a common beginner pitfall is where one forgets that $f$ is zero outside of its support, for example when finding the CDF of an exponential random variable. – Alex R. Jul 31 '19 at 23:49