This is the full question: "If a random variable has density $f(x)= 0.5e^{-|x|}$, for $x\in R$, 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? $-\infty$? so the integration would just be from $-\infty$ to $x$?

  • $\begingroup$ 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 $\endgroup$
    – tmrlvi
    Jul 31 '19 at 22:12
  • $\begingroup$ ($x \in R)$ = ($-\infty < x < \infty)$ $\endgroup$
    – user158565
    Jul 31 '19 at 22:25
  • $\begingroup$ 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. $\endgroup$
    – 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.

  • 1
    $\begingroup$ 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 $[0,\infty).$ $\endgroup$
    – whuber
    Jul 31 '19 at 22:39
  • 2
    $\begingroup$ @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. $\endgroup$
    – Alex R.
    Jul 31 '19 at 23:49
  • 1
    $\begingroup$ That's exactly right. I just wanted to make those points, however obvious they may be to experienced people, a little more explicit. $\endgroup$
    – whuber
    Aug 1 '19 at 12:51

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