10
$\begingroup$

Why "a sum of two absolutely-continuous random variables does not need to be absolutely continuous"? See problem 6.4 on page 6 in https://web.ma.utexas.edu/users/gordanz/notes/basic_probability.pdf

$\endgroup$
18
$\begingroup$

Let $X$ be a standard normal random variable, and let $Y = -X$ (pointwise). Then both are a.c., but $X+Y$ is $0$ everywhere.

$\endgroup$
5
  • 2
    $\begingroup$ Thanks. How to formally argue that degenerate random variable $Z=0$ is not absolutely continuous? Can I say that as a random variable being absolutely continuous implies that it is continuous, thus $Pr(Z=z)$ should be 0 for any $z$ but this is not satisfied here when $z=0$? $\endgroup$ – T34driver Jan 7 at 22:28
  • 3
    $\begingroup$ $Pr$ is a function that assigns a probability for all events in the event space. As such, you can use your knowledge of a.c. functions. Here $Pr(Z = 0) = 1$ and $Pr(Z = z) = 0$ for any $z \neq 0$. Can you use this? $\endgroup$ – Therkel Jan 8 at 8:25
  • $\begingroup$ +1 to Therkel. Two more comments: not all sets have measure, and it’s helpful to introduce notation for the two measures. The second might help you to apply the definition of a.c. $\endgroup$ – Taylor Jan 8 at 15:44
  • 1
    $\begingroup$ @T34driver Absolutely-continuous random variables on $\mathbb R$ have a probability density. $Z$ with $\mathbb P(Z=0)=1$ does not $\endgroup$ – Henry Jan 8 at 16:36
  • 1
    $\begingroup$ @t34driver Absolutely continuous means dominated by Lebesgue measure. The measure of $\{0\}$ in Lebesgue measure is 0. So any measure dominated must also give a measure of 0 on $\{0\}$, which I hope it is clear that $\mu_Z$ does not. $\endgroup$ – Yakk Jan 9 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.