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


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

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    $\begingroup$ 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
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    $\begingroup$ $Pr$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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