# Why the sum of two absolutely-continuous random variables isn't necessarily absolutely continuous?

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

## 1 Answer

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

• 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 $z=0$? – T34driver Jan 7 at 22:28
• $Pr$ 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? – Therkel Jan 8 at 8:25
• +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. – Taylor Jan 8 at 15:44
• @T34driver Absolutely-continuous random variables on $\mathbb R$ have a probability density. $Z$ with $\mathbb P(Z=0)=1$ does not – Henry Jan 8 at 16:36
• @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. – Yakk Jan 9 at 0:07