Suppose I have two random variables $X_i, i=1,2$ distributed on open subsets $U_i$ of a unit ball around $0$ in $\mathbb{R}^d$. Suppose $0\in U_i$ for every $i$. I assume that distribution of each $X_i$ has density $\rho_i$ which is positive ans smooth on $U_i$.

  1. If $X_i$ are independent what would be a simple argument allowing to prove that $X_1+X_2$ has positive density on $U_1+U_2$?
  2. If $X_i$ are not independent, does the previous question make sense? I.e. does $X_1+X_2$ always has positive density on $U_1+U_2$ or at least on an open subset?
  • $\begingroup$ "around in ." ? "for every ." ? $\endgroup$ – user198044 Mar 25 '15 at 16:44
  • $\begingroup$ @Jack What do you mean? "around 0" and "for every i". It looks differently in your web-browser? $\endgroup$ – demitau Mar 25 '15 at 16:48
  • $\begingroup$ Oh they were missing. Now I see them. $\endgroup$ – user198044 Mar 25 '15 at 17:00

Hint: For the first part, use the convolution formula for the pdf of $X_1+X_2$: $\rho_{X_1+X_2}(x)=\int_{\mathbb R^d} \rho_1(t)\rho_2(x-t)\ dt$. For the second part, consider the case where $X_2=-X_1$.

  • $\begingroup$ In the second part if I know in addition that $|X_2|<|X_1|$ is it still possible to construct a similar example? $\endgroup$ – demitau Mar 25 '15 at 22:53
  • $\begingroup$ Yes, for example let $d=1$, let $X_2$ be uniform on [-1,1] (and by modifying $X_2$ on a null set, assume that $X_2$ is never 0), and define $X_1$ as follows: let $A$ be the set $\{\frac{\pm 1}{2^n} : n=1,2,\dots\}$, and set $X_1=\text{sgn}(X_2)\sup\{a\in A : |a|< X_2\}-X_2$; i.e., we round $X_2$ toward zero to the nearest power of $1/2$ (up to sign) and let $X_1$ be the difference. Then $|X_2|<|X_1|$, but $X_1+X_2$ has a discrete distribution, hence does not have a Lebesgue pdf. $\endgroup$ – Brent Kerby Mar 25 '15 at 23:59
  • $\begingroup$ Technically this example does not satisfy the requirement that $X_1$ have a smooth pdf (it has a pdf, but it is discontinuous), but you should be able to modify the definition of $X_1$ to recover this property if desired, say, by using a smooth bump function. $\endgroup$ – Brent Kerby Mar 26 '15 at 0:00
  • $\begingroup$ Your last comment I have not really understood. If I smooth out $X_1$ with a bump function, why the distribution of $X_1+X_2$ still be discrete? $\endgroup$ – demitau Mar 26 '15 at 0:12
  • $\begingroup$ It would not still be discrete, but you could arrange for it to a mixture between a discrete and continuous distribution; hence it would not have a Lebesgue pdf. $\endgroup$ – Brent Kerby Mar 26 '15 at 0:37

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.