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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?
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  • $\begingroup$ "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
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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 $X_2=-X_1$.

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  • $\begingroup$ 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

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