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Let ($\Omega$,F,P) a probability space. Let $A,B$ Borel limitate subsets of $R^n$ with positive lebesgue measure $X: \Omega\rightarrow A$ and $Y:\Omega\rightarrow B$ two independent random variables with distributions absolutely continous with respect to Lebesgue measure of $R^n$: $D=|X-Y|$

Please help proving that $D$ is absolutely continous rispect lebesgue measure and express it in terms of the distributions of $X$ and $Y$.

Thank you

wiskundeliefhebber
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user62138
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    Welcome to Math.SE! Please, consider updating your question to include what you have tried / where you are getting stuck. You will find that people on this site will be significantly faster to help you if you do that; that way, we know exactly what help you need. – Did Aug 07 '13 at 10:26
  • I have tried to solve it by usin the partition function – user62138 Aug 07 '13 at 10:46
  • And what did it yield? Show your work! (Beware though that "partition function" does not mean what you think.) – Did Aug 07 '13 at 10:56
  • $F(t)=P[D\leq t]=P[|X-Y|\leq t]=P[y\in B_{(x,t)}]=\int_{R^n}(\int_{R^n}1_{B(x,t)}(y)g(y)dy)f(x)dx$ – user62138 Aug 07 '13 at 12:00
  • Why the CDF should help to establish the absolute continuity? – Did Aug 07 '13 at 12:37
  • the distribution is absolutely continous if it has a density which is the derivate of the partition function.I must prove that F(t) is $C^1$ – user62138 Aug 07 '13 at 12:40
  • Not $C^1$, since a priori the density is only integrable. But since your computation of the CDF seems stalled, you might try another characterization of AC... (Unrelated, what you call "partition function" is in fact the "cumulative distribution function".) – Did Aug 07 '13 at 12:48
  • but this characterization of AC is the unic that permits us to calculate then the density – user62138 Aug 07 '13 at 18:48
  • can anyone give me another characterization of AC – user62138 Aug 08 '13 at 08:20
  • http://en.wikipedia.org/wiki/Absolute_continuity#Absolute_continuity_of_measures – Did Aug 08 '13 at 08:27
  • but to express the distribution of D in terms of X and Y we must find its density – user62138 Aug 08 '13 at 08:33
  • No. See answer. – Did Aug 08 '13 at 08:49
  • Ok, but how can I now express the distribution of D in terms of the distributions of X and Y – user62138 Aug 08 '13 at 08:54
  • You might want to use the general approach. – Did Aug 08 '13 at 09:02
  • Can you give me a hint ? – user62138 Aug 08 '13 at 10:40
  • I just did... Did you read the answer I linked to? If you did, it should suggest an approach to your question. – Did Aug 08 '13 at 10:47
  • I red the answer but I don't understand how to modificate it in this case because $x,y\in R^n$ and we want $w=|x-y|\in R\quad$ z=? – user62138 Aug 09 '13 at 08:01

1 Answers1

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This is to debunk the claim, repeated several times by the OP, that, to show that some distribution is absolutely continuous, one must first compute its density.

Consider the case $n=1$. For $y$ in $\mathbb R$ and $C\subseteq\mathbb R$, let $C(y)=\{x\in\mathbb R\mid|x-y|\in C\}$, thus, $C(y)=(y+C)\cup(y-C)$. The Lebesgue measure is invariant by the translations and the reflections hence, for every measurable $C$, $\mathrm{Leb}(y+C)=\mathrm{Leb}(y-C)=\mathrm{Leb}(C)$ and $\mathrm{Leb}(C(y))\leqslant2\,\mathrm{Leb}(C)$. In particular, if $C$ is negligible, so is $C(y)$, for each $y$.

Now, let $C$ such that $\mathrm{Leb}(C)=0$. Then $P[D\in C]=P[X\in C(Y)]$ and, by independence, $$ P[X\in C(Y)]=\int_{\mathbb R} P[X\in C(y)]\,\mathrm dP_Y(y). $$ If the distribution of $X$ is absolutely continuous, $P[X\in C(y)]=0$ for every $y$ since $\mathrm{Leb}(C(y))=0$ for every $y$. Thus, $P[D\in C]=0$. This holds for every negligible set $C$ hence the distribution of $D$ is absolutely continuous.

Did
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