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Let $(X_i)_{i\in\mathbb{N}}$ be a real valued stationary sequence where each $X_i$ has the same Lebesgue density. Does \begin{align*} \sum_{i=1}^n X_i\end{align*} have a Lebesgue density, too? I tried to show it with Radon-Nikodym, but it led to the following question: Let $E\subset \mathbb{R}^n$ be a set of Lebesgue measure $0$. Why has \begin{align*} \cup_{e\in E}H_e \end{align*} with $H_e:=\{\omega\in\Omega:\sum_{i=1}^n X_i(\omega) = e\}$ Lebesgue measure $0$? Which I couldn't answer quickly. Any ideas?

Thanks!

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    $\begingroup$ Isn't the density of the sum is just a convolution of the densities? I also think that you can just focus on the case $n=2$ since the general one follows by induction. $\endgroup$ – Ilya Jul 18 '13 at 8:17
  • $\begingroup$ I thought it's only the convolution if $X_1$ and $X_2$ are independent (in the case n=2). But I'll check this. Thanks for your quick answer! $\endgroup$ – stroem Jul 18 '13 at 8:40
  • $\begingroup$ But in case they are not independent shall not it be false? E.g. take any symmetric distribution, then $X_1$ and $X_2 = -X_1$ will have the same density, but $X_1+X_2 = 0$ which does not have a Lebesgue density. I'm not sure which role does stationarity play here, though. $\endgroup$ – Ilya Jul 18 '13 at 8:43
  • $\begingroup$ You're right. Would it help to assume, that the $X_i$ are centered and that the correctly normalized sum, let's say $v_n\sum_{i=1}^n X_i\xrightarrow{d}X $, (in distribution) and $X\sim N(0,1)$. Can we get a Lebesque density, let's say if $n$ is big enough? $\endgroup$ – stroem Jul 18 '13 at 10:00
  • $\begingroup$ with $v_n=\sqrt{\operatorname{Var}\left(\sum_{i=1}^n X_i\right)}$. $\endgroup$ – stroem Jul 18 '13 at 10:06
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Of course not. Try $X_n=(-1)^nX_0$, for some $X_0$ "with Lebesgue density" (whatever that means), then the distribution of each $\sum\limits_{k=1}^{2n}X_k=0$ has no density.

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