Suppose we have i.i.d. random variables $X_1, X_2$.
If $\text{Law}(X_1 + X_2)$ is absolutely continuous with respect to the Lebesgue measure $\lambda$, can we infer that each $X_i$ is absolutely continuous with respect to $\lambda$ also?
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@MichaelHardy Doesn't the convolution formula assume that $f_{X_1}$ and $f_{X_2}$ exist (equivalently that $X_1, X_2 \ll \lambda$)? I am trying to go the other way - essentially to infer that $f_{X_1}$ and $f_{X_2}$ exist from the fact $f_{X_1+X_2}$ does. – 12qu Jul 15 '16 at 14:53
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The claim is false. The second answer to this question describes the construction of a singular measure $\mu$ on $[-\pi,\pi]$ such that the convolution $\mu *\mu$ is absolutely continuous. So to find a counterexample to the claim, it suffices to take $X_1$ and $X_2$ iid with distribution $\mu$.
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