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Problem:

I am self-studying measure theory, Lebesgue integration, and related topics. I have come across this interesting example, which I "know" to be a mixture distribution, but because I have little experience in proofs, I am not sure how to show the result rigorously.

Define:

$F(x)=\begin{cases} 2-3^{-x}-2^{\left\lfloor{x}\right\rfloor} & \text{ if } x\ge 0 \\ 0 & \text{ if } x< 0 \end{cases}$

Show that $F(x)$ is a mixture of discrete and absolutely continuous distributions.

Then, evaluate:

$\int_{\mathbb{R}}e^{-x}F(dx)$ and $\int_{\mathbb{R}}xF(dx)$

Any help is always appreciated. Thanks!

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    [This](http://translate.google.com/?tab=wT#la/en/bis%20repetita).2012-07-29

2 Answers 2

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The CDF $F$ of the barycenter $p\mu+(1-p)\nu$ of an absolutely continuous distribution $\mu$ with density $f$ and of a discrete distribution $\nu$, is differentiable on its continuity set $C(F)$. Furthermore, $C(F)$ is at most co-countable and $pf=F'$ on $C(F)$. To get a density $f$ defined everywhere, one can define $f$ at will on $C(F)$ (recall that densities are only defined almost everywhere, anyway).

Thus $1-p=J(F)$, where $J(F)=\sum\limits_xF(x)-F(x-)$ is the sum of the jumps. If $p\lt1$, there are some jumps and the weight at jump $x$ is $\nu(\{x\})=\frac1{1-p}(F(x)-F(x-))$. Finally, $p=1-J(F)$ hence $p$, $\mu$ and $\nu$ are known.

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It looks like there's a typo in $F$. If F is supposed to be a CDF, then $F$ needs to have a limit of 1 as x goes to infinity.

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    I don't see where it says $F$ is a cdf; it could be a general (signed) distribution. If it is supposed to be a CDF it would probably need $\lfloor x \rfloor$ to be replaced ith $-\lfloor x \rfloor$ and it would need to be normalized.2012-07-24