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The distribution of any infinitely divisible random variable is itself infinitely divisible. But this link says the converse is not always true. Can you explain?

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That a random variable is infinitely divisible means exactly that its distribution is infinitely divisible. And vice versa. So what is the question? –  Did Jun 23 '13 at 8:09
    
Then there is none. This link says otherwise, which had me confused. –  cimrg.joe Jun 23 '13 at 8:30
    
Thank you very much! Should I delete my "question"? –  cimrg.joe Jun 23 '13 at 8:45
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I'd be happy to have it explained. Voted to reopen. –  Stefan Hansen Jun 23 '13 at 9:42
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OP: I reformulated the question according to your explanations. Now you could ask on meta for its reopening. –  Did Jun 23 '13 at 11:18

1 Answer 1

That a random variable is infinitely divisible should mean exactly that its distribution is infinitely divisible.

The question originates from some assertions on the page about infinite divisibility of encyclopediaofmath. There the authors define a Poisson random variable $X$ on some probability space $\Omega$ and they note that, on this specific probability space $\Omega$, one cannot decompose $X$ as $X=X_1+X_2$ for some i.i.d. random variables $X_i$, nor as $X=X_1+\cdots+X_n$ for some i.i.d. random variables $X_i$, for any $n\geqslant2$.

Unsurprisingly, to observe this phenomenon, the authors define $X$ on the smallest possible probability space $(\Omega,\mathcal F,P)$, namely, they consider $\Omega=\{0,1,2,\ldots\}$, $\mathcal F=\mathcal P(\Omega)$, $P(\{k\})=\mathrm e^{-\lambda}\lambda^k/k!$ for every $k\geqslant0$, and $X:\Omega\to\mathbb R$ defined by $X(k)=k$. Then, roughly speaking, there is not enough space in $\Omega$ to define some Poisson $\frac12\lambda$ random variables hence the decomposition $X=X_1+X_2$ is impossible on this probability space.

This kind of conundrum is a reason why one usually says that a random variable $X$ on a probability space $\Omega$ is infinitely divisible if there exists some (possibly much larger) probability space $(\Omega',\mathcal F',P')$ and some random variables $X'$, $X'_1$ and $X'_2$ defined on $\Omega'$ such that $X'=X'_1+X'_2$, $X'_1$ and $X'_2$ are i.i.d., and the distributions of $X$ and $X'$ coincide (and likewise for every $n\geqslant2$ instead of $n=2$). Then a random variable is infinitely divisible if and only if its distribution is infinitely divisible.

Ultimately, all this confirms that infinite divisibility, to be useful, should concern distributions rather than random variables (and in fact I personally never encountered the notion used specifically for random variables).

Edit: Here is a proof that, if $0\lt\lambda\lt.9$, one cannot construct even a single Poisson$(\frac12\lambda)$ random variable $X_1$ on the minimal probability space used above to define the Poisson$(\lambda)$ random variable $X$. Let $p_n=P(X=n)$ and $q=P(X_1=0)$. Assume that $1-p_0\lt q$, then $[X=0]$ must be sent to $[X_1=0]$. Assume that $1-p_1\lt q$, then $[X=1]$ must be sent to $[X_1=0]$. Assume that $p_0+p_1\gt q$, then there is a contradiction since $[X\leqslant1]$ cannot be sent to $[X_1=0]$. Hence the construction is impossible if $1-\mathrm e^{-\lambda}\lt\mathrm e^{-\lambda/2}$, $1-\lambda\mathrm e^{-\lambda}\lt\mathrm e^{-\lambda/2}$ and $(1+\lambda)\mathrm e^{-\lambda}\gt\mathrm e^{-\lambda/2}$. These inequalities hold for every $\lambda$ in $(0,\lambda_*)$ where $\lambda_*$ is the unique positive root of $\mathrm e^{\lambda}-\mathrm e^{\lambda/2}=\lambda$, numerically $\lambda_*=.91$.

The problem of the simultaneous construction of $(X_1,X_2)$ on the minimal probability space used above to define the Poisson$(\lambda)$ random variable $X$ is even simpler since every random variable on this space is a function of $X$. But the only way that some random variables $u_1(X)$ and $u_2(X)$ are independent is that one of them (or both) is constant. This cannot happen if $u_1(X)$ and $u_2(X)$ are Poisson$(\frac12\lambda)$ hence $(X_1,X_2)$ do not exist.

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I guess, in $\Omega$ itself there is enough space to define two independent $\mathrm{Poi}(\frac12\lambda)$ random variables since $\Omega^2$ is isomorphic to $\Omega$, so the only problem may come from the choice of the probability measure $P$. Can we show that it is indeed impossible to define two independent $\mathrm{Poi}(\frac12\lambda)$ random variables on the probability space $(\Omega,\mathscr F,P)$ with $\Omega = \Bbb N_0$ and $P$ being $\mathrm{Poi}(\lambda)$ distribution? –  Ilya Jun 24 '13 at 8:37
    
Even being able to define one Poisson$(\frac12\lambda)$ would require a function $u_\lambda$ such that, if $X_\lambda$ is Poisson$(\lambda)$ then $u_\lambda(X_\lambda)$ is Poisson$(\frac12\lambda)$. When $\lambda$ and $\mathrm e^\lambda$ are algebraically independent, each $u_\lambda^{-1}(\{n\})$ must be infinite since $\mathrm e^{\lambda/2}=\sum\limits_k\lambda^{k-n}2^nn!/k!$ where the sum is over every $k$ in $u_\lambda^{-1}(\{n\})$, for each $n$. Not a proof but. –  Did Jun 24 '13 at 8:57
    
I do agree that level sets of $u_\lambda$ may have to be infinite - e.g. in the case you've described - so I thought that some modification of the Riemann rearrangement argument may be of help here. –  Ilya Jun 24 '13 at 9:10

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