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Falling factorial, $(x)_n$, is the product of biggest $n$ terms in factorial, $(x)_n = x(x-1)(x-2)\cdot \ldots \cdot (x-n+1)$. Or the number of ways to color the set of $n$ objects into different colors if you have $x$ possible colors.

Using formula for probabilities of Poisson random variable $X$ one can find that $E((X)_n)=\lambda^n$. Derivation with $\lambda=1$. One may also use probability generating function. But I guess the answer is too beautiful to obtain it by boring summation :)

Starting from three axioms one may find $E(X)=\lambda$ without calculating probabilities. Just divide the interval $[0;1]$ into $n$ parts and make $n$ goes to infinity.

I wonder whether it is possible to find all the expected values $E((X)_n)$ directly from the defining axioms of Poisson process (without derivation of probabilities)???

not a homework — just curiosity :)

parsiad
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Roah
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    In your argument for $E(X)=\lambda,$ one can write $E(X)=nE(X_1)$ but how do you get $E(X_1)=\lambda/n+o(1/n)$? The axioms only say $\mathbb P(X_1=1)=\lambda/n+o(1/n)$ and $\mathbb P(X_1>1)=o(1/n).$ – Dap Feb 25 '19 at 06:51
  • The value of $E((X)n)$ is impossible to find from the axioms of the Poisson process, since $(X)_n$ is not a Poisson random variable. It takes value $0$ with the probability $\sum{i=0}^{n-1}{\lambda^i \over i!}e^{-\lambda}$ and values ${(n+j)! \over j!}$, $j\ge n$, with the probability ${\lambda^j \over j!}e^{-\lambda}$. – rrv Mar 02 '19 at 10:26

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