Continuous version of a Poisson R.V.
I am wondering if there is a continuous version of a Poisson random variable, that has the following two features:
1) Has a CDF that agrees with the discrete Poisson distribution on the integers, and 2) Has moments that agree with those of the Poisson distribution.
Thanks!
$\endgroup$ 93 Answers
$\begingroup$The moment generating function of a Poisson random variable $X$ with parameter $\lambda$ is $$ \mathbb{E}\left[e^{tX}\right]=\sum_{n=0}^\infty \frac{e^{-\lambda}e^{nt}\lambda^n}{n!}=e^{\lambda(e^t-1)} $$ which converges for all $t$, and in fact is the restriction to $\mathbb{R}$ of an entire function. In particular, the power series for the moment generating function centered at zero has a positive radius of convergence. Therefore any random variable with the same moments as $X$ must also be Poisson distributed with parameter $\lambda$.
$\endgroup$ 4 $\begingroup$Somewhat more generally, let $X$ be any random variable whose moment generating function $M(z) = \mathbb E[e^{tX}]$ is analytic in a neighbourhood of $0$. This says that the series $\sum_{j=0}^\infty \mathbb E[X^j] t^j/j! $ has positive radius of convergence, i.e. $|\mathbb E[X^j]| \le C D^j j!$ for some constants $C, D$. Then we have uniqueness in the Hamburger moment problem with these moments: there is no other finite (signed) measure on $\mathbb R$ with the same moments as $X$.
$\endgroup$ $\begingroup$Here's my take on answering the question using Did's response above. The nice thing about his proof is that it only relies on equality of the CDFs at the integers and equality of the first moments.
Let $X$ be Poisson with CDF $F_X$ and $Y$ be any random variable with CDF $F_Y$ such that $F_Y(n)=F_X(n)$ for every nonnegative integer n. Then, $F_X⩾F_Y$ everywhere. This implies that one can couple X and Y in such a way that $Pr(X⩽Y)=1$.
To see this, define consider a new random variable $Z$ that rests on the same probability space as $X$. Conditional on $X=k$, let $Z$ have the distribution of $Y|Y \in [k,k+1).$ Notice that $Pr(X⩽Z)=1.$ Moreover, $$F_Z(x)\\=\sum_{j=0}^{\lfloor{x} \rfloor-1}f_X(\lfloor{j} \rfloor)+f_X(\lfloor{x} \rfloor)*F_Z(x|X=\lfloor{x} \rfloor) \\=F_X(\lfloor{x} \rfloor)+f_X(\lfloor{x} \rfloor)*F_Y(x|Y \in[\lfloor{x} \rfloor,\lfloor{x} \rfloor+1)) \\=F_Y(\lfloor{x} \rfloor)+f_Y(\lfloor{x} \rfloor)*F_Y(x|Y \in[\lfloor{x} \rfloor,\lfloor{x} \rfloor+1)) \\=F_Y(\lfloor{x} \rfloor)+f_Y(\lfloor{x} \rfloor)* \frac{F_Y(x)-F_Y(\lfloor{x} \rfloor)}{f_Y(\lfloor{x} \rfloor)} \\=F_Y(x).$$
Moreover, $E(x)=E(y) (=E(z))$ implies that $Pr(Z=Y)=1.$ Thus, the answer to my question was 'No".
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