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Let $\{X_n\}_{n\in \mathbb{N}}$ be a sequence of nonnegative discrete random variables, and $X$ be a nonnegative discrete random variable. The probability generating function $\psi_X(z)=\sum_{k=0}^\infty z^k \mathbb{P}(X=k)$ is convergent when $z \leq 1$.

Now assume that $\psi_{X_n}(z) \to \psi_X(z)$ for all $z\leq 1$. Does this implies that $X_n\to X$ in distribution?

I googled about this topic, and found an article,

http://ferrari.dmat.fct.unl.pt/personal/mle/PUBL-rdf/GPGF04VII19.pdf

Here, theorem 2.10 says that if we can find a neighborhood of 1 such that $\psi_{X_n}(z)\to \psi_{X}(z)$ for all $z$ in that neighborhood, $X_n\to X$ in distribution. However, in my case, $\psi_X(z)$ is convergent when $z\leq 1$, so we cannot find a neighborhood of 1 satisfy the condition of theorem 2.10. Does this mean that $X_n$ may not converge to $X$ even if the PGF converges?

Thanks in advance.

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  • $\begingroup$ Yes, this follows, because $\psi_n\to\psi$ locally uniformly on $|z|<1$, so the derivatives at $z=0$ converge. (You don't say this very explicitly, but I'm assuming your rv's take values in $0,1,2, \ldots$.) $\endgroup$
    – Christian Remling
    Commented Jan 28, 2017 at 18:08
  • $\begingroup$ You're right, they take values in 0.1,2,... Thank you very much! $\endgroup$
    – user3141978
    Commented Jan 29, 2017 at 7:31
  • $\begingroup$ because $\psi_n\to\psi$ locally uniformly on $|z|<1$, so the derivatives at $z=0$ converge. -> why this is true? Would be appreciated for references. $\endgroup$
    – user3141978
    Commented Feb 15, 2017 at 17:25

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