A doubt about 'convergence in distribution implies convergence in probability generating function'

Question

Suppose $X$ and $(X_n)_{n\ge1}$ are discrete random variables having non-negative integer values such that $$ \lim_{n\to \infty}P(X_n=k)=P(X=k)$$ holds for each $k\ge0.$ And suppose PGF of $X$ is $f$ and that of $(X_n)_{n\ge1}$ are $(f_n)_{n\ge1}.$ Then I am supposed to prove that $$ \lim_{n\to \infty}f_n(s)=f(s)$$ for any $s\in [-1,1].$ Using standard analysis, I can deal with the case $|s|<1$ and the case $s=1$ is trivial. But I am painfully stuck for $s=(-1).$ How to show $$\lim_{n\to\infty} f_n(-1)=f(-1) ?$$

Answer 1

What you need is $E(-1)^{X_n} \to E(-1)^{X}$. This is same as $P\{X_n$even$\}$-$P\{X_n$odd$\}$ $\to$ $P\{X$even$\}$-$P\{X$odd$\}$. It is enough to show that $P\{X_n$even$\}$ $\to$ $P\{X$even$\}$. Given $\epsilon >0$ choose $N$ such that $P\{X>N\}<\epsilon$. Since $P\{X_n=k\} \to P\{X=k\}$ for eack $k \leq N$ we get $P\{X_n \leq N\} \to P\{X \leq N\}$. Hence $P\{X_n > N\} \to P\{X > N\} < \epsilon$. It follows that $P\{X_n > N\} <\epsilon $ for $n$ sufficiently large, say $n \geq m$. Now, for $n \geq m$, (denoting sum over even values of $i$ by $\sum') $$|P\{X_n \in \{0,2,4,...\}\}-P\{X \in \{0,2,4,...\}\}| \leq \sum_{i=1}^{N}' |P\{X_n=i\}-P\{X=i\}|+2\epsilon$ since $|\sum_{i=N+1}^{\infty}' P\{X_n=i\}-P\{X=i\}|=|P\{X_n >N\}-P\{X>N\}| <P\{X_n>N\}+P\{X>N\}<2\epsilon$. Since the finite sum here also tends to 0 as $n \to \infty$ the proof is complete.