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I recently came across the following lemma while learning harmonic analysis, but don't know how to prove it by using analytical methods.

Lemma: For all $t \geq 0, x>0$, we have that \begin{equation} \lim _{\lambda \rightarrow \infty} e^{-\lambda t} \sum_{k \leq \lambda x} \frac{(\lambda t)^{k}}{k !}=\mathbf{1}_{[0, x)}(t)+\frac{1}{2} \mathbf{1}_{\{x\}}(t), \end{equation} where $\mathbf{1}_{[0, x)}(t)$ is the indicator function on set $[0, x)$.

If anyone can provide the proof, I would like to thank you here in advance.

xiuhua
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1 Answers1

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To a probabilist this is an easy (first-semester) exercise, but the probability facts used need to be translated into classical analysis terms.

Let $X_\mu$ denote a Poisson random variable with parameter $\mu$, so that $P(X_\mu=k)=e^{-\mu} \mu^n/n!$. The question posed, is to evaluate $\lim_{\lambda\to\infty}P(X_{\lambda t}\le \lambda x)$. It is a simple consequence of the weak law of large numbers that the random variable $X_\mu/\mu$ converges in distribution to the constant 1, as $\mu\to\infty$; this implies that $\lim_{\lambda\to\infty}P(X_{\lambda t}\le\lambda x)=0$ if $t>x$ and equals $1$ if $t<x$. The case $t=x$ is covered by the central limit theorem: the random variable $(X_\mu-\mu)/\sqrt\mu$ converges in distribution to a standard Gaussian. This means that for each $s$, $\lim_{\mu\to\infty}P( (X_\mu-\mu)/\sqrt \mu \le s)=\Phi(s)$, where $\Phi$ is the c.d.f. of a standard Gaussian r.v.: the choice $x=t$ corresponds to $s=0$. In this case the limit is $1/2$.

kimchi lover
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