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In the second volume of Feller's "An Introduction to Probability Theory and Its Applications", section VII.6 on Inversion formulas for Laplace transform starts with the limit

$\displaystyle{\lim_{\lambda\to\infty}e^{-\lambda\theta}\sum_{k\leq\lambda x}\frac{(\lambda\theta)^k}{k!}}$ or, in another form, $\displaystyle{\lim_{n\to\infty}e^{-n}\sum_{k=0}^{an}\frac{n^k}{k!}}$.

Feller says that the above limit is $0$ if $a<1$ and $1$ if $a>1$, and refers to previous chapters.

I am interested in direct computation of the limit (related question here does not seen illuminating in this aspect), without references to Central Limit theorem (like seen in other similar questions; usage of Poisson variables is possible).

StubbornAtom
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user35953
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    It provides the answer for the case $a=1$. It is not very helpful when $a\neq 1$. – user35953 Dec 17 '19 at 22:20
  • The overall approach should be the same. Start with $\int_n^\infty t^{an}e^{-t}~\mathrm dt$ and integrate by parts to relate to the sum. Then take the limit. – Simply Beautiful Art Dec 17 '19 at 22:31
  • If $a\gt 1$, series still behaves like $e^n$ for large n. – herb steinberg Dec 17 '19 at 22:59
  • The same can be said for the sum running only up to $n$, but the limit in that case turns out to be 1/2. – user35953 Dec 18 '19 at 01:13
  • And I wonder what is close-voters motivation. At least initially math.stackexchange was exactly for this kind of questions. It is much easier to rewrite definition of derivative in a few different ways, and those questions are not closed, despite obviously zero value... Perhaps it is because answering those make people feel very smart and knowledgeable, unlike some tough limits... – user35953 Dec 18 '19 at 01:18
  • math.se is still for this kind of question; in fact the closure for duplication, rather than as being outside the scope of the site, reinforces that notion. Anyway the motivation specified by the closers is literally written out in the notice before the question body. If you still object to the reason for closing, it would useful if you specified more precisely than you have why the answers to the putatively duplicated question do not adequately answer your question as written. – Travis Willse Dec 18 '19 at 06:14

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This is in a chapter called "Laws of Large Numbers. Applications in Analysis"

The CLT has very little to do with this problem. On the other hand the Strong (and Weak) Law of Large numbers has a lot to do with the limit.

What Feller has shown is an example of a 0-1 law. I'd suggest considering increasing sequences and positive integer valued $\lambda$.

You may interpret $X_{\lambda}$ as the sum of $\lambda$ iid poisson random variables, each having mean $\theta$, hence $X_{\lambda}$ is poisson with mean $\lambda \theta$.

Feller then says that the probability of the event
$\big \vert X_{\lambda} - \lambda \theta \big \vert \gt \lambda \epsilon $ is small for large enough $\lambda$

i.e.
$P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert \gt \lambda \epsilon\Big) \lt \epsilon' $
for any $\epsilon \gt 0$ and $\epsilon' \gt 0$ by selecting large enough $\lambda$

a more conventional form for this is $P\Big(\big \vert \frac{X_{\lambda}}{\lambda} - \theta \big \vert \gt \epsilon\Big) \lt \epsilon' $
which is something the SLLN tells you. (You could actually use WLLN here if you wanted.)

crude note: this implies that $P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert = \lambda \epsilon\Big)$ may be made arbitrarily small

But for any positive $x \lt \theta$
$e^{-\lambda \theta}\sum_{k\leq \lambda x} \frac{(\lambda \theta)^k}{k!} $
$= P\big(X_{\lambda}\leq \lambda x\big) $
$\leq P\Big( X_{\lambda} \leq \lambda (\theta -\epsilon)\Big) $ (by selecting small enough $\epsilon$, making use of monotone non-decreasing nature of CDFs)
$= P\Big(\lambda\epsilon \leq \lambda \theta - X_{\lambda})\Big) $
$= P\Big(\lambda \theta - X_{\lambda}\geq \lambda\epsilon \Big) $
$\leq P\Big(\lambda \theta - X_{\lambda}\geq \lambda\epsilon \Big) + P\Big(X_{\lambda} -\lambda \theta \geq \lambda\epsilon \Big) $
$= P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert \gt \lambda \epsilon\Big) + P\Big(\big \vert X_{\lambda} - \lambda \theta \big \vert = \lambda \epsilon\Big) $
$ \lt \frac{\epsilon'}{2} + \frac{\epsilon'}{2} $
$= \epsilon'$
where of course probabilities are bounded below by zero, and in this case bounded above by $\epsilon'$ which may be made arbitrarily small by selecting sufficiently large $\lambda$. This gives the zero, whenever $x \lt \theta$

A near identical argument on the complement gives the 1 when $\theta \lt x$

This sequential result then implies the desired result on tthe next page (233 of the second edition) where Feller shows, in effect, the injectivity of the laplace transform of a real non-negative random variables' distribution (with the usual caveats: assuming the transform exists, and equivalence up to a set of measure zero)

user8675309
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