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I'm seeking theses asymptotic expansions :

$$ e^{-an}\sum_{k=n+1}^{\infty} \frac{(na)^k}{k!} $$

$$ e^{-an}\sum_{k=1}^{n-1} \frac{(na)^k}{k!} $$

in terms of n going to the infinity.

Thanks you for your help.

Fabian
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Pagode
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  • Have tried Taylor Laplace (just in order to get a limit or majoration with $ \mathcal{O} $ which is not the question there) . I've tried to reindex sum , to substract them. – Pagode May 22 '18 at 21:30
  • How much accuracy do you need? An estimate accurate to within a constant factor can be obtained using the Lagrange remainder. – Ian May 22 '18 at 22:19
  • I need first to show the substraction of the two sequence goes to zero. Then I would like to get first the exact limit of both quantities then an asymptotic expansion (a bit more précise). – Pagode May 23 '18 at 12:11
  • Furthermore , i don't see the efficiency of Lagrange remainder because of n in the sum even with reindex – Pagode May 23 '18 at 12:13
  • The $n$ in the sum doesn't actually matter, the only catch here is that you happen to be sending the argument of the exponential to infinity at the same time as you send the number of terms in the finite sum to infinity. But the statement still holds. – Ian May 23 '18 at 16:27
  • (Well, it does matter, but the way it matters depends on $a$...) – Ian May 23 '18 at 16:33
  • Relevant: https://math.stackexchange.com/q/160248/5531 – Antonio Vargas May 24 '18 at 00:07

1 Answers1

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I assume $a>0$.

Notice that your tail sum is the probability that a Poisson($an$) variable exceeds $n$. This is the probability that the average of $n$ Poisson($a$) variables exceeds $1$. For $a \neq 1$ the result of the limit comes from the weak law of large numbers: if $a<1$ then the result is $0$, if $a>1$ then the result is $1$. For $a=1$ you can use the central limit theorem: the average behaves asymptotically like $N(1,1/n)$ so the limit is $1/2$, from symmetry of the normal distribution (even though the underlying Poisson distribution isn't symmetric). This is highly non-obvious from the analysis point of view.

Since the Poisson distribution has all cumulants, you can try to use Edgeworth series to get higher order asymptotics.

Ian
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  • I have to calculate this limits from a exercise of probability. Thanks you for your answer but how I could deal with without central limit theorem ? I will use your advices for my personal exploration of the problems but I just need there more simple theory to expose it to persons who don't know about central limit theorem and deeper uses. – Pagode May 23 '18 at 18:13
  • @Pagode Honestly I am skeptical about how you would get that the leading order asymptotic in the $a=1$ case is $1/2$ in any other way. The other cases are really just the weak law of large numbers, not the central limit theorem, so they can be done using simpler estimates. – Ian May 23 '18 at 18:15