Estimate for tail of power series of exponential function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:39:30Z http://mathoverflow.net/feeds/question/20113 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20113/estimate-for-tail-of-power-series-of-exponential-function Estimate for tail of power series of exponential function? Greg Martin 2010-04-01T23:15:07Z 2011-05-10T07:15:23Z <p>I would like to have an estimate for the infinite series $$\sum_{k=B}^\infty \frac{A^k}{k!},$$ where $A$ is a large positive quantity and $B$ is just a little bit bigger than $A$, namely, $B = A + C \sqrt A$ for some fixed large positive constant $C$. (In my application, $A$ and thus $B$ are increasing functions of some other variable, but $C$ really will stay fixed.)</p> <p>I expect that the answer should look something like $$?\ \sum_{k=B}^\infty \frac{A^k}{k!} \ll e^{-C^2/2} \ \ ?$$ uniformly in $A$, $B$, and $C$. (Possibly there should even be an asymptotic formula.) It would be great to be able to just quote such an estimate "off the shelf". I've only been able to find such estimates when $B$ is substantially larger than $A$, such as $B > 5A$.</p> <p>Does anyone know of a bound of this type in the literature? Many thanks.</p> http://mathoverflow.net/questions/20113/estimate-for-tail-of-power-series-of-exponential-function/20118#20118 Answer by Michael Lugo for Estimate for tail of power series of exponential function? Michael Lugo 2010-04-01T23:50:11Z 2010-04-02T14:13:45Z <p>Let's instead consider the sum</p> <p>$$\sum_{k = A + C \sqrt{A}}^\infty {e^{-A} A^k \over k!}$$ which of course differs from yours just by a factor of $e^{-A}$.</p> <p>Then this sum is the probability that a Poisson random variable of mean $A$ is at least $A + C\sqrt{A}$.</p> <p>A Poisson with mean $A$ has standard deviation $\sqrt{A}$, and as $A \to \infty$ the Poissons become asymptotically normal. So we have</p> <p>$$\sum_{k = A + C \sqrt{A}}^\infty {e^{-A} A^k \over k!} \to \Phi(C)$$</p> <p>as $A \to \infty$, where $\Phi$ is the CDF of the standard normal. So your sum is asymptotic to $e^A \Phi(C)$.</p> <p>Alternatively, if you'd like an explicit inequality, your sum can be bounded above by the geometric series with first term $A^B/B!$ and common term ratio $A/B$. Therefore, your sum is less than $${A^B \over B!} {1 \over 1-A/B}$$ and this can be rewritten as $${A^B \over B!} \left( 1 + {\sqrt{A} \over C} \right)$$ The product $A^B/B!$ is, as $A \to \infty$ with $B = A + C \sqrt{A}$, $${1 \over \sqrt{2\pi}} e^{-C^2/2} A^{-1/2} e^A (1+o(1))$$ by Stirling's formula. In the factor $1 + \sqrt{A}/C$ we can neglect $1$ as $A \to \infty$, so we get that </p> <p>$$\sum_{k = A+C\sqrt{A}}^\infty {e^{-A} A^k \over k!} \le {1 \over \sqrt{2\pi}} C e^{-C^2/2} e^A (1 + o(1))$$</p> <p>By, say, <a href="http://mathworld.wolfram.com/StirlingsApproximation.html" rel="nofollow">the double inequality (26) here</a> it should be possible to get explicit bounds.</p> http://mathoverflow.net/questions/20113/estimate-for-tail-of-power-series-of-exponential-function/20170#20170 Answer by Didier Piau for Estimate for tail of power series of exponential function? Didier Piau 2010-04-02T17:16:47Z 2010-04-05T16:50:11Z <p>The exponential inequality yields explicit bounds as follows. For every $B\ge A$, consider the series $$S(A,B)=\sum_{k=B}^{+\infty}\frac{A^k}{k!}.$$ Then, as Michael Lugo noticed, $\mathrm{e}^{-A}S(A,B)=P(N_A\ge B)$, where $N_A$ is a Poisson random variable with parameter $A$. </p> <p>For every positive $t$, $N_A\ge B$ if and only if $\mathrm{e}^{tN_A-tB}\ge1$, and for every nonnegative random variable $X$, $P(X\ge1)\le E(X)$. Using this for $X=\mathrm{e}^{tN_A-tB}$, one gets $$\mathrm{e}^{-A}S(A,B)\le E(X)=E(\mathrm{e}^{tN_A})\ \mathrm{e}^{-tB}.$$ To go further, one uses the fact that the Laplace transform $E(\mathrm{e}^{tN_A})$ of a Poisson distribution is $\mathrm{e}^{A(\mathrm{e}^t-1)}$ and one optimizes the upper bound with respect to $t\ge0$. That is, one plugs in the inequality the value of $t$ such that $\mathrm{e}^t=B/A$, which yields $$S(A,B)\le\mathrm{e}^{B-B\log(B/A)}.$$ Note that this upper bound is not asymptotic (which is nice) but that, in general, it is not optimal in the regime of the central limit theorem you are interested in (which is not so nice).</p> <p>Turning to the case where $A\to\infty$ and $B=A+C\sqrt{A}$ with $C>0$ fixed, the expansion of $\log(B/A)$ up to the order $o(1/A)$ yields $$S(A,A+C\sqrt{A})\le\mathrm{e}^{A-C^2/2+o(1)}=\mathrm{e}^{-C^2/2}\mathrm{e}^A(1+o(1)).$$ Which is a (very odd) way to check that $\Phi(C)\le\mathrm{e}^{-C^2/2}$...</p> <p><b>Edit</b> : Non asymptotic upper bound : $$S(A,A+C\sqrt{A})\le\mathrm{e}^{-C^2/2}\mathrm{e}^{A}\exp(C^3/(2\sqrt{A})).$$</p> http://mathoverflow.net/questions/20113/estimate-for-tail-of-power-series-of-exponential-function/64474#64474 Answer by danil bykov for Estimate for tail of power series of exponential function? danil bykov 2011-05-10T07:15:23Z 2011-05-10T07:15:23Z <p>Hi Michael. I like your solution of this task. Can I know who author of this idea (replacing of sum by probability)? Maybe you can specify some book or paper?</p>