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Though the question here (Partial sums of exponential series - Stack Exchange) is similar, it is more specialized and I rather need a general approximation for an arbitrary partial sum.

Essentially, I am trying to approximate the probability mass function of a particular random variable and I ended up with a Poisson random variable's CDF in the mix. Hence, for my purpose, I need to figure out a reasonable approximation of the sum:

$\displaystyle\sum_{k = 0}^{r} \frac{\lambda^k}{k!}$ OR the tail, i.e. $\displaystyle\sum_{k = r}^{\infty} \frac{\lambda^k}{k!}$

Does someone know some approximations for this? Also, if there exist conditions for those approximations to be valid, I'd like to know them as well.

Thanks in advance!

Addendum: There appears to be a closed form expression for such a partial sum:

$\displaystyle\sum_{k = 0}^{r} \frac{\lambda^k}{k!} = e^\lambda \frac{\Gamma(r + 1, \lambda)}{\Gamma(r + 1)}$,

where $\Gamma(a, x)$ is defined as: $\displaystyle \Gamma(a, x) = \int_x^\infty t^{a - 1} e^{-t} \,dt$ and $\displaystyle \Gamma(a) = \Gamma(a, 0)$.

Is there a simple closed form approximation for the Gamma function? At the end of the day, somehow or the other, I either end up with a summation sign or an integral. I just want to be able to pin down this partial sum as a numeric quantity, that is reasonably approximate.

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    $\begingroup$ What language/environment are you using for computations? Chances are someone has already implemented the Gamma function more efficiently and better than you could in a few hours. Unless the approximation is central to the research you're attempting to perform, use a library. For instance, in C++: boost.org/doc/libs/1_42_0/libs/math/doc/sf_and_dist/html/… $\endgroup$
    – dls
    Jul 15, 2012 at 21:28
  • $\begingroup$ The idea is to not use exact values but come up with approximations that are suitable for analytical interpretation. I am attempting to mathematically model a system that has been implemented in real life. In order for the model to be of value, I need to make it as simple as possible. Therefore, to reduce complexity, I am using the mean values rather than the whole PMF, justified by ideas such as mean dynamics to model the dynamics of the system. To be clear, I am asking for an analytical approximation of aforementioned partial sum. Thanks for the response however (May help me in the future) $\endgroup$
    – lite-whowantstoknow
    Jul 15, 2012 at 21:39
  • $\begingroup$ Gotcha, I misinterpreted your phrase 'numeric quantity'. $\endgroup$
    – dls
    Jul 15, 2012 at 23:07
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    $\begingroup$ How large are your $r$ and $\lambda$ ? $\endgroup$
    – user65203
    Dec 14, 2015 at 14:17

2 Answers 2

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Use Taylor's series with remainder. We know that $$ e^\lambda = \sum_{k=0}^r \frac{\lambda^k}{k!} + \frac{e^{c\lambda}\lambda^{r+1}}{(r+1)!}, $$ for some $c \in [0,1]$. Therefore $$ \frac{\lambda^{r+1}}{(r+1)!} \leq e^\lambda - \sum_{k=0}^r \frac{\lambda^k}{k!} \leq e^\lambda \frac{\lambda^{r+1}}{(r+1)!}. $$ You can also get these estimates using more elementary means: $$ \sum_{k=r+1}^\infty \frac{\lambda^k}{k!} = \frac{\lambda^{r+1}}{(r+1)!} \left[ 1 + \frac{\lambda}{r+2} + \frac{\lambda^2}{(r+2)(r+3)} + \cdots \right] < \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \frac{\lambda^t}{t!} = \frac{\lambda^{r+1}}{(r+1)!} e^\lambda. $$ We can get a different upper bound by comparison to a geometric series, when $\lambda < r+2$: $$ \sum_{k=r+1}^\infty \frac{\lambda^k}{k!} \leq \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \left(\frac{\lambda}{r+2}\right)^t = \frac{\lambda^{r+1}}{(r+1)!} \frac{r+2}{r+2-\lambda}. $$

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  • $\begingroup$ Thank you for giving me an approximation AND a starting point to think in the right direction. Hopefully this approximation in itself is good enough for my purposes. At least, now I know which direction to look for answers. Thanks again! $\endgroup$
    – lite-whowantstoknow
    Jul 15, 2012 at 21:42
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    $\begingroup$ Edit: The upper bound didn't seem to be all that tight! Lower bound was better, but still needed tweaking. I then realized I could use the quantity $c$ in $\displaystyle e^{c\lambda}$ as a parameter that can be used for fitting the analytically computed mean value (obtained from the accurate PMF). It seems to work perfectly for a well chosen, small value of $c$. I am quite glad that now, my analysis seems systematic and not ad-hoc as it was earlier! Thanks for the help by the way. $\endgroup$
    – lite-whowantstoknow
    Jul 15, 2012 at 22:38
  • $\begingroup$ How do you get the reminder of the exponential Taylor series approximation? You write $$e^{c\lambda} \frac{ \lambda^{r+1} }{(r+1)!}$$ which can be no other than the reminder term. How do you get to that? $\endgroup$
    – Dr Potato
    Apr 22, 2020 at 20:43
  • $\begingroup$ thanks for the answer! Is there any meaningful way to pick c as a function of r? That is, can one justify for example the choice c=1/sqrt(r)? Otherwise the bound becomes very lose. Thanks :) $\endgroup$
    – Jonasson
    Nov 19, 2020 at 14:12
  • $\begingroup$ Taylor's theorem only guarantees the existence of some $c$. If you want more accuracy, take more terms in the series. $\endgroup$
    – Yuval Filmus
    Nov 19, 2020 at 14:13
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matlab has a built-in function gammainc($\lambda, a$, 'upper') = ${\Gamma(a, \lambda)\over \Gamma(\lambda)}$.

Also see "A computational procedure for incomplete Gamma functions" by Walter Gautschi.

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