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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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    How large are your $r$ and $\lambda$ ?2015-12-14

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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    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.2012-07-15
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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.