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  • I need to compute the probability of getting more than $x$ "successes" in a large number of trials $\left(\,10^{11}\,\right)$ of an event with a small probability $\left(\,10^{-7}\,\right)$.
  • Exact Binomial won't work, and the Poisson approximation does not seem appropriate.

Thanks.

Felix Marin
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Dmitri
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  • Why won't "exact binomial" work? – Bobson Dugnutt May 03 '17 at 21:15
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    Use Normal approximation – David Quinn May 03 '17 at 21:16
  • The Poisson approximation seems perfectly appropriate to me. Or in this case, since the quotient $10^{11}/10^7$ is so big, a normal approximation could serve. – Michael Hardy May 03 '17 at 21:20
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    Or use R: e.g. pbinom(9876, size=10^11, prob=10^-7, lower.tail=FALSE) gives about $0.8917494$. This compares with the normal approximation with continuity correction of being above $9876.5$ having a probability of about $0.8915848$ – Henry May 03 '17 at 21:26
  • @Henry - This is exactly what I needed. Thank you. If you make it into an answer, I will be happy to accept it. Curious how sharp the transition is from 1 to 0 probability; there's got to be an analytical expression for cumulative probability >=0.5, say.I can simply plot it in R though. – Dmitri May 03 '17 at 21:51

2 Answers2

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As requested in comments:

You could use R: for example the probability of being strictly more than $9876$ could be about 

> pbinom(9876, size=10^11, prob=10^-7, lower.tail=FALSE)
[1] 0.8917494

This compares with the normal approximation with continuity correction of being above $9876.5$ giving the close

> 1 - pnorm((9876.5 - 10^11 * 10^-7)/sqrt(10^11 * 10^-7 * (1-10^-7)))
[1] 0.8915848

In either case the standard deviation is close to $100$ so here only values close to something like $10000 \pm 300$ will give interesting probabilities

Henry
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I'm guessing that if the Poisson approximation does not seem appropriate, it's because the expected value and the variance (which, for the Poisson distribution, is the same as the expected value) are so big. In that case, approximating it by a normal distribution can serve.

If $X\sim\operatorname{Poisson}(\lambda)$ then the distribution of $\displaystyle \frac{X-\lambda}{\sqrt{\lambda}}$ approaches the standard normal distribution (i.e. the normal with expected value $0$ and standard deviation $1$) as $\lambda\to\infty$. In your case, you have $\lambda=10^4,$ which is plenty. If $X\sim\operatorname{Poisson}(10^4)$ then, for example, $$ \Pr(9910 \le X \le 10050) = \Pr\left( \frac{9910-10000}{\sqrt{10000}} \le \frac{X-10000}{\sqrt{10000}} \le \frac{10050-10000}{\sqrt{10000}} \right) $$ $$ \approx \Pr\left( \frac{9910-10000}{\sqrt{10000}} \le Z \le \frac{10050-10000}{\sqrt{10000}} \right) = \Pr(-0.9 < Z < 0.5) $$ where $Z\sim N(0,1).$

Michael Hardy
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    I would have thought that the real issue with the Poisson approximation here was that you still have to add up a lot of numbers if you wanted the probability of "more than $x$". Meanwhile when the numbers involved are smaller, the Poisson can have the "wrong" variance – Henry May 03 '17 at 21:28