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This question is related to this question but I see that one part is really not a statistics question.

That $\lim_{n \to \infty} (1 - 1/n)^n = 1/e $ is clear.

What is not clear to me is under what circumstances

$$\lim_{n \to \infty} \prod_{i=1}^n (1 - F_i) = 1/e. $$

Let me give some examples of candidates for $F_i$:

  1. Subdivide the standard normal curve into (say) 100 subintervals from, say, Z = -5 to Z = 5. With each subinterval associate a number $F_i$ equal to the area under the curve in that subinterval.

  2. [removed]

  3. For $\frac{1}{2}\int_{-\pi/2}^{\pi/2}\cos x ~dx$ subdivide the interval $[ -\pi/2 < x < \pi/2]$ and find $F_i $ in a similar way.

So either I am making a consistent mistake in calculating (not out of the question) or there is something general that I am missing.

Can someone suggest why this is true, if it is, and how it generalizes?

Thank you.

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daniel
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  • When you say "subdivide the standard normal curve into...", do you mean that $F_i=\int_{i/n}^{(i+1)/n}f(x),\mathrm dx$, where $f$ is the density of a standard Gaussian distribution? Since the Poisson distribution is discrete, how do you generalize this idea to include point 2.? – Ian May 12 '14 at 03:24
  • @Ian: Fair enough. I only have time to remove it at the moment. I think my computer was using a $\Gamma$ function or something to interpolate and I will have to go back and check. – daniel May 12 '14 at 09:29

2 Answers2

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I am not sure it helps in a statistical context, but I think the precise condition is that $\sum_{n=1}^{\infty} \frac{1}{n}\left( \sum_{i=1}^{\infty} F_{i}^{n} \right) = 1.$ You get this by taking the log, and looking at the Taylor series for $\log(1-x).$

In response to comment: The thing to note is that $\log(1 - x ) = -\sum_{k=1}^{\infty} \frac{x^{k}}{k}$ when $|x| < 1.$ Assuming each $F_{i}$ has absolute value less that $1,$ you require $-1 = \lim_{i \to \infty} \sum_{i=1}^{n} - \log(1-F_{i}) = 1.$ In other words, you want $\sum_{i=1}^{\infty} - \log(1-F_{i}) = 1$, assuming the left hand series converges. Hence you want $\sum_{i=1}^{\infty} \sum_{n=1}^{\infty} \frac{F_{i}^{n}}{n} =1$ assuming all relevant series converge. If $0 \leq F_{i} <1$ for each $i,$ and all partial sums $\sum_{i=1}^{M} \sum_{n=1}^{N} \frac{F_{i}^{n}}{n}$ are less then or equal to $1,$ then it is permissible to change the order of summation and in that case you require precisely that the original double sum converges to $1$ (in which case the order of summation does not matter).

Geoff Robinson
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  • Do you think for $|F_i|< 1|$ we could also just say $\log(1 + (-F_i)) \approx -F_i,~$ ...etc.? And so $\sum (-F_i) \approx -1$ and exactly in the limit as n gets large? – daniel May 13 '14 at 01:19
  • Well, in that case, you can say that if $\sum_{i=1}^{n} F_{i}$ is around $1$, then the product $\prod_{i=1}^{n} (1-F_{i})$ will be close to $\frac{1}{e},$ but even if $\sum_{i=1}^{\infty } F_{i} = 1,$ the product will definitely not be exactly $\frac{1}{e}.$ – Geoff Robinson May 13 '14 at 06:50
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There's nothing special about $1/e$, just use a product for $e^x$ function and plug $x=-1$ into the expression. A relevant topic that answers this particular question is

Exponential Function as an Infinite Product

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orion
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