I have a (seemingly) simple question. How can I see (rigorously) that \begin{equation} \prod_{i=1}^n (1-p_i) = \exp{\left(-\sum_{i=1}^n p_i\right)} + O\left(\sum_{i=1}^n p_i^2\right) \end{equation} for $\max_{i}p_i\rightarrow 0$ and $p_i\geq 0,\, \forall i$. For $n=1$, the above is a just a Taylor approximation. But, for $n>1$, I have trouble combining the individual error terms (for the individual $i$'s) in the product to obtain the $O(\sum p_i^2)$ error term. Any ideas? Every hint/help is much appreciated. Thanks in advance!
How to show $\prod_{i=1}^n (1-p_i) = \exp{(-\sum_{i=1}^n p_i)} + O(\sum_{i=1}^n p_i^2)$
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0but ma$n$y thanks for all the help, it is much appreciated! – 2012-12-05
2 Answers
I think that the problem is now correct.
Suppose $\forall i,\;|1-p_i| \leq 1$ (for instance if $\forall i,\;0\leq p_i \leq 1$). We have $ \left|\prod_{i=1}^n (1-p_i) - \prod_{i=1}^n e^{-p_i}\right| \leq \sum_{i=1}^n |1 - p_i - e^{-p_i}| \leq \frac{1}{2} \sum_{i=1}^n p_i^2. $
Hence, $ \prod_{i=1}^n (1-p_i) = \exp\left(-\sum_{i=1}^n p_i\right) + O\!\left(\sum_{i=1}^n p_i^2\right). $
The first inequality comes from the fact that if $a_1,\dots,a_n$ and $b_1,\dots,b_n$ are complex numbers with modulus lesser than $1$, then $|\prod a_i - \prod b_i| \leq \sum |a_i - b_i|$ (this follows from an easy recurrence).
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0great answer, many thanks! I really appreciate your support! – 2012-12-05
Consider
$ \log \prod_{i=1}^n (1-p_i) = \sum_{i=1}^n \log(1-p_i).$
Since $\max_i p_i \to 0$, use the Taylor series expansion of $\log(1-p_i)$, then it follows that
$ \log \prod_{i=1}^n (1-p_i) = -\sum_{i=1}^n p_i + O\left( \sum_{i=1}^n p_i^2 \right). $
Take exponents of both sides and use $e^x=1+ O(x)$, $x$ around $0$, to obtain
$ \prod_{i=1}^n (1-p_i)= exp(-\sum_{i=1}^n p_i) + 1 + O\left( \sum_{i=1}^n p_i^2 \right). $
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0Your solution is not sharp enough : when taking $p_i=0$, it writes $1 = 2 + O(0)$. – 2012-12-04