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2005 Paper 2 Q13
D: 1600.0 B: 1500.0
Poisson distribution binomial distribution conditional probability approximation small parameter expansion Taylor series hypothesis testing

The number of printing errors on any page of a large book of \(N\) pages is modelled by a Poisson variate with parameter \(\lambda\) and is statistically independent of the number of printing errors on any other page. The number of pages in a random sample of \(n\) pages (where \(n\) is much smaller than \(N\) and \(n\ge2\)) which contain fewer than two errors is denoted by \(Y\). Show that \(\P(Y=k) = \binom n k p^kq^{n-k}\) where \(p=(1+\lambda)e^{-\lambda}\) and \(q=1-p\,\). Show also that, if \(\lambda\) is sufficiently small,

  1. \(q\approx \frac12 \lambda^2\,\);
  2. the largest value of \(n\) for which \(\P(Y=n)\ge 1-\lambda\) is approximately \(2/\lambda\,\);
  3. \(\P(Y>1 \;\vert\; Y>0) \approx 1-n(\lambda^2/2)^{n-1}\;.\)

Solution: First notice that the the probability a page contains fewer than two errors is \(\mathbb{P}(X < 2)\) where \(X \sim Po(\lambda)\), ie \(\mathbb{P}(X<2) = e^{-\lambda} + \lambda e^{-\lambda} = (1+\lambda)e^{-\lambda}\). Therefore the number of pages \(Y\) with fewer than two errors out of our sample of \(n\) is \(Bin(n, p)\) where \(p\) is as before. ie \(\mathbb{P}(Y = k) = \binom{n}{k} p^kq^{n-k}\).

  1. \(\,\) \begin{align*} && q &= 1- p = 1-(1+\lambda)e^{-\lambda} \\ &&&= 1 - (1+ \lambda)(1 - \lambda + \tfrac12 \lambda^2 + o(\lambda^3)) \\ &&&= 1 - 1+ \lambda - \lambda+\lambda^2 - \tfrac12 \lambda^2 + o(\lambda^3) \\ &&&= \tfrac12 \lambda^2 + o(\lambda^3) \end{align*}
  2. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= p^n \\ &&&= (1+\lambda)^ne^{-\lambda n} \\ &&&= (1 + n \lambda + \frac{n(n-1)}{2} \lambda^2 + \cdots)(1 - \lambda n + \frac{\lambda^2 n^2}{2} + \cdots) \\ &&&= 1 + 0 \lambda + \left ( \frac{n(n-1)}{2} + \frac{n^2}{2} - n^2 \right) \lambda^2 + o(\lambda^3) \\ &&&= 1 - \frac{n}{2} \lambda^2 + o(\lambda^3) \end{align*} So if \(\frac{n}{2} \lambda \leq 1\) or \(n \leq \frac{2}{\lambda}\) \(\mathbb{P}(Y = n) \leq 1- \lambda\).
  3. \(\,\) \begin{align*} && \mathbb{P}(Y > 1 | Y > 0) &= \frac{1-(q^n + npq^{n-1})}{1-q^n} \\ &&&= 1 - \frac{npq^{n-1}}{1-q^n} \\ &&&= 1 -n \frac{(1+ \lambda)e^{-\lambda} (\tfrac12 \lambda^2 + o(\lambda^3))^{n-1}}{1-(\tfrac12 \lambda^2 + o(\lambda^3))^n} \\ &&&= 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} \frac{(1+ \lambda)(1-\lambda + \lambda^2/2 - \cdots)(1+o(\lambda)^{n-1}}{1-(\tfrac12 \lambda^2 + o(\lambda^3))^n} \\ &&&= 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} (1 + o(\lambda)) \\ &&&\approx 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} \end{align*}

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