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2007 Paper 3 Q12
D: 1700.0 B: 1487.4
expectation variance covariance conditional expectation random sum discrete uniform distribution law of total expectation law of total variance bivariate data

I choose a number from the integers \(1, 2, \ldots, (2n-1)\) and the outcome is the random variable \(N\). Calculate \( \E(N)\) and \(\E(N^2)\). I then repeat a certain experiment \(N\) times, the outcome of the \(i\)th experiment being the random variable \(X_i\) (\(1\le i \le N\)). For each \(i\), the random variable \(X_i\) has mean \(\mu\) and variance \(\sigma^2\), and \(X_i\) is independent of \(X_j\) for \(i\ne j\) and also independent of \(N\). The random variable \(Y\) is defined by \(Y= \sum\limits_{i=1}^NX_i\). Show that \(\E(Y)=n\mu\) and that \(\mathrm{Cov}(Y,N) = \frac13n(n-1)\mu\). Find \(\var(Y) \) in terms of \(n\), \(\sigma^2\) and \(\mu\).

Solution: \begin{align*} && \E[N] &= \sum_{i=1}^{2n-1} \frac{i}{2n-1} \\ &&&= \frac{2n(2n-1)}{2(2n-1)} = n\\ && \E[N^2] &= \sum_{i=1}^{2n-1} \frac{i^2}{2n-1} \\ &&&= \frac{(2n-1)(2n)(4n-1)}{6(2n-1)} \\ &&&= \frac{n(4n-1)}{3} \\ && \var[N] &= \frac{n(4n-1)}{3} - n^2 \\ &&&= \frac{n^2-n}{3} \end{align*} \begin{align*} && \E[Y] &= \E \left [ \E \left [ \sum_{i=1}^N X_i | N = k\right] \right]\\ &&&= \E \left[ N\mu \right] = n\mu \\ \\ && \mathrm{Cov}(Y,N) &= \mathbb{E}[XY] - \E[X]\E[Y] \\ &&&= \E \left [ \E \left [N \sum_{i=1}^N X_i | N = k\right] \right] - n^2 \mu \\ &&&= \E[N^2\mu] - n^2 \mu \\ &&&= \left ( \frac{n^2(4n-1)}{3} - n^2 \right) \mu \\ &&&= \frac{n^2-n}{3}\mu \\ \\ && \E[Y^2] &= \E \left [ \E \left [ \left ( \sum_{i=1}^N X_i \right) ^2\right ] \right] \\ &&&= \E \left [ \E \left [ \sum_{i=1}^N X_i ^2 + 2\sum_{i,j} X_iX_j\right ] \right] \\ &&&= \E \left [ \sum_{i=1}^N \left ( \E[X_i ^2] + 2\sum_{i,j} \E[X_i]\E[X_j]\right ) \right] \\ &&&= \E \left [ N(\sigma^2 + \mu^2) + (N^2-N)\mu^2\right] \\ &&&= n(\sigma^2+\mu^2) + \left ( \frac{n^2-n}{3}-n \right)\mu^2 \\ &&&= n\sigma^2 + \frac{n^2-n}{3} \mu^2 \\ \Rightarrow && \var[Y] &= n\sigma^2 + \frac{n^2-n}{3} \mu^2 - n^2\mu^2 \\ &&&= n\sigma^2 - \frac{2n^2+n}{3} \mu^2 \end{align*}

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2006 Paper 1 Q13
D: 1484.0 B: 1468.0
Poisson distribution conditional probability expectation law of total expectation law of total probability geometric distribution geometric series binomial expansion

A very generous shop-owner is hiding small diamonds in chocolate bars. Each diamond is hidden independently of any other diamond, and on average there is one diamond per kilogram of chocolate.

  1. I go to the shop and roll a fair six-sided die once. I decide that if I roll a score of \(N\), I will buy \(100N\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6} \l \frac{1 - \e^{-0.6} }{ 1 - \e^{-0.1}} \r \] Show also that the expected number of diamonds I find is 0.35.
  2. Instead, I decide to roll a fair six-sided die repeatedly until I score a 6. If I roll my first 6 on my \(T\)th throw, I will buy \(100T\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6 - 5\e^{-0.1}} \] Calculate also the expected number of diamonds that I find. (You may find it useful to consider the the binomial expansion of \(\l 1 - x \r^{-2}\).)

Solution: Not that the number of diamonds per kilogram is \(1\) so we are assuming it is \(Po(M)\) where \(M\) is the mass in kg. In particular \(\E[X] = M\) and \(\mathbb{P}(X = 0) = e^{-M}\)

  1. \(\,\) \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^6\mathbb{P}(\text{no diamonds and roll }n) \\ &&&= \sum_{n=1}^6 \tfrac16 e^{-\frac{n}{10}} \\ &&&= \frac{e^{-0.1}}6 \left ( \frac{1-e^{-0.6}}{1-e^{-0.1}}\right) \\ && \E[\text{diamonds}] &= \sum_{n=1}^6 \E(\text{diamonds}|N = n)\mathbb{P}(N = n) \\ &&&= \sum_{n=1}^6 0.1n \cdot \frac16 \\ &&&= 0.1 \cdot \frac{7}{2} = 0.35 \end{align*}
  2. \(\mathbb{P}(T = k) = \left ( \frac56 \right)^{t-1} \frac16\), so \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^\infty\mathbb{P}(\text{no diamonds and }T=n) \\ &&&= \sum_{n=1}^\infty e^{-0.1n} \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{e^{-0.1}}{6} \frac1{1- \frac56 e^{-0.1}} \\ &&&= \frac{e^{-0.1}}{6 - 5e^{-0.1}} \\ \\ && \E[\text{diamonds}] &= \sum_{n=1}^\infty \E(\text{diamonds}|T = n)\mathbb{P}(T = n) \\ &&&= \sum_{n=1}^\infty 0.1n \cdot \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{0.1}{6} \sum_{n=1}^\infty n \cdot \left ( \frac56 \right)^{n-1} \\ &&&= \frac{1}{60} \frac{1}{(1- \tfrac56)^2} \\ &&&= \frac{6}{10} = \frac35 \end{align*}

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