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2017 Paper 3 Q12
D: 1700.0 B: 1500.2
probability joint distribution marginal distribution independence covariance discrete random variables bivariate data summation

The discrete random variables \(X\) and \(Y\) can each take the values \(1\), \(\ldots\,\), \(n\) (where \(n\ge2\)). Their joint probability distribution is given by \[ \P(X=x, \ Y=y) = k(x+y) \,, \] where \(k\) is a constant.

  1. Show that \[ \P(X=x) = \dfrac{n+1+2x}{2n(n+1)}\,. \] Hence determine whether \(X\) and \(Y\) are independent.
  2. Show that the covariance of \(X\) and \(Y\) is negative.

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(X = x) &= \sum_{y=1}^n \mathbb{P}(X=x,Y=y) \\ &&&= \sum_{y=1}^n k(x+y) \\ &&&= nkx + k\frac{n(n+1)}2 \\ \\ && 1 &= \sum_{x=1}^n \mathbb{P}(X=x) \\ &&&= nk\frac{n(n+1)}{2} + kn\frac{n(n+1)}2 \\ &&&= kn^2(n+1) \\ \Rightarrow && k &= \frac{1}{n^2(n+1)} \\ \Rightarrow && \mathbb{P}(X = x) &= \frac{nx}{n^2(n+1)} + \frac{n(n+1)}{2n^2(n+1)} \\ &&&= \frac{n+1+2x}{2n(n+1)} \\ \\ && \mathbb{P}(X=x)\mathbb{P}(Y=y) &= \frac{(n+1)^2+2(n+1)(x+y)+4xy}{4n^2(n+1)^2} \\ &&&\neq \frac{x+y}{n^2(n+1)} \end{align*} Therefore \(X\) and \(Y\) are not independent.
  2. \(\,\) \begin{align*} && \E[X] &= \sum_{x=1}^n x \mathbb{P}(X=x) \\ &&&= \sum_{x=1}^n x \mathbb{P}(X=x)\\ &&&= \sum_{x=1}^n x \frac{n+1+2x}{2n(n+1)} \\ &&&= \frac{1}{2n(n+1)} \left ( (n+1) \sum x + 2\sum x^2\right)\\ &&&= \frac{1}{2n(n+1)} \left ( \frac{n(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{3} \right) \\ &&&= \frac{1}{2} \left ( \frac{n+1}{2} + \frac{2n+1}{3} \right)\\ &&&= \frac{1}{2} \left ( \frac{7n+5}{6} \right)\\ &&&= \frac{7n+5}{12} \\ \\ && \textrm{Cov}(X,Y) &= \mathbb{E}\left[XY\right] - \E[X] \E[Y] \\ &&&= \sum_{x=1}^n \sum_{y=1}^n xy \frac{x+y}{n^2(n+1)} - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \sum \sum (x^2 y+xy^2) - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \left (\sum y \right )\left (\sum x^2\right ) - \E[X]^2 \\ &&&=\frac{(n+1)(2n+1)}{12} - \left ( \frac{7n+5}{12}\right)^2 \\ &&&= \frac1{144} \left (12(2n^2+3n+1) - (49n^2+70n+25) \right)\\ &&&= \frac{1}{144} \left (-25n^2-34n-13 \right) \\ &&& < 0 \end{align*} since \(\Delta = 34^2 - 4 \cdot 25 \cdot 13 = 4(17^2-25 \times 13) = -4 \cdot 36 < 0\)

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1990 Paper 3 Q15
D: 1700.0 B: 1482.6
probability geometric distribution marginal distribution independence joint distribution expected value summation discrete random variables

An unbiased twelve-sided die has its faces marked \(A,A,A,B,B,B,B,B,B,B,B,B.\) In a series of throws of the die the first \(M\) throws show \(A,\) the next \(N\) throws show \(B\) and the \((M+N+1)\)th throw shows \(A\). Write down the probability that \(M=m\) and \(N=n\), where \(m\geqslant0\) and \(n\geqslant1.\) Find

  1. the marginal distributions of \(M\) and \(N\),
  2. the mean values of \(M\) and \(N\).
Investigate whether \(M\) and \(N\) are independent. Find the probability that \(N\) is greater than a given integer \(k\), where \(k\geqslant1,\) and find \(\mathrm{P}(N > M).\) Find also \(\mathrm{P}(N=M)\) and show that \(\mathrm{P}(N < M)=\frac{1}{52}.\)

Solution: \begin{align*} \mathbb{P}(M = m, N = n) &= \left ( \frac{3}{12} \right)^m \left ( \frac{9}{12} \right)^n \frac{3}{12} \\ &= \frac{3^n}{4^{m+n+1}} \end{align*}

  1. \begin{align*} \mathbb{P}(M = m) &= \sum_{n = 1}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{n = 1}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{1}{4^{m+1}} \sum_{n = 1}^{\infty} \left ( \frac34\right)^n \\ &= \frac{1}{4^{m+1}} \frac{3/4}{1/4} \\ &= \frac{3}{4^{m+1}} \\ \\ \mathbb{P}(N = n) &= \sum_{m = 0}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{m = 0}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{3^n}{4^{n+1}} \sum_{m = 0}^{\infty} \left ( \frac14\right)^n \\ &= \frac{3^n}{4^{n+1}} \frac{1}{3/4} \\ &= \frac{3^{n-1}}{4^{n}} \\ \end{align*}
  2. \(M+1 \sim Geo(\frac34) \Rightarrow \mathbb{E}(M) = \frac43 -1 = \frac13\) \(N \sim Geo(\frac14) \Rightarrow \mathbb{E}(N) = 4\)
\(M,N\) are independent since \(\mathbb{P}(M = m, N =n ) = \mathbb{P}(M=m)\mathbb{P}(N=n)\) \begin{align*} \mathbb{P}(N > k) &= \sum_{n=k+1}^{\infty} \mathbb{P}(N = n) \\ &= \sum_{n=k+1}^{\infty} \frac{3^{n-1}}{4^{n}} \\ &= \frac{3^k}{4^{k+1}} \sum_{n = 0}^{\infty} \left ( \frac34\right)^n \\ &= \frac{3^k}{4^{k+1}} \frac{1}{1/4} \\ &= \frac{3^k}{4^k} \end{align*} \begin{align*} \mathbb{P}(N > M) &= \sum_{m=0}^{\infty} \mathbb{P}(N > m) \mathbb{P}(M = m) \\ &= \sum_{m=0}^{\infty} \left (\frac34 \right)^m \frac{3}{4^{m+1}}\\ &=\sum_{m=0}^{\infty} \frac{3^{m+1}}{4^{2m+1}}\\ &= \frac{3}{4} \frac{1}{13/16} \\ &= \frac{12}{13} \\ \\ \mathbb{P}(N=M) &= \sum_{m=1}^{\infty} \mathbb{P}(N=m, M=m) \\ &= \sum_{m=1}^{\infty} \frac{3^m}{4^{2m+1}} \\ &= \frac{3}{64} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m \\ &= \frac{3}{64} \frac{1}{13/16} \\ &= \frac{3}{52}\\ \\ \mathbb{P}(N < M) &= 1 - \frac34 - \frac3{52} \\ &= 1 - \frac{48}{52} - \frac{3}{52} \\ &= 1 - \frac{51}{52} \\ &= \frac{1}{52} \end{align*}

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