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1990 Paper 2 Q15
D: 1600.0 B: 1500.0
probability cumulative distribution function order statistics probability density function expectation geometric probability disc target maximum of random variables

A target consists of a disc of unit radius and centre \(O\). A certain marksman never misses the target, and the probability of any given shot hitting the target within a distance \(t\) from \(O\) it \(t^{2}\), where \(0\leqslant t\leqslant1\). The marksman fires \(n\) shots independently. The random variable \(Y\) is the radius of the smallest circle, with centre \(O\), which encloses all the shots. Show that the probability density function of \(Y\) is \(2ny^{2n-1}\) and find the expected area of the circle. The shot which is furthest from \(O\) is rejected. Show that the expected area of the smallest circle, with centre \(O\), which encloses the remaining \((n-1)\) shots is \[ \left(\frac{n-1}{n+1}\right)\pi. \]

Solution: Another way to describe \(Y\) is the maximum distance of any shot from \(O\). Let \(X_i\), \(1 \leq i \leq n\) be the \(n\) shots then, \begin{align*} F_Y(y) &= \mathbb{P}(Y \leq y) \\ &= \mathbb{P}(X_i \leq y \text{ for all } i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i \leq y) \tag{each shot independent}\\ &= \prod_{i=1}^n y^2\\ &= y^{2n} \end{align*} Therefore \(f_Y(y) = \frac{\d}{\d y} (y^{2n}) = 2n y^{2n-1}\). \begin{align*} \mathbb{E}(\pi Y^2) &= \int_0^1\pi y^2 \f_Y(y) \d y \\ &=\pi \int_0^1 2n y^{2n+1} \d y \\ &=\left ( \frac{n}{n+1} \right )\pi \end{align*}. Let \(Z\) be the distance of the second furthest shot, then: \begin{align*} && F_Z(z) &= \mathbb{P}(Z \leq z) \\ &&&= \mathbb{P}(X_i \leq z \text{ for at least } n - 1\text{ different } i) \\ &&&= n\mathbb{P}(X_i \leq z \text{ for all but 1}) + \mathbb{P}(X_i \leq z \text{ for all } i) \\ &&&= n \left ( \prod_{i=1}^{n-1} \mathbb{P}(X_i \leq z) \right) \mathbb{P}(X_n > z) + z^{2n} \\ &&&= nz^{2n-2}(1-z^2) + z^{2n} \\ &&&= nz^{2n-2} -(n-1)z^{2n} \\ \Rightarrow && f_Z(z) &= n(2n-2)z^{2n-3}-2n(n-1)z^{2n-1} \\ \Rightarrow && \mathbb{E}(\pi Z^2) &= \int_0^1 \pi z^2 \left (n(2n-2)z^{2n-3}-2n(n-1)z^{2n-1} \right) \d z \\ &&&= \pi \left ( \frac{n(2n-2)}{2n} - \frac{2n(n-1)}{2n+2}\right) \\ &&&= \left ( \frac{n-1}{n+1} \right) \pi \end{align*}

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1989 Paper 3 Q15
D: 1700.0 B: 1503.8
probability cumulative distribution function uniform distribution order statistics median probability density function integration expectation variance

The continuous random variable \(X\) is uniformly distributed over the interval \([-c,c].\) Write down expressions for the probabilities that:

  1. \(n\) independently selected values of \(X\) are all greater than \(k\),
  2. \(n\) independently selected values of \(X\) are all less than \(k\),
where \(k\) lies in \([-c,c]\). A sample of \(2n+1\) values of \(X\) is selected at random and \(Z\) is the median of the sample. Show that \(Z\) is distributed over \([-c,c]\) with probability density function \[ \frac{(2n+1)!}{(n!)^{2}(2c)^{2n+1}}(c^{2}-z^{2})^{n}. \] Deduce the value of \({\displaystyle \int_{-c}^{c}(c^{2}-z^{2})^{n}\,\mathrm{d}z.}\) Evaluate \(\mathrm{E}(Z)\) and \(\mathrm{var}(Z).\)

Solution:

  1. \begin{align*} \mathbb{P}(n\text{ independent values of }X > k) &= \prod_{i=1}^n \mathbb{P}(X > k) \\ &= \left ( \frac{c-k}{2c}\right)^n \end{align*}
  2. \begin{align*} \mathbb{P}(n\text{ independent values of }X < k) &= \prod_{i=1}^n \mathbb{P}(X < k) \\ &= \left ( \frac{k+c}{2c}\right)^n \end{align*}
\begin{align*} &&\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta) &= \mathbb{P}(n\text{ values } < z - \delta \text{ and } n \text{ values} > z + \delta) \\ &&&= \binom{2n+1}{n,n,1} \left ( \frac{c-(z+\delta)}{2c}\right)^n\left ( \frac{(z-\delta)+c}{2c}\right)^n \frac{2 \delta}{2 c} \\ &&&= \frac{(2n+1)!}{n! n!} \frac{((c-(z+\delta))(c+(z-\delta)))^n 2\delta}{2^n c^n} \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-(z+\delta))(c+(z-\delta)))^n 2\delta \\ \Rightarrow && \lim_{\delta \to 0} \frac{\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta)}{2 \delta} &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-z)(c+z))^n \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2) \\ \end{align*} \begin{align*} && 1 &= \int_{-c}^c \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ \Rightarrow && \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!} &= \int_{-c}^c (c^2-z^2)^n \d z \end{align*} \begin{align*} \mathbb{E}(Z) &= \int_{-c}^c z \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z (c^2-z^2)^n \d z \\ &= 0 \end{align*} \begin{align*} \mathrm{Var}(Z) &= \mathbb{E}(Z^2) - \mathbb{E}(Z)^2 \\ &= \mathbb{E}(Z^2) \\ &= \int_{-c}^c z^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z^2 (c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \left ( \left [ -\frac{1}{2(n+1)}z(c^2-z^2)^{n+1} \right]_{-c}^c + \frac{1}{2(n+1)}\int_{-c}^c (c^2-z^2)^{n+1} \d z \right) \\ &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \frac{1}{2(n+1)} \frac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} \\ &= \frac{(n+1)^2(2c)^2}{(n+1)(2n+2)(2n+3)} \\ &= \frac{2c^2}{2n+3} \end{align*}

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