Problems

Filters
Clear Filters

6 problems found

2025 Paper 3 Q11
D: 1500.0 B: 1500.0
cumulative distribution function exponential distribution order statistics probability density function logarithmic approximation median confidence interval independent random variables

  1. Let \(\lambda > 0\). The independent random variables \(X_1, X_2, \ldots, X_n\) all have probability density function $$f(t) = \begin{cases} \lambda e^{-\lambda t} & t \geq 0 \\ 0 & t < 0 \end{cases}$$ and cumulative distribution function \(F(x)\). The value of random variable \(Y\) is the largest of the values \(X_1, X_2, \ldots, X_n\). Show that the cumulative distribution function of \(Y\) is given, for \(y \geq 0\), by $$G(y) = (1 - e^{-\lambda y})^n$$
  2. The values \(L(\alpha)\) and \(U(\alpha)\), where \(0 < \alpha \leq \frac{1}{2}\), are such that $$P(Y < L(\alpha)) = \alpha \text{ and } P(Y > U(\alpha)) = \alpha$$ Show that $$L(\alpha) = -\frac{1}{\lambda}\ln(1 - \alpha^{1/n})$$ and write down a similar expression for \(U(\alpha)\).
  3. Use the approximation \(e^t \approx 1 + t\), for \(|t|\) small, to show that, for sufficiently large \(n\), $$\lambda L(\alpha) \approx \ln(n) - \ln\left(\ln\left(\frac{1}{\alpha}\right)\right)$$
  4. Hence show that the median of \(Y\) tends to infinity as \(n\) increases, but that the width of the interval \(U(\alpha) - L(\alpha)\) tends to a value which is independent of \(n\).
  5. You are given that, for \(|t|\) small, \(\ln(1 + t) \approx t\) and that \(e^3 \approx 20\). Show that, for sufficiently large \(n\), there is an interval of width approximately \(4\lambda^{-1}\) in which \(Y\) lies with probability \(0.9\).

Solution:

  1. Note that \(\displaystyle F(y) = \mathbb{P}(X_i < y) = \int_0^y \lambda e^{-\lambda t} \d t = 1-e^{-\lambda y}\). Notice also that \begin{align*} G(y) &= \mathbb{P}(Y < y) \\ &= \mathbb{P}(\max_i(X_i) < y) \\ &= \mathbb{P}(X_i < y \text{ for all }i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i < y) \\ &= \prod_{i=1}^n (1-e^{-\lambda y})\\ &= (1-e^{-\lambda y})^n \end{align*} as required.
  2. \begin{align*} && \mathbb{P}(Y < L(\alpha)) &= \alpha \\ \Rightarrow && (1-e^{-\lambda L(\alpha)})^n &= \alpha \\ \Rightarrow && 1-e^{-\lambda L(\alpha)} &= \alpha^{\tfrac1n} \\ \Rightarrow && L(\alpha) &= -\frac{1}{\lambda}\ln \left (1-\alpha^{\tfrac1n} \right) \end{align*} Notice also: \begin{align*} && \mathbb{P}(Y > U(\alpha)) &= \alpha \\ \Rightarrow && 1 - (1-e^{-\lambda U(\alpha)})^n &= \alpha \\ \Rightarrow && U(\alpha) &= -\frac{1}{\lambda}\ln \left ( 1-(1-\alpha)^{\tfrac1n} \right) \end{align*}
  3. \begin{align*} \lambda L(\alpha) &= -\ln \left (1-\alpha^{\tfrac1n} \right) \\ &= -\ln \left (1-e^{\tfrac1n \ln \alpha} \right) \\ &\approx - \ln \left ( 1 - 1 - \frac1n \ln \alpha\right) \tag{\(e^t \approx 1 + t\)} \\ &= -\ln \left ( \frac{1}{n} \ln \frac{1}\alpha \right) \\ &= - \ln \frac{1}{n} - \ln \left ( \ln \frac{1}{\alpha} \right )\\ &= \ln n - \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \end{align*} since if \(n\) is large, \(\frac{\ln \alpha}{n}\) is small.
  4. The median is the value where \(\mathbb{P}(Y < M) = \frac12\), or in other words \(L(\frac12)\), but this is \(\approx \frac{\ln n - \ln (\ln 2)}{\lambda} \to \infty\). \begin{align*} && \lambda U(\alpha) &\approx \ln n - \ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right) \\ \Rightarrow && \lambda(U(\alpha) - L(\alpha)) &\approx -\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right)+ \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \\ \Rightarrow && U(\alpha) - L(\alpha) &\to \frac{1}{\lambda} \left ( \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right)-\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right ) \right) \end{align*} which doesn't depend on \(n\).
  5. Suppose \(\alpha = \frac{1}{20}\) then \begin{align*} U(\alpha) - L(\alpha) &\approx \frac{1}{\lambda} \left (\ln \ln 20 - \ln \ln \frac{20}{19} \right) \\ &= \lambda^{-1} \left (\ln \ln 20 - \ln \ln (1 + \frac{1}{19}) \right) \\ &\approx \lambda^{-1} \left (\ln 3 - \ln \frac{1}{19} \right) \tag{\(\ln(1+t) \approx t\)} \\ &\approx \lambda^{-1} \ln 3 \cdot 19 \\ &\approx \lambda^{-1} (1 + 3) \\ &\approx 4\lambda^{-1} \end{align*} [Note that \(\ln \ln 20 - \ln \ln \frac{20}{19} = 4.0673\ldots\)]

View
2016 Paper 3 Q13
D: 1700.0 B: 1500.0
kurtosis normal distribution integration expectation summation independent random variables fourth moment proof

Given a random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), we define the kurtosis, \(\kappa\), of \(X\) by \[ \kappa = \frac{ \E\big((X-\mu)^4\big)}{\sigma^4} -3 \,. \] Show that the random variable \(X-a\), where \(a\) is a constant, has the same kurtosis as \(X\).

  1. Show by integration that a random variable which is Normally distributed with mean 0 has kurtosis 0.
  2. Let \(Y_1, Y_2, \ldots, Y_n\) be \(n\) independent, identically distributed, random variables with mean 0, and let \(T = \sum\limits_{r=1}^n Y_r\). Show that \[ \E(T^4) = \sum_{r=1}^n \E(Y_r^4) + 6 \sum_{r=1}^{n-1} \sum_{s=r+1}^{n} \E(Y^2_s) \E(Y^2_r) \,. \]
  3. Let \(X_1\), \(X_2\), \(\ldots\)\,, \(X_n\) be \(n\) independent, identically distributed, random variables each with kurtosis \(\kappa\). Show that the kurtosis of their sum is \(\dfrac\kappa n\,\).

Solution: \begin{align*} &&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\ &&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\ &&&= \kappa_X \end{align*}

  1. \(\,\) \begin{align*} && \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\ &&&= \mathbb{E}(Z^4)-3\\ &&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\ &&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\ &&&= 0 + 3 \textrm{Var}(Z) - 3 =0 \end{align*}
  2. \(\,\) \begin{align*} && \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\ &&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right] \end{align*}
  3. Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with \(T\) and \(Y_i\)s. Note that \(\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)\) and \(\textrm{Var}(T) = n \sigma^2\) \begin{align*} && \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\ &&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\ &&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\ &&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\ &&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\ &&&= \frac{\kappa}{n} \end{align*}

View
2010 Paper 1 Q13
D: 1484.0 B: 1516.0
Poisson distribution exponential waiting time quadratic in exponential probability independent random variables inequality roots of equations

The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean \(\lambda\) texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is \(p\), show that \[ p\e^{2\lambda}-\e^{\lambda}+1=0. \] Given that \(4p<1\), show that there are two positive values of \(\lambda\) that satisfy this equation. The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means \(\lambda_{1}\) and \(\lambda_{2}\) texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also \(p\), find an expression for \(\lambda_{1}+\lambda_{2}\) in terms of \(p\). Find the probability, in terms of \(p\), that she waits between 1 and 2 hours in the morning to receive her first text.

Solution: Let \(X_t\) be the number of texts he recieves before \(t\) hours. So \(X_t \sim P(t\lambda)\) \begin{align*} &&\mathbb{P}(X_1 = 0 \, \cap \, X_2 > 0) &= e^{-\lambda} \cdot \left ( 1-e^{-\lambda}\right) = p \\ \Rightarrow && e^{2\lambda}p &= e^{\lambda} - 1 \\ \Rightarrow && 0 &= pe^{2\lambda}-e^{\lambda} + 1 \\ \Rightarrow && e^{\lambda} &= \frac{1 \pm \sqrt{1-4p}}{2p} \end{align*} Which clearly has two positive roots if \(4p < 1\). We need to show both roots are \(>1\). So considering the smaller one we are looking at: \begin{align*} && \frac{1-\sqrt{1-4p}}{2p} & > 1 \\ \Leftrightarrow && 1-\sqrt{1-4p} &> 2p \\ \Leftrightarrow && 1-2p&> \sqrt{1-4p} \\ \Leftrightarrow && (1-2p)^2&> 1-4p \\ \Leftrightarrow && 1-4p+4p^2&> 1-4p \\ \end{align*} which is clearly true. We must have \(e^{\lambda_1}\cdot e^{\lambda_2} = \frac{1}{p}\), so \(\lambda_1 + \lambda_2 = -\ln p\) by considering the product of the roots in our quadratic. (Vieta). Therefore the probability she waits between 1 and 2 hours in the morning is \(e^{-(\lambda_1 + \lambda_2)} \cdot ( 1- e^{-(\lambda_1+\lambda_2)}) = p \cdot (1-p)\)

View
2010 Paper 3 Q13
D: 1700.0 B: 1516.0
correlation covariance bivariate data linear combinations of random variables independent random variables expectation variance Cholesky decomposition

In this question, \({\rm Corr}(U,V)\) denotes the product moment correlation coefficient between the random variables \(U\) and \(V\), defined by \[ \mathrm{Corr}(U,V) \equiv \frac{\mathrm{Cov}(U,V)}{\sqrt{\var(U)\var(V)}}\,. \] The independent random variables \(Z_1\), \(Z_2\) and \(Z_3\) each have expectation 0 and variance 1. What is the value of \(\mathrm{Corr} (Z_1,Z_2)\)? Let \(Y_1 = Z_1\) and let \[ Y_2 = \rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2\,, \] where \(\rho_{12}\) is a given constant with $-1<\rho _{12}<1$. Find \(\E(Y_2)\), \(\var(Y_2)\) and \(\mathrm{Corr}(Y_1, Y_2)\). Now let \(Y_3 = aZ_1 + bZ_2 + cZ_3\), where \(a\), \(b\) and \(c\) are real constants and \(c\ge0\). Given that \(\E(Y_3) = 0\), \(\var(Y_3) = 1\), \( \mathrm{Corr}(Y_1, Y_3) =\rho^{{2}}_{13} \) and \( \mathrm{Corr}(Y_2, Y_3)= \rho^{{2}} _{23}\), express \(a\), \(b\) and \(c\) in terms of \(\rho^{2} _{23}\), \(\rho^{2}_{13}\) and \(\rho^{2} _{12}\). Given constants \(\mu_i\) and \(\sigma_i\), for \(i=1\), \(2\) and \(3\), give expressions in terms of the \(Y_i\) for random variables \(X_i\) such that \(\E(X_i) = \mu_i\), \(\var(X_i) = \sigma_ i^2\) and \(\mathrm{Corr}(X_i,X_j) = \rho_{ij}\).

Solution: \begin{align*} \mathrm{Corr} (Z_1,Z_2) &= \frac{\mathrm{Cov}(Z_1,Z_2)}{\sqrt{\var(Z_1)\var(Z_2)}} \\ &= \frac{\mathbb{E}(Z_1 Z_2)}{\sqrt{1 \cdot 1}} \\ &= \frac{\mathbb{E}(Z_1)\mathbb{E}(Z_2)}{\sqrt{1 \cdot 1}} \\ &= \frac{0}{1} \\ &= 0 \end{align*} \begin{align*} && \mathbb{E}(Y_2) &= \mathbb{E}(\rho_{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \mathbb{E}(\rho_{12} Z_1) + \mathbb{E}( (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \rho_{12}\mathbb{E}( Z_1) + (1 - {\rho_{12}^2})^{ \frac12}\mathbb{E}( Z_ 2) \\ &&&= 0\\ \\ && \textrm{Var}(Y_2) &= \textrm{Var}(\rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \textrm{Var}(\rho_{12} Z_1)+\textrm{Cov}(\rho_{12} Z_1,(1 - {\rho_{12}^2})^{ \frac12} Z_ 2 ) + \textrm{Var}((1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \rho_{12}^2\textrm{Var}( Z_1)+\rho_{12} (1 - {\rho_{12}^2})^{ \frac12} \textrm{Cov}(Z_1, Z_ 2 ) + (1 - {\rho_{12}^2})\textrm{Var}(Z_ 2) \\ &&&= \rho_{12}^2 + (1-\rho_{12}^2) = 1 \\ \\ && \textrm{Cov}(Y_1, Y_2) &= \mathbb{E}((Y_1-0)(Y_2-0)) \\ &&&= \mathbb{E}(Z_1 \cdot (\rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2)) \\ &&&= \rho_{12} \mathbb{E}(Z_1^2) + (1-\rho_{12}^2)^{\frac12}\mathbb{E}(Z_1, Z_2) \\ &&&= \rho_{12} \\ \Rightarrow && \textrm{Corr}(Y_1, Y_2) &= \frac{\textrm{Cov}(Y_1, Y_2)}{\sqrt{\textrm{Var}(Y_1)\textrm{Var}(Y_2)}} \\ &&&= \frac{\rho_{12}}{1 \cdot 1} = \rho_{12} \end{align*} Suppose \(Y_3 =aZ_1 +bZ_2+cZ_3\) with \(\mathbb{E}(Y_3) = 0\) (must be true), \(\textrm{Var}(Y_3) = 1 = a^2+b^2+c^2\) and \(\textrm{Corr}(Y_1, Y_3) = \rho_{13}, \textrm{Corr}(Y_2, Y_3) = \rho_{23}\). \begin{align*} && \textrm{Corr}(Y_1,Y_3) &= \textrm{Cov}(Y_1, Y_3) \\ &&&= \textrm{Cov}(Z_1, aZ_1 +bZ_2+cZ_3) \\ &&&= a \\ \Rightarrow && a &= \rho_{13} \\ \\ && \textrm{Corr}(Y_2,Y_3) &= \textrm{Cov}(Y_2, Y_3) \\ &&&= \textrm{Cov}(\rho_{12}Z_1+(1-\rho_{12}^2)^\frac12Z_2, \rho_{13}Z_1 +bZ_2+cZ_3) \\ &&&= \rho_{12}\rho_{13}+(1-\rho_{12}^2)^\frac12b \\ \Rightarrow && \rho_{23} &= \rho_{12}\rho_{13}+(1-\rho_{12}^2)^\frac12b \\ \Rightarrow && b &= \frac{\rho_{23}-\rho_{12}\rho_{13}}{(1-\rho_{12}^2)^\frac12} \\ && c &= \sqrt{1-\rho_{13}^2-\frac{(\rho_{23}-\rho_{12}\rho_{13})^2}{(1-\rho_{12}^2)}} \end{align*} Finally, let \(X_i = \mu_i + \sigma_i Y_i\)

View
2005 Paper 3 Q12
D: 1700.0 B: 1516.0
bivariate data linear combination variance minimisation unbiased estimator optimisation confidence interval independent random variables

Five independent timers time a runner as she runs four laps of a track. Four of the timers measure the individual lap times, the results of the measurements being the random variables \(T_1\) to \(T_4\), each of which has variance \(\sigma^2\) and expectation equal to the true time for the lap. The fifth timer measures the total time for the race, the result of the measurement being the random variable \(T\) which has variance \(\sigma^2\) and expectation equal to the true race time (which is equal to the sum of the four true lap times). Find a random variable \(X\) of the form \(aT+b(T_1+T_2+T_3+T_4)\), where \(a\) and \(b\) are constants independent of the true lap times, with the two properties:

  1. whatever the true lap times, the expectation of \(X\) is equal to the true race time;
  2. the variance of \(X\) is as small as possible.
Find also a random variable \(Y\) of the form \(cT+d(T_1+T_2+T_3+T_4)\), where \(c\) and \(d\) are constants independent of the true lap times, with the property that, whatever the true lap times, the expectation of \(Y^2\) is equal to \(\sigma^2\). In one particular race, \(T\) takes the value 220 seconds and \((T_1 + T_2 + T_3 + T_4)\) takes the value \(220.5\) seconds. Use the random variables \(X\) and \(Y\) to estimate an interval in which the true race time lies.

Solution: Let the expected total time for the race be \(\mu\). Let \(X = aT + b(T_1 + T_2+T_3+T_4)\) then \(\E[X] = a\E[T] + b\E[T_1+\cdots+T_4] = a \mu + b \mu = (a+b)\mu\). So \(a+b=1\). \begin{align*} && \var[X] &= a^2\var[T] + b^2(\var[T_1] + \var[T_2] + \var[T_3] + \var[T_4]) \\ &&&= a^2\sigma^2 + 4b^2 \sigma^2 \\ &&& = \sigma^2 (a^2 + 4(1-a)^2 ) \\ &&&= \sigma^2 (5a^2 - 8a + 4) \\ &&&= \sigma^2 \left ( 5 \left ( a - \frac45 \right)^2 - \frac{16}{5}+4 \right)\\ &&&= \sigma^2 \left ( 5 \left ( a - \frac45 \right)^2 + \frac{4}{5}\right) \end{align*} Therefore variance is minimised when \(a = \frac45, b = \frac15\). Let \(Y = cT + d(T_1 + T_2+T_3+T_4)\) then \begin{align*} && \E[Y^2] &= \E \left [c^2T^2 + 2cd T(T_1+T_2+T_3+T_4) + d^2(T_1+T_2+T_3+T_4)^2 \right] \\ &&&= c^2 (\mu^2 + \sigma^2) + 2cd \mu^2 + d^2 (\var[T_1 + \cdots + T_4] + \mu^2) \\ &&&= c^2(\mu^2+\sigma^2) + 2cd \mu^2 + d^2(4\sigma^2 + \mu^2) \\ &&&= (c^2 + 2cd + d^2) \mu^2 + (c^2+4d^2) \sigma^2 \\ &&&= (c+d)^2 \mu^2 + (c^2+4d^2) \sigma^2 \\ \\ \Rightarrow && d &= -c \\ && 1 &= c^2 + 4d^2 \\ \Rightarrow && c &= \pm \frac{1}{\sqrt5} \\ && d &= \mp \frac{1}{\sqrt5} \end{align*} Given our results, our best estimate for \(\mu\) is \(\frac45 \cdot 220 + \frac15 220.5 = 220.1\). Our estimate for \(\sigma^2 = \left( \frac{1}{\sqrt{5}}(220.5-220) \right)^2 = \frac{1}{20}\). Note that \(\var[X] = \frac45\sigma^2 \approx \frac{1}{25}\) so we are looking at an interval \((220.1 - 0.4, 220.1 + 0.4) = (219.7, 220.5)\) using an interval of two standard errors.

View
1987 Paper 2 Q15
D: 1500.0 B: 1500.0
geometric probability uniform distribution level crossing collision inequality area calculation independent random variables optimisation

A train of length \(l_{1}\) and a lorry of length \(l_{2}\) are heading for a level crossing at speeds \(u_{1}\) and \(u_{2}\) respectively. Initially the front of the train and the front of the lorry are at distances \(d_{1}\) and \(d_{2}\) from the crossing. Find conditions on \(u_{1}\) and \(u_{2}\) under which a collision will occur. On a diagram with \(u_{1}\) and \(u_{2}\) measured along the \(x\) and \(y\) axes respectively, shade in the region which represents collision. Hence show that if \(u_{1}\) and \(u_{2}\) are two independent random variables, both uniformly distributed on \((0,V)\), then the probability of a collision in the case when initially the back of the train is nearer to the crossing than the front of the lorry is \[ \frac{l_{1}l_{2}+l_{2}d_{1}+l_{1}d_{2}}{2d_{2}\left(l_{2}+d_{2}\right)}. \] Find the probability of a collision in each of the other two possible cases.

View