4 problems found
A continuous random variable is said to have an exponential distribution with parameter \(\lambda\) if its density function is \(\f(t) = \lambda \e ^{- \lambda t} \; \l 0 \le t < \infty \r\,\). If \(X_1\) and \(X_2\), which are independent random variables, have exponential distributions with parameters \(\lambda_1\) and \(\lambda_2\) respectively, find an expression for the probability that either \(X_1\) or \(X_2\) (or both) is less than \(x\). Prove that if \(X\) is the random variable whose value is the lesser of the values of \(X_1\) and \(X_2\), then \(X\) also has an exponential distribution. Route A and Route B buses run from my house to my college. The time between buses on each route has an exponential distribution and the mean time between buses is 15 minutes for Route A and 30 minutes for Route B. The timings of the buses on the two routes are independent. If I emerge from my house one day to see a Route A bus and a Route B bus just leaving the stop, show that the median wait for the next bus to my college will be approximately 7 minutes.
The life times of a large batch of electric light bulbs are independently and identically distributed. The probability that the life time, \(T\) hours, of a given light bulb is greater than \(t\) hours is given by \[ \P(T>t) \; = \; \frac{1}{(1+kt)^\alpha}\;, \] where \(\alpha\) and \(k\) are constants, and \(\alpha >1\). Find the median \(M\) and the mean \(m\) of \(T\) in terms of \(\alpha\) and \(k\). Nine randomly selected bulbs are switched on simultaneously and are left until all have failed. The fifth failure occurs at 1000 hours and the mean life time of all the bulbs is found to be 2400 hours. Show that \(\alpha\approx2\) and find the approximate value of \(k\). Hence estimate the probability that, if a randomly selected bulb is found to last \(M\) hours, it will last a further \(m-M\) hours.
Solution: The median \(M\) is the value such that \begin{align*} && \frac12 &= \mathbb{P}(T > M) \\ &&&= \frac1{(1+kM)^\alpha} \\ \Rightarrow && 2 &= (1+kM)^{\alpha} \\ \Rightarrow && M &= \frac{2^{1/\alpha}-1}{k} \end{align*} The distribution of \(T\) is \(f_T(t) = \frac{k \alpha}{(1+kt)^{\alpha+1}}\) and so \begin{align*} && m &= \int_0^\infty t f_T(t) \d t \\ &&&= \int_0^\infty \frac{tk \alpha}{(1+kt)^{\alpha+1}} \d t \\ &&&= \int_0^\infty \frac{\alpha+tk \alpha-\alpha}{(1+kt)^{\alpha+1}} \d t \\ &&&= \alpha \int_0^\infty (1+kt)^{-\alpha} \d t - \alpha \int_0^\infty (1+kt)^{-(\alpha+1)} \d t \\ &&&= \alpha \left [ -\frac1{k(\alpha-1)}(1+kt)^{-\alpha+1}\right]_0^\infty- \alpha \left [ -\frac1{k\alpha}(1+kt)^{-\alpha}\right]_0^\infty \\ &&&= \frac{\alpha}{k(\alpha-1)} - \frac{1}{k} \\ &&&= \frac{1}{k(\alpha-1)} \end{align*} \begin{align*} && \frac{2^{1/\alpha}-1}{k} &= 1000 \\ && \frac{1}{k(\alpha-1)} &= 2400 \\ \Rightarrow && \frac{\alpha-1}{2^{1/\alpha}-1} &\approx 2.4 \\ && \frac{2-1}{\sqrt2-1} &= \sqrt{2}+1 \approx 2.4 \\ \Rightarrow && \alpha &\approx 2 \\ && k &= \frac{1}{2400} \end{align*} \begin{align*} && \mathbb{P}(T > m | T > M) &= \frac{\mathbb{P}(T > m)}{\mathbb{P}(T > M)} \\ &&&= \frac{2}{(1+km)^{\alpha}} \\ &&&= \frac{2}{(1 + \frac{1}{\alpha-1})^\alpha} \\ &&&\approx \frac{2}{4} =\frac12 \end{align*}
A random variable \(X\) has the probability density function \[ \mathrm{f}(x)=\begin{cases} \lambda\mathrm{e}^{-\lambda x} & x\geqslant0,\\ 0 & x<0. \end{cases} \] Show that $${\rm P}(X>s+t\,\vert X>t) = {\rm P}(X>s).$$ The time it takes an assistant to serve a customer in a certain shop is a random variable with the above distribution and the times for different customers are independent. If, when I enter the shop, the only two assistants are serving one customer each, what is the probability that these customers are both still being served at time \(t\) after I arrive? One of the assistants finishes serving his customer and immediately starts serving me. What is the probability that I am still being served when the other customer has finished being served?
Solution: \begin{align*} && \mathbb{P}(X > t) &= \int_t^{\infty} \lambda e^{-\lambda x} \d x\\ &&&= \left[ -e^{-\lambda x} \right]_t^\infty \\ &&&= e^{-\lambda t}\\ \\ && \mathbb{P}(X > s + t | X > t) &= \frac{\mathbb{P}(X > s + t)}{\mathbb{P}(X > t)} \\ &&&= \frac{e^{-(s+t)\lambda}}{e^{-t\lambda}} \\ &&&= e^{-s\lambda} = \mathbb{P}(X > s) \end{align*} The probability both are still being served (independently) is \(\mathbb{P}(X > t)^2 = e^{-2\lambda t}\). The probability is exactly \(\frac12\). The property we proved in the first part of the questions shows the distribution is memoryless, ie we are both experiencing samples from the same distribution. Therefore we are equally likely to finish first.
During his performance a trapeze artist is supported by two identical ropes, either of which can bear his weight. Each rope is such that the time, in hours of performance, before it fails is exponentially distributed, independently of the other, with probability density function \(\lambda\exp(-\lambda t)\) for \(t\geqslant0\) (and 0 for \(t < 0\)), for some \(\lambda > 0.\) A particular rope has already been in use for \(t_{0}\) hours of performance. Find the distribution for the length of time the artist can continue to use it before it fails. Interpret and comment upon your result. Before going on tour the artist insists that the management purchase two new ropes of the above type. Show that the probability density function of the time until both ropes fail is \[ \mathrm{f}(t)=\begin{cases} 2\lambda\mathrm{e}^{-\lambda t}(1-\mathrm{e}^{-\lambda t}) & \text{ if }t\geqslant0,\\ 0 & \text{ otherwise.} \end{cases} \] If each performance lasts for \(h\) hours, find the probability that both ropes fail during the \(n\)th performance. Show that the probability that both ropes fail during the same performance is \(\tanh(\lambda h/2)\).
Solution: This is the memoryless property of the exponential distribution so it has the same distribution as when \(t = 0\). Let \(T\) be the time the rope fails, then \begin{align*} && \mathbb{P}(T > t | T > t_0) &= \frac{\mathbb{P}(T > t)}{\mathbb{P}(T > t_0)} \\ &&&= \frac{e^{-\lambda t}}{e^{-\lambda t_0}} \\ &&&= e^{-\lambda(t-t_0)} \end{align*} This means that each rope (as long as it hasn't broken) can be considered "as good as new". Suppose \(T_1, T_2 \sim Exp(\lambda)\) are the time to failures for each rope, then \begin{align*} && \mathbb{P}(\max(T_1, T_2) < t) &= \mathbb{P}(T_1 < t, T_2 < t) \\ &&&= (1-e^{-\lambda t})^2 \\ \Rightarrow && f(t) &= 2(1-e^{-\lambda t}) \cdot (\lambda e^{-\lambda t}) \\ &&&= 2\lambda e^{-\lambda t}(1-e^{-\lambda t}) \end{align*} Therefore \(\max(T_1, T_2) \sim Exp(2\lambda)\) and the pdf is as described. \begin{align*} && \mathbb{P}(\text{both fail during the }n\text{th}) &= \left ( \int_{(n-1)h}^{nh} \lambda e^{-\lambda t} \d t \right)^2 \\ &&&=\left (\left [ -e^{-\lambda t}\right]_{(n-1)h}^{nh} \right)^2 \\ &&&= \left ( e^{-\lambda (n-1)h}( 1-e^{-\lambda h}) \right)^2 \\ &&&= e^{-2(n-1)h\lambda}(e^{-\lambda h}-1)^2 \\ \\ && \mathbb{P}(\text{both fail in same performance}) &= \sum_{n=1}^{\infty} \mathbb{P}(\text{both fail during the }n\text{th}) \\ &&&= \sum_{n=1}^{\infty}e^{-2(n-1)h\lambda}(e^{-\lambda h}-1)^2 \\ &&&= (e^{-\lambda h}-1)^2 \frac{1}{1-e^{-2h\lambda}} \\ &&&= \frac{e^{-\lambda h}-1}{1+e^{-h\lambda}} \\ &&&= \tanh(\lambda h/2) \end{align*}