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2014 Paper 2 Q12
D: 1600.0 B: 1484.8
probability hazard function cumulative distribution function probability density function conditional probability exponential distribution curve sketching differential equation

The lifetime of a fly (measured in hours) is given by the continuous random variable \(T\) with probability density function \(f(t)\) and cumulative distribution function \(F(t)\). The hazard function, \(h(t)\), is defined, for \(F(t) < 1\), by \[ h(t) = \frac{f(t)}{1-F(t)}\,. \]

  1. Given that the fly lives to at least time \(t\), show that the probability of its dying within the following \(\delta t\) is approximately \(h (t) \, \delta t\) for small values of \(\delta t\).
  2. Find the hazard function in the case \(F(t) = t/a\) for \(0< t < a\). Sketch \(f(t)\) and \(h(t)\) in this case.
  3. The random variable \(T\) is distributed on the interval \(t > a\), where \(a>0\), and its hazard function is \(t^{-1}\). Determine the probability density function for \(T\).
  4. Show that \(h(t)\) is constant for \(t > b\) and zero otherwise if and only if \(f(t) =ke^{-k(t-b)}\) for \(t > b\), where \(k\) is a positive constant.
  5. The random variable \(T\) is distributed on the interval \(t > 0\) and its hazard function is given by \[ h(t) = \left(\frac{\lambda}{\theta^\lambda}\right)t^{\lambda-1}\,, \] where \(\lambda\) and \(\theta\) are positive constants. Find the probability density function for \(T\).

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
  2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
    TikZ diagram
  3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
  4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
  5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}

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