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2010 Paper 2 Q12
D: 1600.0 B: 1486.8
probability cumulative distribution functions piecewise PDF expectation median inequality continuous random variable

The continuous random variable \(X\) has probability density function \(\f(x)\), where \[ \f(x) = \begin{cases} a & \text {for } 0\le x < k \\ b & \text{for } k \le x \le 1\\ 0 & \text{otherwise}, \end{cases} \] where \(a > b > 0\) and \(0 < k < 1\). Show that \(a > 1\) and \(b < 1\).

  1. Show that \[ \E(X) = \frac{1-2b+ab}{2(a-b)}\,. \]
  2. Show that the median, \(M\), of \(X\) is given by \(\displaystyle M=\frac 1 {2a}\) if \(a+b\ge 2ab\) and obtain an expression for the median if \(a+b\le 2ab\).
  3. Show that \(M < \E(X)\,\).

Solution: \begin{align*} && 1 &= \int_0^1 f(x) \d x \\ &&&= ak + b(1-k) \\ &&&= b + (a-b)k \\ \Rightarrow && k &= \frac{1-b}{a-b} \\ \Rightarrow && b & < 1 \tag{\(0 < k, \,a > b\)} \\ && k &> 1 \\ \Rightarrow && a-b & > 1-b \\ \Rightarrow && a > 1 \end{align*}

  1. \(\,\) \begin{align*} && \E[X] &= \int_0^1 x \cdot f(x) \d x \\ &&&= \int_0^k ax \d x + \int_k^1 b x \d x \\ &&&= a \frac{k^2}{2} + b \frac{1-k^2}{2} \\ &&&= \frac12b + (a-b) \frac{(1-b)^2}{2(a-b)^2} \\ &&&= \frac{(1-b)^2+b(a-b)}{2(a-b)} \\ &&&= \frac{1-2b+ab}{2(a-b)} \end{align*}
  2. \(\,\) The median \(M\) satisfies \[\frac12 = \int_0^M f(x) \d x \] If \(ka = \frac{a-ab}{a-b} \leq \frac12 \Leftrightarrow 2a-2ab \leq a-b \Leftrightarrow a+b \leq 2ab\) then \(M > k\) otherwise \(M < k\). In the latter case: \begin{align*} && \frac12 &= Ma \\ \Rightarrow && M &= \frac{1}{2M} \end{align*} In the former case \begin{align*} && \frac12 &= ka + (M-k)b \\ &&&= k(a-b) + Mb \\ &&&= 1-b + M b \\ \Rightarrow && M &= 1-\frac1{2b} \end{align*}

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