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2012 Paper 3 Q13
- The random variable \(Z\) has a Normal distribution with mean \(0\) and variance \(1\). Show that the expectation of \(Z\) given that \(a < Z < b\) is \[ \frac{\exp(- \frac12 a^2) - \exp(- \frac12 b^2) } {\sqrt{2\pi\,} \,\big(\Phi(b) - \Phi(a)\big)}, \] where \(\Phi\) denotes the cumulative distribution function for \(Z\).
- The random variable \(X\) has a Normal distribution with mean \(\mu\) and variance \(\sigma^2\). Show that \[ \E(X \,\vert\, X>0) = \mu + \sigma \E(Z \,\vert\,Z > -\mu/\sigma). \] Hence, or otherwise, show that the expectation, \(m\), of \(\vert X\vert \) is given by \[ m= \mu \big(1 - 2 \Phi(- \mu / \sigma)\big) + \sigma \sqrt{2 / \pi}\; \exp(- \tfrac12 \mu^2 / \sigma^2) \,. \] Obtain an expression for the variance of \(\vert X \vert\) in terms of \(\mu \), \(\sigma \) and \(m\).
Solution:
- \(\,\) \begin{align*} && \mathbb{E}(Z| a < Z < b) &= \mathbb{E}(Z\mathbb{1}_{(a,b)}) /\mathbb{E}(\mathbb{1}_{(a,b)}) \\ &&&= \int_a^b z \phi(z) \d z \Big / (\Phi(b) - \Phi(a)) \\ &&&= \frac{\int_a^b \frac{1}{\sqrt{2 \pi}}z e^{-\frac12 z^2} \d z}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left [-e^{-\frac12 z^2} \right]_a^b}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left (e^{-\frac12 a^2}-e^{-\frac12 b^2} \right)}{\Phi(b) - \Phi(a)} \\ \end{align*}
- \(\,\) \begin{align*} && \mathbb{E}(X |X > 0) &= \mathbb{E}(\mu + \sigma Z | \mu + \sigma Z > 0) \\ &&&= \mathbb{E}(\mu + \sigma Z | Z > -\tfrac{\mu}{\sigma}) \\ &&&= \mathbb{E}(\mu| Z > -\tfrac{\mu}{\sigma})+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ &&&= \mu+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ \end{align*} Hence \begin{align*} &&\mathbb{E}(|X|) &= \mathbb{E}(X | X > 0)\mathbb{P}(X > 0) - \mathbb{E}(X | X < 0)\mathbb{P}(X < 0) \\ &&&=\left ( \mu+ \sigma \mathbb{E}(Z | Z > -\mu /\sigma)\right)(1-\Phi(-\mu/\sigma)) - \left ( \mu+ \sigma \mathbb{E}(Z | Z < -\mu /\sigma)\right)\Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2\pi}(1-\Phi(-\mu/\sigma))}(1-\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2 \pi} \Phi(-\mu/\sigma)} \Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \sqrt{\frac{2}{\pi}} \exp(-\tfrac12 \mu^2/\sigma^2) \end{align*} Finally, \begin{align*} && \textrm{Var}(|X|) &= \mathbb{E}(|X|^2) - [\mathbb{E}(|X|)]^2 \\ &&&= \mu^2 + \sigma^2 - m^2 \end{align*}
2005 Paper 2 Q14
The probability density function \(\f(x)\) of the random variable \(X\) is given by $$\f(x) = k\left[{\phi}(x) + {\lambda}\g(x)\right]$$ where \({\phi}(x)\) is the probability density function of a normal variate with mean 0 and variance 1, \(\lambda \) is a positive constant, and \(\g(x)\) is a probability density function defined by \[ \g(x)= \begin{cases} 1/\lambda & \mbox{for \(0 \le x \le {\lambda}\)}\,;\\ 0& \mbox{otherwise} . \end{cases} \] Find \(\mu\), the mean of \(X\), in terms of \(\lambda\), and prove that \(\sigma\), the standard deviation of \(X\), satisfies. $$\sigma^2 = \frac{\lambda^4 +4{\lambda}^3+12{\lambda}+12} {12(1 + \lambda )^2}\;.$$ In the case \(\lambda=2\):
- draw a sketch of the curve \(y=\f(x)\);
- express the cumulative distribution function of \(X\) in terms of \(\Phi(x)\), the cumulative distribution function corresponding to \(\phi(x)\);
- evaluate \(\P(0 < X < \mu+2\sigma)\), given that \(\Phi (\frac 23 + \frac23 \surd7)=0.9921\).
Solution: \begin{align*} && 1 &= \int_{-\infty}^{\infty} f(x) \d x \\ &&&= k[1 + \lambda] \\ \Rightarrow && k &= \frac{1}{1+\lambda} \\ \\ && \mu &= \int_{-\infty}^\infty x f(x) \d x \\ &&&= k \int_{-\infty}^\infty x \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x g(x) \d x \\ &&&= k \cdot 0 + k \lambda \cdot \frac{\lambda}{2} \\ &&&= \frac{\lambda^2}{2(1+\lambda)} \\ \\ && \E[X^2] &= \int_{-\infty}^\infty x^2 f(x) \d x \\ &&&= k \int_{-\infty}^\infty x^2 \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x^2 g(x) \d x \\ &&&= k \cdot 1 + k \lambda \int_0^{\lambda} \frac{x^2}{\lambda} \d \lambda \\ &&&= k + \frac{k \lambda^3}{3} \\ &&&= \frac{3+\lambda^3}{3(1+\lambda)} \\ && \var[X] &= \frac{3+\lambda^3}{3(1+\lambda)} - \frac{\lambda^4}{4(1+\lambda)^2} \\ &&& = \frac{(3+\lambda^3)4(1+\lambda) - 3\lambda^4}{12(1+\lambda)^2} \\ &&&= \frac{\lambda^4+4\lambda^3+12\lambda + 12}{12(1+\lambda)^2} \end{align*}
- \(\,\)
- \(\,\) \begin{align*} && \mathbb{P}(X \leq x) &= \int_{-\infty}^x f(x) \d x \\ &&&= \begin{cases} \frac13 \Phi(x) & \text{if } x < 0 \\ \frac13\Phi(x) + \frac13x & \text{if } 0 \leq x \leq 2 \\ \frac13 \Phi(x) + \frac23 & \text{if } 2 < x \end{cases} \end{align*} When \(\lambda = 2\), \(\mu = \frac{4}{6} = \frac23\), \(\sigma^2 = \frac{16+32+24+12}{12 \cdot 9} = \frac{7}{9}\), so \(\mu + 2 \sigma = \frac23 + \frac{2\sqrt7}{3}>2\). Therefore \begin{align*} && \P(0 < X < \mu + 2\sigma) &= \frac13 \Phi\left (\frac{2+2\sqrt{7}}{3} \right) + \frac23 - \Phi(0) \\ &&&= \tfrac13 \cdot 0.9921 +\tfrac23 - \tfrac12 \\ &&&= 0.4974 \end{align*}