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1997 Paper 3 Q13
D: 1700.0 B: 1500.0

Let \(X\) and \(Y\) be independent standard normal random variables: the probability density function, \(\f\), of each is therefore given by \[ \f(x)=\left(2\pi\right)^{-\frac{1}{2}}\e^{-\frac{1}{2}x^{2}}. \]

  1. Find the moment generating function \(\mathrm{E}(\e^{\theta X})\) of \(X\).
  2. Find the moment generating function of \(aX+bY\) and hence obtain the condition on \(a\) and \(b\) which ensures that \(aX+bY\) has the same distribution as \(X\) and \(Y\).
  3. Let \(Z=\e^{\mu+\sigma X}\). Show that \[ \mathrm{E}(Z^{\theta})=\e^{\mu\theta+\frac{1}{2}\sigma^{2}\theta^{2}}, \] and hence find the expectation and variance of \(Z\).

Solution:

  1. \(\,\) \begin{align*} && \E[e^{\theta X}] &= \int_{-\infty}^{\infty} e^{\theta x} \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2 } \d x\\ &&&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2+\theta x} \d x\\ &&&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x^2-2\theta x)} \d x\\ &&&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x-\theta )^2+\frac12\theta^2 } \d x\\ &&&= e^{\frac12\theta^2 }\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x-\theta )^2 } \d x\\ &&&=e^{\frac12\theta^2 } \end{align*}
  2. \begin{align*} && M_{aX+bY} (\theta) &= \mathbb{E}[e^{\theta (aX+bY)}] \\ &&&= e^{\frac12(a\theta)^2} \cdot e^{\frac12(b\theta)^2} \\ &&&= e^{\frac12(a^2+b^2)\theta^2} \end{align*} Therefore we need \(a^2+b^2 = 1\)
  3. \(\,\) \begin{align*} && \E[Z^\theta] &= \E[e^{\mu \theta + \sigma \theta X}] \\ &&&= e^{\mu \theta}e^{\frac12 \sigma^2 \theta^2} \\ &&&=e^{\mu \theta + \frac12 \sigma^2 \theta^2} \\ \end{align*} \begin{align*} \mathbb{E}(Z) &= \mathbb{E}[Z^1] \\ &= e^{\mu + \frac12 \sigma^2} \\ \var[Z] &= \E[Z^2] - \left ( \E[Z] \right)^2 \\ &= e^{2 \mu+ 2\sigma^2} - e^{2\mu + \sigma^2} \\ &= e^{2\mu+\sigma^2} \left (e^{\sigma^2}-1 \right) \end{align*} [NB: This is the lognormal distribution]

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1997 Paper 3 Q14
D: 1700.0 B: 1516.0

An industrial process produces rectangular plates of mean length \(\mu_{1}\) and mean breadth \(\mu_{2}\). The length and breadth vary independently with non-zero standard deviations \(\sigma_{1}\) and \(\sigma_{2}\) respectively. Find the means and standard deviations of the perimeter and of the area of the plates. Show that the perimeter and area are not independent.

Solution: Let \(L \sim N(\mu_1, \sigma_1^2)\), \(B \sim N(\mu_2, \sigma_2)^2\), so \begin{align*} && \mathbb{E}(\text{perimeter}) &= \E(2(L+B)) \\ &&&= 2\E[L]+2\E[B] \\ &&&= 2(\mu_1+\mu_2) \\ &&\var[\text{perimeter}] &= \E\left [ (2(L+B))^2 \right] - \left ( \E[2(L+B)] \right)^2 \\ &&&= 4\E[L^2+2LB+B^2] - 4(\mu_1+\mu_2)^2 \\ &&&= 4(\sigma_1^2+\mu_1^2+2\mu_1\mu_2+\sigma_2^2+\mu_2^2) - 4(\mu_1+\mu_2)^2\\ &&&= 4(\sigma_1^2+\sigma_2^2) \\ &&\text{sd}[\text{perimeter}] &= 2\sqrt{\sigma_1^2+\sigma_2^2} \\ \\ && \E[\text{area}] &= \E[LB] \\ &&&= \E[L]\E[B] \\ &&&= \mu_1\mu_2 \\ && \var[\text{area}] &= \E[(LB)^2] - \left (\E[LB] \right)^2 \\ &&&= \E[L^2]\E[B^2]-\mu_1^2\mu_2^2 \\ &&&= (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) -\mu_1^2\mu_2^2 \\ &&&= \sigma_1^2\mu_2^2 + \sigma_2^2\mu_1^2 + \sigma_1^2\sigma_2^2\\ && \text{sd}(\text{area}) &= \sqrt{\sigma_1^2\mu_2^2 + \sigma_2^2\mu_1^2 + \sigma_1^2\sigma_2^2} \\ \\ && \E[\text{perimeter} \cdot \text{area}] &= \E[2(L+B)LB] \\ &&&= 2\E[L^2]\E[B] + 2\E[L]\E[B^2] \\ &&&= 2(\sigma_1^2+\mu_1^2)\mu_2 + 2(\sigma_2^2+\mu_2^2)\mu_1 \\ && \E[\text{perimeter}] \E[\text{area}] &= 2(\mu_1+\mu_2) \cdot \mu_1\mu_2 \end{align*} Since the latter does not depend on \(\sigma_i\) but the former does they cannot be equal in general, therefore they cannot be independent. [See also STEP 2006 Paper 3 Q14]

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