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2008 Paper 3 Q12
D: 1700.0 B: 1516.0
moment generating functions Laplace distribution curve sketching variance central limit theorem normal approximation sum of independent random variables hypothesis testing

Let \(X\) be a random variable with a Laplace distribution, so that its probability density function is given by \[ \f(x) = \frac12 \e^{-\vert x \vert }\;, \text{ \(-\infty < x < \infty \)}. \tag{\(*\)} \] Sketch \(\f(x)\). Show that its moment generating function \({\rm M}_X(\theta)\) is given by \({\rm M}_X(\theta)= (1-\theta^2)^{-1}\) and hence find the variance of \(X\). A frog is jumping up and down, attempting to land on the same spot each time. In fact, in each of \(n\) successive jumps he always lands on a fixed straight line but when he lands from the \(i\)th jump (\(i=1\,,2\,,\ldots\,,n\)) his displacement from the point from which he jumped is \(X_i\,\)cm, where \(X_i\) has the distribution \((*)\). His displacement from his starting point after \(n\) jumps is \(Y\,\)cm (so that \(Y=\sum\limits_{i=1}^n X_i\)). Each jump is independent of the others. Obtain the moment generating function for \(Y/ \sqrt {2n}\) and, by considering its logarithm, show that this moment generating function tends to \(\exp(\frac12\theta^2)\) as \(n\to\infty\). Given that \(\exp(\frac12\theta^2)\) is the moment generating function of the standard Normal random variable, estimate the least number of jumps such that there is a \(5\%\) chance that the frog lands 25 cm or more from his starting point.

Solution:

TikZ diagram
\begin{align*} && M_X(\theta) &= \E \left [ e^{\theta X} \right] \\ &&&= \int_{-\infty}^{\infty} e^{\theta x} f(x) \d x \\ &&&= \int_{-\infty}^0 e^{\theta x}\frac12 e^{x} \d x+ \int_0^{\infty} e^{\theta x} \frac12 e^{-x} \d x \\ &&&= \frac12 \left [ \frac{1}{1+\theta}e^{(1+\theta)x} \right]_{-\infty}^0 +\frac12 \left [ \frac{1}{\theta-1}e^{(\theta-1)x} \right]_{0}^{\infty} \\ &&&= \frac12 \left ( \frac{1}{1+ \theta} + \frac{1}{1-\theta} \right) \\ &&&= \frac{1}{1-\theta^2} = (1-\theta^2)^{-1} \\ &&&= 1 + \theta^2 + \theta^4 + \cdots \\ \\ \Rightarrow && \E[X] &= 0 \\ && \E[X^2] &= 2 \\ \Rightarrow && \var[X] &= 2 \end{align*} \begin{align*} && M_{Y/\sqrt{2n}}(\theta) &= \E \left [ \exp \left ( \theta \frac{Y}{\sqrt{2n}} \right) \right] \\ &&&= \E \left [ \exp \left ( \frac{\theta }{\sqrt{2n}} \sum_{i=1}^n X_i \right) \right] \\ &&&= \E \left [ \prod_{i=1}^n \exp \left ( \frac{\theta }{\sqrt{2n}} X_i \right) \right] \\ &&&= \prod_{i=1}^n \E \left [\exp \left ( \frac{\theta }{\sqrt{2n}} X_i \right) \right] \tag{independence}\\ &&&= \prod_{i=1}^n M_{X_i} \left ( \frac{\theta }{\sqrt{2n}} \right)\\ &&&= \prod_{i=1}^n M_{X} \left ( \frac{\theta }{\sqrt{2n}} \right)\\ &&&= M_{X} \left ( \frac{\theta }{\sqrt{2n}} \right)^n\\ &&&= \left (1 - \frac{\theta^2}{2n} \right)^{-n} \to \exp(\tfrac12 \theta^2) \end{align*} Given that \(M_{Y/\sqrt{2n}} \to M_Z\) we assume that \(Y/\sqrt{2n} \to Z\) or \(Y/\sqrt{2n} \approx Z\). \begin{align*} && 5\% &\approx \mathbb{P}(|Z| > 2) \\ &&&\approx \mathbb{P} \left (|Y| > 2\sqrt{2n} \right) \end{align*} So we wish to choose \(n\) such that \(2\sqrt{2n} = 25\) or \(n = \frac{625}8 \approx 78\) so take \(n = 79\)

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