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2001 Paper 3 Q14
A random variable \(X\) is distributed uniformly on \([\, 0\, , \, a\,]\). Show that the variance of \(X\) is \({1 \over 12} a^2\). A sample, \(X_1\) and \(X_2\), of two independent values of the random variable is drawn, and the variance \(V\) of the sample is determined. Show that \(V = {1 \over 4} \l X_1 -X_2 \r ^2\), and hence prove that \(2 V\) is an unbiased estimator of the variance of X. Find an exact expression for the probability that the value of \(V\) is less than \({1 \over 12} a^2\) and estimate the value of this probability correct to one significant figure.
Solution: \begin{align*} && \E[X] &= \frac{a}{2}\tag{by symmetry} \\ &&\E[X^2] &= \int_0^a \frac{1}{a} x^2 \d x \\ &&&= \frac{a^3}{3a} = \frac{a^2}{3} \\ \Rightarrow && \var[X] &= \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12} \\ \end{align*} \begin{align*} && V &=\frac{1}{2} \left ( \left ( X_1 - \frac{X_1+X_2}{2} \right )^2+\left ( X_2- \frac{X_1+X_2}{2} \right )^2 \right ) \\ &&&= \frac{1}{8} ((X_1 - X_2)^2 + (X_2 - X_1)^2 ) \\ &&&= \frac14 (X_1-X_2)^2 \\ \\ && \E[2V] &= \E \left [ \frac12 (X_1 - X_2)^2 \right] \\ &&&= \frac12 \E[X_1^2] - \E[X_1X_2] + \frac12 \E[X_2^2] \\ &&&= \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12} \end{align*} Therefore \(2V\) is an unbiased estimator of the variance of \(X\).
