MATH103 April 2005
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Question 02

A certain random variable X takes values in the interval $0\leq x\leq 2$, with probability density
 $\displaystyle p(x)=1x/2$
Find the mean and variance of X.

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Hint 2

For a continuous probability density p(x) with mean $\mu$ the variance is given by
 $V=\int _{a}^{b}(x\mu )^{2}p(x)dx$
Alternatively, the variance can also be calculated by
 $V=\int _{a}^{b}x^{2}p(x)dx\mu ^{2}$

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Solution

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Given the probability density p(x)=1x/2 over the interval $0\leq x\leq 2$ we plug this into the formula for the mean to get
 ${\begin{aligned}\mu &=\int _{0}^{2}x\left(1{\frac {x}{2}}\right)dx\\&=\int _{0}^{2}\left(x{\frac {x^{2}}{2}}\right)dx\\&=\left[{\frac {x^{2}}{2}}{\frac {x^{3}}{6}}\right]_{0}^{2}\\&=\left({\frac {2^{2}}{2}}{\frac {2^{3}}{6}}\right)\left({\frac {0^{2}}{2}}{\frac {0^{3}}{6}}\right)\\&=\mathbf {\frac {2}{3}} \end{aligned}}$
Using the first formula for the variance with $\mu ={\frac {2}{3}}$,
 ${\begin{aligned}V&=\int _{0}^{2}\left(x{\frac {2}{3}}\right)^{2}\left(1{\frac {x}{2}}\right)dx\\&=\int _{0}^{2}\left(x^{2}{\frac {4}{3}}x+{\frac {4}{9}}\right)\left(1{\frac {x}{2}}\right)dx\\&=\int _{0}^{2}\left({\frac {x^{3}}{2}}+{\frac {5}{3}}x^{2}{\frac {14}{9}}x+{\frac {4}{9}}\right)dx\\&=\left[{\frac {x^{4}}{8}}+{\frac {5}{9}}x^{3}{\frac {7}{9}}x^{2}+{\frac {4}{9}}x\right]_{0}^{2}\\&=\left({\frac {2^{4}}{8}}+{\frac {5}{9}}(2)^{3}{\frac {7}{9}}(2)^{2}+{\frac {4}{9}}(2)\right)\left({\frac {0^{4}}{8}}+{\frac {5}{9}}(0)^{3}{\frac {7}{9}}(0)^{2}+{\frac {4}{9}}(0)\right)\\&={\frac {16}{8}}+{\frac {40}{9}}{\frac {28}{9}}+{\frac {8}{9}}0=2+{\frac {20}{9}}\\&=\mathbf {\frac {2}{9}} \end{aligned}}$
Alternatively, using the other formula for the variance, we get the same value:
 ${\begin{aligned}V&=\int _{0}^{2}x^{2}\left(1{\frac {x}{2}}\right)dx\left({\frac {2}{3}}\right)^{2}\\&=\int _{0}^{2}\left(x^{2}{\frac {x^{3}}{2}}\right)dx{\frac {4}{9}}\\&=\left[{\frac {x^{3}}{3}}{\frac {x^{4}}{8}}\right]_{0}^{2}{\frac {4}{9}}\\&=\left({\frac {2^{3}}{3}}{\frac {2^{4}}{8}}\right)\left({\frac {0^{3}}{3}}{\frac {0^{4}}{8}}\right){\frac {4}{9}}={\frac {8}{3}}2{\frac {4}{9}}\\&=\mathbf {\frac {2}{9}} \end{aligned}}$

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