Science:Math Exam Resources/Courses/MATH103/April 2016/Question 09 (d)

MATH103 April 2016

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Question 09 (d)
Let $x$ be a continuous random variable taking values in $[0,100]$ with probability density function $p(x)$, mean value $\mu$, variance $\sigma ^{2}$, median value $x_{\text{med}}$, and cumulative function $F(x)$. Show that $\mu =\int _{0}^{100}(1-F(x))\,dx$. Hint. The formula for $\mu$ in (a) is a good starting point.

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Apply integration by parts to the answer in part (a).

Hint 2
Remember that for a probability density function $p(x),x\in [0,100],$ $\int _{0}^{100}p(x)\,dx=1,$ and the cumulative density function is given by $F(x)=\int _{0}^{x}p(t)\,dt.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By part (a), we have $\mu =\int _{0}^{100}xp(x)\,dx.$ Following the hint, we integrate by parts. Recalling the integration by parts formula, $\int _{a}^{b}u\,dv=uv|_{a}^{b}-\int _{a}^{b}v\,du,$ we take ${\color {OrangeRed}u=x}$ and $dv=p(x)\,dx$, so that $du=dx,{\color {red}v}=\int _{0}^{x}p(t)\,dt={\color {red}F(x)}$ (i.e., $F$ is an antiderivative of $p$; see hint 2). Then $\mu =\int _{0}^{100}xp(x)\,dx=\left.xF(x)\right|_{0}^{100}-\int _{0}^{100}F(x)\,dx.$ Now $\left.xF(x)\right|_{0}^{100}=100F(100)-0,$ but $p$ is a probability density function, so $F(100)=\int _{0}^{100}p(x)\,dx=1.$ Therefore $\mu =100-\int _{0}^{100}F(x)\,dx=\int _{0}^{100}1\,dx-\int _{0}^{100}F(x)\,dx=\color {blue}\int _{0}^{100}(1-F(x))\,dx.$