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2011 Paper 2 Q13
D: 1600.0 B: 1500.0

What property of a distribution is measured by its skewness?

  1. One measure of skewness, \(\gamma\), is given by \[ \displaystyle \gamma= \frac{ \E\big((X-\mu)^3\big)}{\sigma^3}\,, \] where \(\mu\) and \(\sigma^2\) are the mean and variance of the random variable \(X\). Show that \[ \gamma = \frac{ \E(X^3) -3\mu \sigma^2 - \mu^3}{\sigma^3}\,. \] The continuous random variable \(X\) has probability density function \(\f\) where \[ \f(x) = \begin{cases} 2x & \text{for } 0\le x\le 1\,, \\[2mm] 0 & \text{otherwise}\,. \end{cases} \] Show that for this distribution \(\gamma= -\dfrac{2\sqrt2}{5}\).
  2. The decile skewness, \(D\), of a distribution is defined by \[D= \frac { {\rm F}^{-1}(\frac9{10}) - 2{\rm F} ^{-1}(\frac12) + {\rm F}^{-1} (\frac1{10}) } {{\rm F}^{-1}(\frac9{10}) - {\rm F} ^{-1} (\frac1{10})}\,, \] where \({\rm F}^{-1}\) is the inverse of the cumulative distribution function. Show that, for the above distribution, \( D= 2 -\sqrt5\,.\) The Pearson skewness, \(P\), of a distribution is defined by \[ P = \frac{3(\mu-M)}{\sigma} \,,\] where \(M\) is the median. Find \(P\) for the above distribution and show that \(D > P > \gamma\,\).


Solution: Skewness is a measure of the symmetry (specifically the lack-thereof) in the distribution. How much mass is there on one side rather than another.

  1. \(\,\) \begin{align*} && \gamma &= \frac{\E \left [ (X - \mu)^3 \right ]}{\sigma^3} \\ &&&= \frac{\E \left [ X^3 - 3\mu X^2 + 3\mu^2 X - \mu^3 \right ]}{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu \E[X^2] + 3\mu^2 \E[X] - \mu^3 }{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu (\mu^2 + \sigma^2) + 3\mu^2\cdot \mu- \mu^3 }{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu \sigma^2 - \mu^3 }{\sigma^3} \\ \end{align*} \begin{align*} && f(x) &= \begin{cases} 2x & \text{for } 0\le x\le 1\,, \\[2mm] 0 & \text{otherwise}\,. \end{cases} \\ && \E[X] &= \int_0^1 2x^2 \d x \\ &&&= \frac23 \\ && \E[X^2] &= \int_0^1 2x^3 \d x \\ &&&= \frac12 \\ && \E[X^3] &= \int_0^1 2x^4 \d x \\ &&&= \frac25 \\ \\ && \mu &= \frac23 \\ && \sigma^2 &= \frac12 - \frac49 = \frac{1}{18} \\ && \gamma &= \frac{\frac25 - 3 \cdot \frac23 \cdot \frac1{18} - \frac8{27}}{\frac{1}{54\sqrt2}} \\ &&&= -\frac{2\sqrt2}{5} \end{align*}
  2. First note that \(\displaystyle F(x) = \int_0^x 2t \d t = x^2\) for \(x \in [0,1]\). In particular, \(F^{-1}(x) = \sqrt{x}\), so \begin{align*} && D &= \frac { {\rm F}^{-1}(\frac9{10}) - 2{\rm F} ^{-1}(\frac12) + {\rm F}^{-1} (\frac1{10}) } {{\rm F}^{-1}(\frac9{10}) - {\rm F} ^{-1} (\frac1{10})} \\ &&&= \frac{\sqrt{\frac9{10}} - 2 \sqrt{\frac5{10}} + \sqrt{\frac1{10}}}{\sqrt{\frac9{10}}-\sqrt{\frac1{10}}} \\ &&&= \frac{3-2\sqrt5+1}{3 - 1} \\ &&&= \frac{4-2\sqrt5}{2} = 2-\sqrt5 \end{align*} \begin{align*} && P &= \frac{3(\mu - M)}{\sigma} \\ &&&= \frac{3(\frac23 - \sqrt{\frac12})}{\frac{1}{3\sqrt2}} \\ &&&= 6 \sqrt2 - 9 \end{align*} First we compare \(P\) and \(D\), \(6\sqrt2-9\) and \(2-\sqrt5\) \begin{align*} && D & > P \\ \Leftrightarrow && 2-\sqrt5 &> 6\sqrt2 - 9 \\ \Leftrightarrow && 11 -6\sqrt2 &> \sqrt 5 \\ \Leftrightarrow && (121 + 72 - 132\sqrt2) & > 5 \\ \Leftrightarrow && 188 & > 132\sqrt2 \\ \Leftrightarrow && 47 & > 33 \sqrt 2\\ \Leftrightarrow && 2209 & > 2178 \end{align*} also \begin{align*} && P &> \gamma \\ \Leftrightarrow && 6\sqrt2 - 9 &> -\frac{2\sqrt2}{5} \\ \Leftrightarrow && 30\sqrt2 - 45 & > -2\sqrt2 \\ \Leftrightarrow && 32 \sqrt 2 &> 45 \\ \Leftrightarrow && 2048 &> 2025 \end{align*}