- #1

mathmari

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For a random variable $X$ the skewness is defined by \begin{equation*}\eta (X):=E\left (\left (\frac{X-\mu }{\sigma}\right )^3\right )\end{equation*} where $E(X)=\mu$ and $\text{Var}(X)=\sigma^2$.

I want to show that \begin{equation*}\eta (X)=\frac{E(X^3)-3E(X)E(X^2)+2E(X)^3}{\sigma^3}\end{equation*} and that \begin{equation*}a>0 \Rightarrow \eta (aX+b)=\eta (X)\end{equation*} I have done the following:

Is $\sigma$ a real constant? So by linearity of the expected value we can write the term $\frac{1}{\sigma^3}$ outside of $E$, right? (Wondering)

\begin{align*}\eta (X)&=E\left (\left (\frac{X-\mu }{\sigma}\right )^3\right ) \\ & = E\left ( \frac{(X-\mu)^3 }{\sigma^3} \right ) \\ & = \frac{1}{\sigma^3}\cdot E\left ( (X-\mu)^3 \right )\\ & = \frac{E\left (X^3-3X^2\mu+3X\mu^2-\mu^3 \right )}{\sigma^3} \\ & =\frac{E\left ( X^3-3X^2(E(X))+3X(E(X))^2-(E(X))^3 \right )}{\sigma^3} \\ & = \frac{E[ X^3]-E[3X^2(E(X))]+E[3X(E(X))^2]-E[(E(X))^3]}{\sigma^3} \\ & = \frac{E[ X^3]-3E[X^2(E(X))]+3E[X(E(X))^2]-E[(E(X))^3]}{\sigma^3} \end{align*}

Is everything correct so far? How could we continue? (Wondering)