If Y=aX+b, show that Y has the same coefficient of skewness?
The skewness of a random variable X is the third standardized moment, defined as "\\tilde{\\mu}^3(X)=\\sigma(X)^{-3}E[(X-\\mu(X))^3]" , where "\\mu(X),\\ \\sigma(X)" are the mean and the standard deviation of X. If "Y=aX+b", then
"\\mu(Y)=E(aX+b)=a\\mu(X)+b"
"Var(Y)=E((Y-\\mu(Y))^2)=E((aX+b-(a\\mu(X)+b))^2)=a^2E((X-\\mu(X))^2)=a^2Var(X)"
"\\sigma(Y)=\\sqrt{Var(Y)}=\\sqrt{a^2Var(X)}=|a|\\sigma(X)"
"\\tilde{\\mu}^3(Y)=\\sigma(Y)^{-3}E[(Y-\\mu(Y))^3]=|a|^{-3}\\sigma(X)^{-3}E[(aX+b-(a\\mu(X)+b))^3]=\\frac{a^3}{|a|^3}E((X-\\mu(X))^3)=\\mathrm{sign}(a) \\tilde{\\mu}^3(X)"
Therefore, if a>0, then Y has the same coefficient of skewness as X. If a<0 then X, Y have the same absolute values of the coefficient of skewness, but their signs are different.
If a=0, then the coefficient of skewness of Y is undefined.
Comments
Leave a comment