Question : Prove that the variance of cX+d is c^2 V(X)+d where X is a continuous random variable.
Solution :
We know that variance of a continuous random variable is \[Var\left(X\right)=\int_{-\infty}^{\infty}{x^2\ f\left(x\right)dx}-\left(\int_{-\infty}^{\infty}{x\ f\left(x\right)dx}\right)^2\] And variance of a function g(X) of continuous random variable is \[Var\left(g\left(X\right)\right)=\int_{-\infty}^{\infty}{g\left(x\right)^2f\left(x\right)dx}-\left(\int_{-\infty}^{\infty}{g(x)\ f\left(x\right)dx}\right)^2\] Let g(X) = cX+d \[Var\left(cX+d\right)=\int_{-\infty}^{\infty}{\left(cx+d\right)^2f\left(x\right)dx}-\left(\int_{-\infty}^{\infty}{\left(cx+d\right)\ f\left(x\right)dx}\right)^2\] \[=c^2\int_{-\infty}^{\infty}{x^2f\left(x\right)dx}+d^2\int_{-\infty}^{\infty}f\left(x\right)dx+2cd\int_{-\infty}^{\infty}xf\left(x\right)dx-\left(c\int_{-\infty}^{\infty}{x\ f\left(x\right)dx}+d\int_{-\infty}^{\infty}{\ f\left(x\right)dx}\right)^2\] When X is a continuus random variable then \[\int_{-\infty}^{\infty}{\ f\left(x\right)dx}=1, \mu=\int_{-\infty}^{\infty}xf\left(x\right)dx\] Hence \[Var\left(cX+d\right)=c^2\int_{-\infty}^{\infty}{x^2f\left(x\right)dx}+d^2+2cd\ \mu-\left(c\ \mu+d\right)^2\] \[=c^2\int_{-\infty}^{\infty}{x^2f\left(x\right)dx}+d^2+2cd\ \mu-c^2\ \mu^2-d^2-2cd\mu\ =c^2\left(\int_{-\infty}^{\infty}{x^2f\left(x\right)dx}-\mu^2\right)\] \[Var\left(cX+d\right)=c^2Var\left(X\right)\] Therefore variance of cX+d is c^2 Var(X)+d .