0

Suppose that $X$ is a discrete random variable with $E(X) = 1$ and $E[X(X −2)]= 3$. Find $Var(−3X + 5)$.

$$E[X(X-2)]=E(X^2-2X)=3$$ $$Var(-3X+5)=E[(-3X+5-1)^2 ]=E(9X^2-24X+16)$$ $$E(9X^2-24X+16)=9E(X^2-2X)-6E(X)+16=37$$

I found the answer to be $27-6+16=37$, but the solution manual says the answer is 36. I've triple-checked the last calculation and don't believe I've made an error, so I'm not sure why my answer is wrong.

LadymadeStar
  • 71

1 Answers1

2

$var (-3X+5)=E(-3X+5-2)^{2}$ since $E(-3X+5)=-3+5=2$. So your second line is wrong. The correct answer is $36$.

Kavi Rama Murthy
  • 311,013
  • Note also that some confusion could have been avoided by using $\mathsf{Var}(aZ+b)=a^2\mathsf{Var}(Z)$:$$\small\begin{align}\mathsf{Var}(-3X+5)&=9~\mathsf {Var}(X)\&=9~\mathsf E((X-\mathsf E(X))^2)\&=9~\mathsf E((X-1)^2)\&=9~\mathsf E(X^2-2X+1)\end{align}$$ – Graham Kemp Jan 15 '20 at 01:37