Question Number 6605 by WAI LIN last updated on 05/Jul/16
![Let c be a constant. Using Var(X) = E [(X− E [X])^2 ], show that (a) Var(c X) = c^2 Var(X); (b) Var ( c+X) = Var(X).](https://www.tinkutara.com/question/Q6605.png)
$${Let}\:{c}\:{be}\:{a}\:{constant}.\:{Using}\:{Var}\left({X}\right)\:=\:{E}\:\left[\left({X}−\:{E}\:\left[{X}\right]\right)^{\mathrm{2}} \right],\:{show}\:{that} \\ $$$$\left({a}\right)\:{Var}\left({c}\:{X}\right)\:=\:{c}^{\mathrm{2}} \:{Var}\left({X}\right); \\ $$$$\left({b}\right)\:{Var}\:\left(\:{c}+{X}\right)\:=\:{Var}\left({X}\right). \\ $$
Commented by prakash jain last updated on 05/Jul/16
![E(cX)=cE(X) E(c+X)=c+E(X) Var(cX)=E[(cX−cE(X))^2 ] =E[c^2 (X−E(X))^2 ]=c^2 E[(X−E(X))^2 ]=c^2 Var(X) Var(c+X)=E[(c+X−c+E(X))^2 ]=Var(X)](https://www.tinkutara.com/question/Q6618.png)
$${E}\left(\mathrm{c}{X}\right)={cE}\left({X}\right) \\ $$$${E}\left({c}+{X}\right)={c}+{E}\left({X}\right) \\ $$$${Var}\left({cX}\right)={E}\left[\left({cX}−{cE}\left({X}\right)\right)^{\mathrm{2}} \right] \\ $$$$={E}\left[{c}^{\mathrm{2}} \left({X}−{E}\left({X}\right)\right)^{\mathrm{2}} \right]={c}^{\mathrm{2}} {E}\left[\left({X}−{E}\left({X}\right)\right)^{\mathrm{2}} \right]={c}^{\mathrm{2}} {Var}\left({X}\right) \\ $$$${Var}\left({c}+{X}\right)={E}\left[\left({c}+{X}−{c}+{E}\left({X}\right)\right)^{\mathrm{2}} \right]={Var}\left({X}\right) \\ $$$$ \\ $$