Hi, can someone help me prove this theorem:
If X has the variance σ^2 then var(aX+b)=[a^2]*[σ^2]
Many thanks!
You will find the desired proof in many statistics textbooks (a good reason to visit the library of your institute) and I'm sure Google will turn up plenty of websites that give a proof.
And I'm sure there's a proof somewhere on MHF. The Search toll might uncover it (as well as other interesting things).
What attempt have you made on the proof?
This is what I've worked on
Var(aX +b) = E[(aX +b)^2] − (E[aX +b])^2
= E[a^2X^2+2abX +b^2] − (aE[X] + b)^2
= a^2E[X2] + 2abE[X] + b^2
−(a^2(E[X])2+2abE[X] + b^2)
= a^2(E[X^2] − (E[X])^2)
= a^2Var(X)
I'm not the best with proofs so I'd like to see whether its the same as someone else's
That looks ok
BUT using the definition is easier, than the short cut formula.
LET $\displaystyle Y=aX+b$
Then $\displaystyle E(Y)=aE(X)+b$
and $\displaystyle V(Y)=E[(Y-E(Y))^2]$
$\displaystyle =E[(aX+b-aE(X)-b)^2]$
$\displaystyle =a^2E[(X-E(X))^2]=a^2V(X)$