variance of $aX+b$
How do I show that $$\text{Var}(aX+b)=a^2\text{Var}(X)$$. Since, I am reading statistic first time, I don't have any idea how to start.
Thanks for helping me.
$\endgroup$ 02 Answers
$\begingroup$See the solution is easy but at least you have to try once. Just applying the definition of variance you will get the desired result. Although I am writing the solution but please try by yourself.
$\endgroup$ 1 $\begingroup$$$\begin{align*}\mathsf{Var}(X)& =\mathsf E[X^2]-\mathsf E[X]^2\\[2ex]\implies \mathsf{Var}(aX+b) &=\mathsf E[(aX+b)^2]-\mathsf E[aX+b]^2\\[1ex]& = \mathsf E[a^2X^2+2abX+b^2]-(a\mathsf E[X]+b)^2 \quad \text{as } \mathsf E[aX]=a\mathsf E[X]\ \text{and}\ \mathsf E[b]=b\\[1ex]& = a^2\mathsf E[X^2]+2ab\mathsf E[X]+b^2-a^2\mathsf E[X]^2-2ab\mathsf E[X]-b^2\\[1ex]& = a^2(\mathsf E[X^2]-\mathsf E[X]^2)\\[1ex]& = a^2\mathsf {Var}(X)\end{align*}$$
Directly from the definition: $Var(aX)=E[(aX)^2] - E[(aX)]^2=E[a^2X^2]-E[(aX)]^2=a^2E[X^2]-(aE[X])^2=a^2E[X^2]-a^2E[X]^2=a^2(E[X^2]-E[X]^2)=a^2Var(X),$ where in the third and fourth equality, I have applied the linearity of Expectation, in the form $E[cX]=cE[X]$.
Next, observe $Var[Y+b]=Var(Y)$, with a similiar proof to the above, using directly the definition of $Var[Y]$, again.
Putting the two together, you have $Var(aX+b)= a^2Var(X)$. q.e.d.
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