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I would like to show that $Var(aX + b) = a^2 Var(X)$ using a different proof from my book

Let Y = aX + b

$Var(Y) = E(Y^2) - E[Y]^2$

= $E[(aX + b)^2] - E(aX + b)^2$

= $E[(a^2 X^2 + b^2 + 2abX] - (a\mu+b)^2$

= $a^2E(X^2) + b^2 + 2abE[X] - (a^2\mu^2 + b^2 + 2ab\mu)$

= $a^2E(X^2) + b^2 + 2ab\mu - (a^2E[X]^2 + b^2 + 2ab\mu)$

= $a^2E[X^2] - a^2E[X]^2$

= $a^2(E[X^2]-E[X]^2)$

= $a^2 Var(X)$

My book does like three lines and they start out with $Var(aX + b) = E[(aX + b - (a\mu + b) ) ^2]$. I don't understand why they substract $(a\mu + b)$ instead of $\mu$ because $Var(X) = E[(X - \mu)^2]$

EDIT: I am basically asking if my proof is correct or not

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2 Answers 2

up vote 1 down vote accepted

As mentioned above, they subtract $a\mu + b$ because that is the mean of $Y$. Every time you find the variance of a random variable from the definition, you need to subtract its mean in the formula.

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Note that $E[Y] = E[aX + b] = aE[X] + b = a\mu + b$.

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Yeah I know, I used that fact for my second line of proof –  sidht Oct 17 '12 at 5:40
    
And what you're trying to find is $Var(Y)$, not $Var(X)$. –  Tunococ Oct 17 '12 at 6:45
    
Yeah and I got my result to be a^2 Var(X) –  sidht Oct 17 '12 at 21:52
1  
I am not sure if you feel that I answered your question or not. I basically said the same thing as Ken Dunn. You asked why they subtract $a\mu + b$ instead of $\mu$. My answer is because you're computing $Var(Y)$, and $E[Y] = a\mu + b$. –  Tunococ Oct 18 '12 at 9:59
    
Is my alternative proof correct though? –  sidht Oct 19 '12 at 1:26

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