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The mainstream way to show $V[aX+b]= a^2 V[X]$ is by using LOTUS. However, LOTUS seems to me too powerful and out-of-reach for a last-year high-school student.

Therefore I was wondering if we could derive it from the definition of Variance:

$V[aX+b]=\sum (x-E[aX+b])^2 p(aX+b=x)$

We know that $E[aX+b] = aE[X]+b$ but I don't know how to handle $p(aX+b=x)$.

I was thinking to do $p(aX+b=x)=p((aX+b)^{-1} (\{ x\}))=p(aX^{-1} (\{ x\}) +b)$

but I don't know if that makes sense to add to a point in the sample space ($X^{-1}(\{ x\})$) a constant $b$.

Do you think it is possible to derive $V[aX+b]=a^2 V[X]$ without the LOTUS formula ?

I am working with finite discrete sample spaces, and a RV is here a function from the sample space to $\mathbb R$.

Moreover, $E$ is defined as $E[X]=\sum_i x_i p(X=x_i) $

Thanks i.a.

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    $\begingroup$ en.wikipedia.org/wiki/Law_of_the_unconscious_statistician for others who wonder $\endgroup$
    – Tommi
    Commented Mar 17 at 15:13
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    $\begingroup$ You need to give the precise definition of random variable and expected value you are using. In high school I assume you are not using the measure theoretic notion, so what definitions are you using? Are you only working with finite discrete event spaces? $\endgroup$ Commented Mar 17 at 16:14
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    $\begingroup$ Yes I am working with finite discrete event spaces. A RV is here a function from a sample space to $\mathbb R$. $E[X]=\sum x_i p(X=x_i) $ $\endgroup$
    – niobium
    Commented Mar 17 at 16:22
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    $\begingroup$ I'm inclined to disagree with your assertion that "LOTUS" is the mainstream way to do this. I think Kostya's answer is better than the one that you accepted. $\endgroup$ Commented Mar 23 at 3:44

3 Answers 3

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I think you will want to start by convincing the audience that $p(aX+b = ax_i + b)$ is equal to $p(X=x_i)$, probably with examples. I am not an expert statistician so please let me know politely if I am confused by what you mean here.

Assuming that introduction is correct, I would then rename that whole mess $p_i$, because it is a distraction from where the action is really happening. So the notation I would be using is

  • $E[X] = \sum_i x_i p_i$
  • $V[X] = \sum_i (x_i-E[X])^2p_i$
  • $V[aX+b] = \sum_i (ax_i + b - E[aX+b])^2p_i$

Then proceed like:

$\begin{align*} V[aX+b] & = \sum_i (ax_i+b - E[aX+b])^2p_i \\ & = \sum_i (ax_i+b - aE[X]-b)^2p_i \\ & = \sum_i (ax_i - aE[X])^2p_i \\ & = a^2 \sum_i (x_i - E[X])^2p_i \\ & = a^2 V[X] \\ \end{align*} $

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This is a consequence of the definition of the variance (1) the linearity of expectation (2) and an algebraic manipulation (3): $$V(aX+b)\stackrel{(1)}{=}\mathbb{E}(aX+b-\mathbb{E}(aX+b))^2\stackrel{(2)}{=}\mathbb{E}(aX+b-a\mathbb{E}(X) - b))^2\stackrel{(3)}{=}\mathbb{E}[a^2(X-\mathbb{E}(X))^2]\stackrel{(2)}{=}a^2\mathbb{E}(X-\mathbb{E}(X))^2\stackrel{(1)}{=}a^2V(X).$$

So, you don't really need to invoke the definition or formula for the expectation to prove this, just use its properties.

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Do it in two easy steps:

  1. Prove that $V[X+b]=V[X]$. This one is easy to prove since variance is a measure of deviation from the mean, hence change of origin will not affect it. Mathematically, $ E({X+b})=E(X)+b$ $$V[X+b] = E[((X+b)-E(X+b))^2] = E[(X-E(X))^2] = V[X] $$

  2. Prove that $V[aX]=a^2V[X]$. This one is easy to prove as well. Start with $E[aX]=aE[X]$. Then, $$ V[aX] = E[((aX)-E(aX))^2] = E[a^2(X-E(X))^2] = a^2V[X] $$

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