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.

  • 8
    $\begingroup$ en.wikipedia.org/wiki/Law_of_the_unconscious_statistician for others who wonder $\endgroup$
    – Tommi
    Mar 17 at 15:13
  • 4
    $\begingroup$ 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? $\endgroup$ Mar 17 at 16:14
  • 1
    $\begingroup$ 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
    Mar 17 at 16:22
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 23 at 3:44

3 Answers 3


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*} $


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.