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Could someone help me to derive/understand how we can derive $E[X^2]-E[X]^2$ from $E[(X-E(X))^2]$ in the variance formula of a random variable $X$?

I am writing the formula below:

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

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    $\begingroup$ What is your particular difficulty? Can you expand the square in $E[(X-E(X))^2]$? $\endgroup$ Oct 30, 2022 at 10:33
  • $\begingroup$ I think many introductory statistics or probability theory textbooks (and lecture notes) derive the equivalence between the two expressions. You just need to find one to look it up. $\endgroup$ Oct 30, 2022 at 12:37
  • $\begingroup$ The answer that I was looking for was solved below, but thank you very much for your comments! $\endgroup$
    – Sera
    Oct 30, 2022 at 19:17

1 Answer 1

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

We FOIL the quadratic,

$$E[X^2 -2XE[X]+(E[X])^2]$$

We apply the expectation to each term,

$$E[X^2] -2E[XE[X]]+(E[X])^2$$

In the middle piece, $E[X]$ is a constant number that can be brought out of the expectation.

$$E[X^2] -2E[X]E[X]+(E[X])^2$$

We have that $E[X]E[X]=(E[X])^2$

$$E[X^2] -2(E[X])^2+(E[X])^2$$

We are done, $$E[X^2] -(E[X])^2$$

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    $\begingroup$ Thank you very much, that's exactly what I was looking for!! $\endgroup$
    – Sera
    Oct 30, 2022 at 19:15

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