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I want to understand something about the derivation of $\text{Var}(X) = E[X^2] - (E[X])^2$

Variance is defined as the expected squared difference between a random variable and the mean (expected value): $\text{Var}(X) = E[(X - \mu)^2]$

Then:

$\operatorname{Var}(X) = E[(X - \mu)^2]$

$\operatorname{Var}(X) = E[(X - E[X])^2]$

$\operatorname{Var}(X) = E[(X - E[X])(X - E[X])]$

$\operatorname{Var}(X) = E[X^2 - 2XE[X] + (E[X])^2]$

$\operatorname{Var}(X) = E[X^2] - 2E[XE[X]] + E[(E[X])^2]$

$\operatorname{Var}(X) = E[X^2] - 2E[E[X]E[X]] + E[(E[X])^2]$

$\operatorname{Var}(X) = E[X^2] - 2(E[X])^2 + (E[X])^2$

$\operatorname{Var}(X) = E[X^2] - (E[X])^2$

What I don't quite understand is the steps that get us from $E[XE[X]]$ to $E[E[X]E[X]]$ to $(E[X])^2$, also $E[(E[X])^2]$ to $(E[X])^2$.

While I'm sure these jumps are intuitive and obvious I would still like to understand how we can (more formally) make these jumps / consider them mathematically equivalent.

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    $\begingroup$ E[c] = c when c is a constant. E[X] is a constant itself, so E[E[X]] = E[X]. $\endgroup$ – o0BlueBeast0o Jul 18 '16 at 20:37
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    $\begingroup$ I prefer to think of it as $E[X~E[X]] = E[X]\cdot E[X]$. The expectation operator $E[~]$ is linear, so $E[X+Y] = E[X]+E[Y]$. Also, $E[\alpha X] = \alpha E[X]$ for constant $\alpha$. As $E[X]$ is a constant, the constant can be pulled out of $E[X~E[X]]$ $\endgroup$ – JMoravitz Jul 18 '16 at 20:39
  • $\begingroup$ Isn't anything a constant assuming we know the answer on the righthand side of an equation? $\endgroup$ – user6596353 Jul 18 '16 at 20:42
  • $\begingroup$ And how can we prove that $E[c] = c$ for constant $c$? $\endgroup$ – user6596353 Jul 18 '16 at 20:43
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    $\begingroup$ If the random variable $W$ is the amount you get from one play of a gambling game, then $E(W)$ is the average amount you get. If $W$ is constant, say $c$, then every time you get $c$, so on average you get $c$. $\endgroup$ – André Nicolas Jul 18 '16 at 20:55
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$\newcommand{\E}{\operatorname{E}}$It should not have been written as $$ \E[X\E[X]] = \E[\E[X]\E[X]]. $$ Instead, it should have said $$ \E[X\E[X]] = \E[X] \E[X]. $$ The justification is this: $$ \E[X\cdot5] = 5\E[X], $$ and similarly for any other constant besides $5$. And in this context, "constant" means "not random". So just treat $\E[X]$ the same way you treat $5$, because it's a constant.

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