If I roll two fair dice and let $Y$ be the number obtained by multipling the scores, the expected value of $Y$ is $E[Y]=\dfrac{49}{4}$. Let $X$ be the score obtained by rolling a single fair die. Then $E[X]=\dfrac{7}{2}$. Is it a coincidence that $E[Y]=(E[X])^2$?
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2no, its not a coincidence. See here. – Masacroso Jan 01 '18 at 21:47
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2No coincidence: see this. – Fabio Somenzi Jan 01 '18 at 21:47
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Theorem $1$. If $A$ and $B$ are independent discrete random variables, then $$E[A]\times E[B]=E[AB].$$
Proof.
$$\begin{aligned}E[AB]&=\sum_{a,b}abP(A=a \cap B=b)\\ &=\sum_{a,b}abP(A=a)P(B=b) \quad\quad\quad \text{ (since $A$ and $B$ are independent)}\\ &=\left(\sum_a aP(A=a)\right)\left(\sum_b bP(B=b)\right)\\ &=E[A]E[B] \end{aligned}$$
Theorem $2$. If $A$ and $B$ are independent continuous random variables, then $$E[A]\times E[B]=E[AB].$$
Proof.
See here.
A. Goodier
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