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In a game of roulette if you bet \$1 you either win \$35 with a probability of 1/38 or loose \$1 with a probability of 37/38. Thus I calculate the chance of winning on a single play as $E(W)=35/38 -37/38 = -1/19$

It has been explained to me that we calculate Variance using

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

In our case $f(X)=W$

and I am told that $E[W^2]=35^2\frac{1}{38}+(-1)^2 \frac{37}{38}$

Why can we not use $E[W^2]=(-1/19)^2$ ?

Kirsten
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    $E\left[ X^2\right]\neq E[X]^2$. For $\textit {independent}$ variables $X,Y$ we have $E[XY]=E[X]E[Y]$ but $X$ is not generally independent of itself. – lulu Oct 01 '20 at 16:06
  • Thanks @lulu I clarified the question – Kirsten Oct 01 '20 at 16:15
  • I don't see where the edit fixed the error. You say you want to use $E\left[W^2\right]=E[W]^2$ but that equality is false. – lulu Oct 01 '20 at 16:16
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    To state the obvious: If $E\left[X^2\right]=E[X^2]$ then $\text{Var}(X)=E\left[X^2\right]-E[X]^2=0$, which really only holds if the variable is constant. – lulu Oct 01 '20 at 16:19
  • That's the answer. When you say it like that it becomes obvious. Thank you. For some reason it was not obvious to me before. – Kirsten Oct 01 '20 at 16:22

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