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Let $X \sim U[-1,1]$. We define :

$Y=\begin{cases} X^2 &\text{for } X > \frac{1}{2} \\ 3 &\text{for } X \le \frac{1}{2} \end{cases}$

Find expected value of $Y$. My solution : $ E(Y)=E\left( X^2 \cdot \hbox{1}_{\left\{ X > \frac{1}{2}\right\} }+ 3 \cdot \hbox{1} _{\left\{ X \le \frac{1}{2}\right\} }\right)$. And I have sum of two integrals : $\int_{ \frac{1}{2} }^{1}x^2\cdot \frac{1}{2}dx + \int_{-1}^{ \frac{1}{2} } \frac{3}{2} dx$.

What do you think? It's correct?

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    $\begingroup$ Yes it is correct. $\endgroup$ – StubbornAtom Dec 15 '18 at 11:27

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