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Let $Y$ be such a random variable, that :

$Y = \begin{cases} 1, & \mbox{if } X \le 1/2 \\ X, & \mbox{if } X>1/2 \end{cases}$

where $X$ has uniform distribution on $[0,1]$.

My solution:

$F_Y(t)=P(Y \le t) = P(1 \le t, X \le1/2)+P(X \le t, X>1/2)=1/2P(1 \le t)+1/2 F_X(t)$.

So, CDF of $Y$:

$F_y(t) = \begin{cases} 0, & \mbox{if } t <0 \\ 1/2 \cdot t, & \mbox{if } t \in [0,1) \\ 1, & \mbox{if } t \ge 1 \end{cases}$

PDF of $Y$ :

$f_y(t) = \begin{cases} 0, & \mbox{if } t <0 \wedge t\ge 1 \\ 1/2 , & \mbox{if } t \in [0,1) \end{cases}$

$E(Y)= \int_{0}^{1}\frac{1}{2}\cdot y dy$

I know, how to finish it, if the above is correct.

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  • $\begingroup$ $Y$ is a mixed random variable, so there is no pdf of $Y$ in the usual sense. Just use $E[g(X)]=\int g(x)f_X(x)\,dx$ for any (measurable) function $g$ where $f_X$ is the pdf of $X$. This is a consequence of this theorem. $\endgroup$ Commented Mar 3, 2020 at 16:16

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There is a mistake when you said that $P(X \le t, X>1/2)=1/2 F_X(t)$. This is not true. Otherwise the architecture of the argument is a good way to go indeed.

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  • $\begingroup$ So, I have $P(X \le t, X>1/2)=P(X \le t, 1-X \le 1/2)=P(X \le t)(1-P(X \le 1/2))=P(X \le t)(1-F_X(1/2))$ $\endgroup$ Commented Dec 5, 2019 at 18:18
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    $\begingroup$ Wow, be careful, you act as if $X$ is independent of itself! obviously this is never the case. Product formula works only when variables are independent. $\endgroup$ Commented Dec 5, 2019 at 18:19
  • $\begingroup$ What do you think about this ? : $E(Y)= \int_{0}^{\frac{1}{2}} 1 dx +\int_{\frac{1}{2}}^{1} x dx$. $\endgroup$ Commented Dec 5, 2019 at 18:24
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    $\begingroup$ Yes this is a good way to proceed too, that is the correct answer for the expectation. $\endgroup$ Commented Dec 5, 2019 at 18:27

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