Distribution of $Y$

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We have that $X\sim U[0,2]$ and $Y=\min (X,1)$.

I want to find the distribution of $Y$.

We have that \begin{align*}F_Y(y)&=P(Y\leq y)=P(\min (X,1)\leq y)=P([X\leq y] \text{ or } \ [1\leq y])=P(\{X\leq y\}\cup \{1\leq y\})\\ & =P(X\leq y)+P(1\leq y)\end{align*}

Is everything correct so far?

Since $X\sim U[0,2]$ do we have that $P(X\leq y)=y$ ? What does it hold for $P(1\leq y)$ ?

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Coul you give me also a hint how we could calculate the expected value of $Y$ ?

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$P\big(\{X\le y\}\cup\{1\le y\}\big)$ need not be equal to $P(X\le y)+P(1\le y)$, unless the two events are disjoint. We have $$P(X\le y)=\begin{cases}0,&y<0\\\frac{y}{2},&0\le y< 2\\1,&2\le y\end{cases}$$ Also trivially, we have $$P(1\le y)=\begin{cases}0,&y<1\\1,&y\ge 1\end{cases}$$ Then we have $\{X\le y\}\cap\{1\le y\}=\begin{cases}\emptyset,&y<1\\\{X\le y\},&y\ge 1\end{cases}$

So $$F_Y(y)=P\big(\{X\le y\}\cup\{1\le y\}\big)=P(X\le y)+P(1\le y)-P\big(\{X\le y\}\cap\{1\le y\}\big)=\begin{cases}P(X\le y)+P(1\le y)=\frac{y}{2},&y<1\\P(1\le y)=1,&y\ge 1\end{cases}$$

Thus $F_Y$ has a jump discontinuity at $y=1$ and we get $$P(Y=1)=P(Y\le 1)-P(Y<1)=F_Y(1)-\lim_{y\to 1,\,y<1}F_Y(y)=1-\frac{1}{2}=\frac{1}{2}.$$

We also have $f_Y(y)=\frac{1}{2},\,y<1$. So $$E[Y]=\int_0^1yf_Y(y)dy+1\cdot P(Y=1)=\int_0^1\frac{y}{2}dy+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}.$$

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Actually, we just have $$f_Y(y)=\frac12$$ for $y<1$, and $P(Y=1)=\frac12$.

Hence, $$E[Y]=\int_0^1\frac{y}{2} \ dy +1\cdot\frac{1}{2}$$

0
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Hint for finding CDF:

$$\begin{aligned}1-F_{Y}\left(y\right) & =\mathsf{P}\left(\min\left(X,1\right)>y\right)\\ & =\mathsf{P}\left(X>y\wedge1>y\right)\\ & =\mathsf{P}\left(X>y\right)\mathsf{P}\left(1>y\right)\\ & =\mathsf{P}\left(X>y\right)1_{\left(y,\infty\right)}\left(1\right)\\ & =\mathsf{P}\left(X>y\right)1_{\left(-\infty,1\right)}\left(y\right) \end{aligned} $$

Here $1$ is treated as a constant random variable.

The third equality because a constant random variable is independent wrt to any random variable.