Finding conditional expectation of two random variables Let $X$ and $Y$ be independent and uniform on $[0,3]$. I want to calculate $E(Y| X<1 \cup Y<1 )$.
Attempt.
First, we calculate the distribution via the cdf:
$$ P(Y \leq y | X<1 \cup Y<1 ) = \frac{ P( \{Y \leq y \} \cap \{X < 1 \cup Y < 1 \} ) }{P(X<1 \cup Y<1) } = $$
$$ \frac{ P( [ \{Y \leq y \} \cap \{ X < 1 \} ] + [ \{Y \leq y \} \cap \{ Y < 1 \} ] - P(\{Y \leq y \} \cap \{X <1\} \cap \{Y < 1\}) }{P(X<1) + P(Y<1) - P(X<1)P(Y<1) } =$$
$$ \frac{ P( [ \{Y \leq y \} \cap \{ X < 1 \} ] \cup [ \{Y \leq y \} \cap \{ Y < 1 \} ] }{P(X<1) + P(Y<1) - P(X<1)P(Y<1) } $$
$$ \frac{ P(Y \leq y)P(X<1) +P(Y \leq \min(y,1) ) - P(Y \leq \min(y,1))P(X<1) }{P(X<1) + P(Y<1) - P(X<1)P(Y<1) } $$
Now, Notice $P(X<1) = P(Y<1) = \frac{1}{3}$ and $P(Y \leq y ) = \frac{ y }{3}$. Thus, the conditional cdf is
$$ \begin{cases}  \frac{y}{5} \; \; \; 0<y<1 \\ \frac{y+2}{5} \; \; \; 1<y<3 \end{cases} $$
now, by taking the derivative we see that the density function is just $\frac{1}{5}$ over $y \in [0,3]$, now thus
$$ E(Y| X<1 \cup Y<1 ) = \int_0^3 y \frac{1}{5} = \frac{9}{10} = 0.9 $$
Now, this is incorrect according to my answer sheet which gives $\boxed{1.1}$ as the answer. What is my mistake?

Here another short solution:


*

*$U = X<1 \cup Y <1 = [0,3]\times [0,1) \cup [0,1)\times [1,3]$ (disjoint union of $5$ squares with probability of $\frac{1}{9}$)

*$P([0,3]\times [0,1)| U) = \frac{3}{5}$, $P([0,1)\times [1,3] |U) = \frac{2}{5}$

*$\Rightarrow$ The mean of $Y$ on $[0,3]\times [0,1)$ is $\frac{1}{2}$ with a weight of $\frac{3}{5}$.

*$\Rightarrow$ The mean of $Y$ on $[0,1)\times [1,3]$ is $2$ with a weight of $\frac{2}{5}$.

*$\Rightarrow$ $E(Y|U)= \frac{1}{2}\cdot \frac{3}{5} + 2 \cdot\frac{2}{5} = \frac{11}{10}$

