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?
 A: I think the two earliest comments are not helpful; the somewhat more helpful later comments by @copperhat and @JimB appeared while I was typing this Answer. You have the correct 
conditional CDF of $Y,$ based on a correct use of products to express
the independence of $X$ and $Y$ in the joint distribution.
However, the conditional PDF of $Y$ is also defined separately
over $(0, 1)$ and $(1,3),$ and so you need to sum two integrals to
get the correct conditional mean of $Y.$ The answer sheet is correct.
I used simulation as a quick way to illustrate the condition (in blue
at the left) and to make a (slightly raggedy) histogram of the conditional 
distribution of $Y$ [where cond was a convenient way to say $(X < 1) \cup (Y < 1)].$ I think it is usually a good idea to make a
quick sketch by hand before starting to work such a problem.

A: Don't attempt to find the conditional CDF.  
By definition, when conditioning over an event with non-zero probability mass we have:
$$\begin{split}\mathsf E(Y\mid X<1\cup Y<1) &= \dfrac{\mathsf E(Y~\mathbf 1_{Y<1\cup X<1})}{\mathsf P(Y<1\cup X<1)}\\&= \dfrac{\mathsf E(Y(\mathbf 1_{Y<1}+\mathbf 1_{X<1\cap Y\geqslant 1}))}{\mathsf P(Y<1)+\mathsf P(X<1\cap Y\geqslant 1)} \\ &= \dfrac{\mathsf E(Y~\mathbf 1_{Y<1})+\mathsf E(\mathbf 1_{X<1})\mathsf E(Y~\mathbf 1_{Y\geqslant 1})}{\mathsf P(Y<1)+\mathsf P(X<1)\mathsf P(Y\geqslant 1)}&\quad{\text{via independence and}\\\text{ Linearity of Expectation}}\\ &=\phantom{\dfrac{\tfrac 16+\tfrac 13\cdot\tfrac 43}{\tfrac 13+\tfrac 13\cdot\tfrac 23}}\end{split}$$
Which is a lot easier to evaluate, and liable to be less error prone.

Note $\displaystyle\mathsf E(Y\mathbf 1_{Y<1}) = \int_0^1 yf_Y(y)~\mathsf d y=\tfrac 13\int_0^1 y~\mathsf d y = \tfrac 16$ et cetera
A: 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 $\frac{2}{5}$.

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

A: Here is a tedious computational answer (this computes the conditional probability):
Let $A = \{ (x,y) | x<1 \text{ or } y  < 1\}$, we can quickly check that $PA = {5 \over 9} >0$.
We would like to compute $P[Y \le y | A]$.
A little work shows that $P[Y \le y | A] = \begin{cases} 0, & y < 0 \\{3 \over 5} y, & y \in [0,1) \\
{3 \over 5} + {1 \over 5} (y-1), & y \in [1,3] \\
1, & y > 3\end{cases} $.
Hence
${dP[Y \le y | A] \over dy} = \begin{cases} 0, & y < 0 \\{3 \over 5}, & y \in [0,1) \\
{1 \over 5}, & y \in (1,3) \\
0, & y > 3\end{cases} $, from which we compute
$E[Y|A] = \int y dP[Y \le y | A] = {3 \over 5} \int_0^1 y dy + {1 \over 5} \int_1^3 y dy = {11 \over 10}$.
