I am building a working model and would like to know whether my computations so far look correct.
I am working with a generally bivariate distribution of two variables $\displaystyle X\sim [ 0,1]$, $\displaystyle Y\sim [ 0,1]$. In general, they might be correlated, but to set ideas I am open to consider extreme cases in which they might be independent. For instance, it could be useful to think about the case in which $X$ and $Y$ are independent and $X\sim U[ 0,1]$, $Y\sim U[ 0,1]$.
I am considering region like this one:
I want to compute the conditional expectation of $\displaystyle X$ and $\displaystyle Y$ conditional on it being in the regions $\displaystyle A$ and $\displaystyle B$. I am thinking about writing this in the following way: \begin{equation*} \mathbb{E}[ Y|A] =\mathbb{E}[ Y|Y >r( x)] =\mathbb{E}[ Y|X< r( x)] \end{equation*} \begin{equation*} \mathbb{E}[ Y|B] =1-\mathbb{E}[ Y|A] \end{equation*}
Conjecture $$ \mathbb{E}[ Y|A] =\begin{cases} 1 & \text{for} \ x< -b/m\\ (1-b-mx)^{-1} & \text{for} \ x\geqslant -b/m \end{cases}$$
Question: does my argument make sense? Are there any other "easy cases" that can allow me to study conditional expectations as the ones above? Below I work with bivariate uniform. I have also been checking this post, which seems like it could be potentially helpful.
Sketch of my argument:
consider:
\begin{equation*} \mathbb{E}[ Y|A] =\mathbb{E}[ Y|X< r( x)] =\int {_{0}}^{1} yf_{Y|X< r( x)}( y|x) dy \end{equation*}
We can compute the expectation by characterizing the marginal distribution:
\begin{equation*} f_{Y|X< r( x)}( y|x) =\frac{f( x,y)}{f_{X< r( x)}( x)} \end{equation*}
The marginal above is decribed by : \begin{equation*} f_{X< r( x)}( x) =\int _{A} f( x,y) \ dy=\int _{A} f( x,y) \ dy=\begin{cases} \int _{0}^{1} f( x,y) dy & \text{for} \ x< -b/m\\ \int _{mx+b}^{1} f( x,y) dy & \text{for} \ x\geqslant -b/m \end{cases} \end{equation*}
Assume that $\displaystyle f(x,y) =1$. Then we have that: \begin{equation*} f_{X< r( x)}( x)=\begin{cases} 1 & \text{for} \ x< -b/m\\ 1-b-mx & \text{for} \ x\geqslant -b/m \end{cases} \end{equation*}
And hence the expected value is given by: \begin{gather*} \int_{0}^{1} yf_{Y|X< r( x)}(y|x) dy\\ =\begin{cases} \int_{0}^{1} y\ dy & \text{for} \ x< -b/m\\ \int_{0}^{1} y( 1-b-mx)^{-1} dy & \text{for} \ x\geqslant -b/m \end{cases}\\ \begin{cases} 1 & \text{for} \ x< -b/m\\ (1-b-mx)^{-1}& \text{for} \ x\geqslant -b/m \end{cases} \end{gather*}
Firstly, what is that lowercase $x$ doing in the first equation?
$A$ is the area above/to the left of the line (and inside the unit square), so it represents the event of $Y>r(X)$ where $r(X)=mX+b$.
$$\begin{align} \mathbb{E}[ Y\mid A] &=\mathbb{E}[ Y\mid Y > mX+b]\\& =\mathbb{E}[ Y\mid X< (Y-b)/m] \end{align}$$
Secondly, your second equation does not hold in general.
Rather: $\mathbb E[Y]=\mathbb P(A)\,\mathbb E[Y\mid A]+\mathbb P(B)\,\mathbb E[Y\mid B]$ by the Law of Total Expectation, so we have:
$$\begin{align}\mathbb E[Y\mid B] &= \dfrac{\mathbb E[Y]-\mathbb P(A)\,\mathbb E[Y\mid A]}{\mathbb P(B)}\\[1ex]&= \dfrac{\mathbb E[Y]-\mathbb P(A)\,\mathbb E[Y\mid A]}{1-\mathbb P(A)}\end{align}$$
Thirdly, you have two random variables, so your expectations shall be evaluated using double integrals.
$$\mathbb E[Y\mid A] = \dfrac{\displaystyle\iint_{A} y\, f_{X,Y}(x, y)\,\mathrm d(x,y)}{\displaystyle\iint_{A} f_{X,Y}(x, y)\,\mathrm d(x,y)}$$
Fourthly, the shapes of $A$ will be either a triangle, quadrilateral, or pentagon, depending on which two sides (or corners) of the square the line intercepts. Determining the bounds of the integrals will need more information than that $x$-intercept.