There are two independent and continuous RVs $X$ and $Y$ and there exists a joint pdf $f_{X,Y}(x,y)$ over the set $M = \{(x,y)| 0 \le x \le 1, 0 \le y \le 1-x \}$. Now there exists a function $g(x) = \frac{1}{2} - cx $ that splits $M$ into two sets, namely $A_1 = \{(x,y)|0 \le x \le 1, 0 \le y \le \frac{1}{2} - cx \}$ and $A_2 = \{(x,y)|0 \le x \le 1, \frac{1}{2} - cx \le y \le 1-x \}$. How do I get $E[Y|A_1] $ and $E[Y|A_2] $ ?
So far I have worked with a simplified problem without $g(x)$. I have calculated $E[Y|X]$ and then next calculated $E[E[Y|X]]$. I have done so by:
- $f_X(x)= \int_0^{1-x} f_{X,Y}(x,y)dy$,
- $f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)} $
- $E[Y|X] = \int_0^{1-x} y f_{Y|X}(y|x) dy $
- Say $E[Y|X]$ is a linear transformation $b(X)$ of $X$ and thereby a RV itself that takes on the values $b(0) < w < b(1)$ (or the other way around, depending on $b(X)$)
- $F_{b(X)} = Prob(b(X)<w)$, from this find $f_{b(X)}(b(X))$
- $E[b(X)] = \int_{b(0)}^{b(1)} b(X) f_{b(X)}(b(X))dx $
- Obviously, in accordance with the law of iterated expectations this leads me to $E[Y]$
But where in my calculations would I add $g(x)$ to calculate $E[Y|A_1]$ and $E[Y|A_2]$?
The conditional expectation of a r.v. $X$ given an event $A$ of positive probability is just $$E[X \mid A] = \frac{E[X \cdot 1_A]}{P(A)}.$$ This is especially useful in your case since your events are just $(X,Y)$ landing in a simple region of the plane. Specifically, translating this into an integral with the joint density, $$ E[Y \mid (X, Y) \in A_1] = \frac{E[Y \cdot 1_{(X, Y) \in A_1}]}{P(A_1)} = \frac{\int_{A_1} y f_{X,Y}(x,y)dxdy}{\int_{A_1} f_{X,Y}(x,y)dxdy} $$ and the result is similar for $E[Y \mid (X, Y) \in A_2]$.