I am having hard time proving (or if it is even provable) whether the conditional expectation E(X|Y=y1)<E(X|Y=y2) if y1>y2. 0<=X<=1, 0<=Y<=1. Assuming that X and Y are continuous random variables and have a joint probability density function.
NB: So basically I want to know if I can do the following. \begin{equation*} \begin{aligned} \begin{split} &E[X|Y=y1]<E[X|Y=y2], \text{if}\ y1>y2\\ &\text{To see this, let us rewrite}\ E[X|Y=y1]<E[X|Y=y2]\ \text{as}\\ &=E[X|Y=y1]-E[X|Y=y2]<0\\ &=\int_0^1 X \cdot f(X|Y=y1)\;\mathrm{d}X - \int_0^1 X \cdot f(X|Y=y2)\;\mathrm{d}X <0\\ &= \int_0^1 X \cdot \tfrac{f(X,Y=y1)}{f(Y=y1)}\;\mathrm{d}X - \int_0^1 X \cdot \tfrac{f(X,Y=y2)}{f(Y=y2)}\;\mathrm{d}X<0 \\ &\text {Since } y1>y2, \text{we can write}\ y1=y2+a, \text{where}\ a>0.\\ & \text{Thus, by substituting for}\ y1\ \text{in the above expression, we have }\\ &=\int_0^1 X\cdot \tfrac{f(X,Y=y2+a)}{f(Y=y2+a)}\;\mathrm{d}X - \int_0^1 X\cdot\tfrac{f(X,Y=y2)}{f(Y=y2)}\;\mathrm{d}X <0\\ &= \int_0^1 X\cdot \tfrac{f(X,Y=y2+a)}{f(Y=y2+a)}\;\mathrm{d}X < \int_0^1 X\cdot\tfrac{f(X,Y=y2)}{f(Y=y2)}\;\mathrm{d}X \\ &\text{and}\\ &E[X|Y=y1]<E[X|Y=y2] \end{split} \end{aligned} \end{equation*}