Expectation of a random variable conditioned on a random upper bound The question has been edited.
Let $X,Y$ be two positive real-valued random variables with continuous densities. Assume that $\mathbb{P}[X>Y]>0$.
Does it always hold that:
\begin{equation}
\mathbb{E}[X|X<Y] <\mathbb{E}[X] 
\end{equation}
If so, how could I formally prove this result?
 A: No, this is not so. And the reason is: it is possible that
$$\operatorname{E}[X|X < Y]>\operatorname{E}[X]$$
As an example, consider random variables $X$ and $Y$ with joint PMF given by
\begin{equation}
p_{X,Y}=
\begin{cases}
0.5 &\text{if }X=2, Y=3\\
0.5 &\text{if }X=1, Y=0\\
0   &\text{otherwise}
\end{cases}
\end{equation}
Evidently
\begin{align}
\operatorname{E}[X]&=1.5\\
\operatorname{E}[X|X < Y]&=2
\end{align}
A: For this proof assume $X$ and $Y$ are independent continuous real-valued random variables with overlapping support. Let $f$ and $F$ denote the probability and cumulative density functions respectively, and let $c = \mathbb{P}(X<Y)^{-1}$.
$$\begin{align}
F_{X|X<Y}(x) &\equiv \mathbb{P}(X \leq x | X < Y) \\
&= c \int_{-\infty}^{x} \int_{x}^{\infty} f_{X}(\hat{x})f_{Y}(\hat{y})d\hat{y} d\hat{x} \\
&= c \int_{-\infty}^{x} f_{X}(\hat{x})(1-F_{Y}(\hat{x})) d\hat{x}
\end{align}$$
$$\Rightarrow f_{X|X<Y}(x) = c f_{X}(x)(1-F_{Y}(x))$$
This rescales $f_{X}(x)$ by $c(1-F_{Y}(x))$ which is a decreasing function of $x$ (strictly where the supports overlap). Thus, $X$ strictly stochastically dominates $X|X<Y$ and we are done!

(The following makes rigorous the strict stochastic dominance.)
Claim: $F_{X}(x) \leq F_{X|X<Y}(x)$ for all $x$.
Proof: Suppose not, then there exists a $\hat{x}$ such that $F_{X}(\hat{x}) > F_{X|X<Y}(\hat{x})$, which implies $c(1-F_{Y}(\hat{x})) < 1$ (otherwise $f_{X}(x) \leq f_{X|X<Y}(x)$ for all $x \leq \hat{x}$ contradicting the assumption). Thus, $f_{X}(x) \geq f_{X|X<Y}(x)$ for all $x \geq \hat{x}$ since $c(1-F_{Y}(\hat{x})) < 1$ by monotonicity of $F_{Y}$.  However, this implies $1 = F_{X}(\infty)>F_{X|X<Y}(\infty) = 1$, which proves the claim by contradiction.
Claim: $c(1-F_{Y}(x))>1$ for some interval of positive support with respect to $X$.
Proof: Suppose not, then $F_{X}(x) \geq F_{X|X<Y}(x)$ for all $x$ in the support of $X$, so $F_{X}(x) = F_{X|X<Y}(x)$ for all $x$ in the support of $X$. This implies $c(1-F_{Y}(x))=1$ for all $x$ in the support of $X$, which contradicts it strictly decreasing on the overlapping support.
Corollary: $F_{X}(x) < F_{X|X<Y}(x)$ over some interval of positive support with respect to $X$.
