Can someone give a clear-cut idea of $E(X|XIf $X$ and $Y$ be two i.i.d. random variables, then what should $E(X|X<Y)$ essentially look like?
P.S.-Is the denominator equal to $\frac12$?
(the two r.v.'s being iid serving as the motivation behind this.)
 A: The i.i.d requirement creates useful symmetries, so indeed there is a factor of $1/2$ in some formulas.  The following is one attempt to interpret "looks like" in a useful and intuitive fashion.

The possibilities for $(X,Y)$ partition into three events: $X\lt Y,$ $X\gt Y,$ and $X=Y.$ 
Therefore, assuming all expectations exist,
$$\mathbb{E}(X+Y) = \mathbb{E}(X+Y\ |\ X\lt Y) \Pr(X\lt Y) + \mathbb{E}(X+Y\ |\ X\gt Y) \Pr(X\gt Y) \\+  \mathbb{E}(X+Y\ |\ X= Y) \Pr(X= Y).$$
Writing $p_0$ for $\Pr(X=Y)$, the symmetry $X\to Y, Y\to X$ shows that the remaining probability of $1-p_0$ is equally split over the events $X\lt Y$ and $X\gt Y$:
$$\Pr(X\lt Y) = \Pr(Y\lt X) = \Pr(X \gt Y) = \frac{1-p_0}{2}.$$
The same symmetry shows the first two conditional expectations are equal while the third reduces to
$$\mathbb{E}(X+Y\ |\ X= Y) \Pr(X=Y) = 2\mathbb{E}(X | X=Y)p_0.$$
Write
$$\mu_0 = \mathbb{E}(X | X=Y).$$
Solving for the conditional expectation yields
$$\mathbb{E}(X+Y\ |\ X \lt Y) = 2 \frac{\mu - \mu_0p_0}{1-p_0}.$$
Similar (but slightly easier) manipulations show that 
$$\mathbb{E}(Y - X \ |\ X \lt Y) = \frac{\mathbb{E}(|Y-X|)}{1-p_0}.$$
Because $X = \frac{1}{2}\left((X+Y) - (Y-X)\right),$
$$\mathbb{E}(X \ |\ X \lt Y) = \frac{\mu - \mu_0p_0 - \mathbb{E}(|Y-X|)/2}{1-p_0}.$$
To begin understanding this, suppose the common distribution of $X$ and $Y$ is continuous: this implies $p_0=0$.  The foregoing expression reduces to
$$\mathbb{E}(X \ |\ X \lt Y) = \mu - \mathbb{E}(|Y-X|)/2,$$
(wherein the expected factor of $1/2$ clearly appears).  That is,

For continuous distributions, the conditional expectation of $X$, given that $X\lt Y$, is less than the unconditional expectation of $X$ by half the expected size of the difference $|Y-X|$.

(This result is intuitively obvious: the amount by which the conditional expectation of $X$ falls below the unconditional expectation must be exactly the amount by which the conditional expectation of $Y$ exceeds the unconditional expectation and those two amounts sum to the expected difference.)
The additional terms $\mu_0$ and $p_0$ can now be understood as corrections for the possibility that $X=Y$.  They are a little harder to obtain intuitively, which is why some care was taken with the foregoing analysis.
