Background

1. Consider an random variable $X$, which can take on a value in $[0, w]$. $[0, w]$ is often referred to as the support of the distribution.
2. Cumulative distribution: $F(x) = Prob[X \leq x]$. We assume $f(x) \geq 0$, for $\forall x \in [0, w]$
3. Expectation: $E[X] = \int_{0}^{w} x f(x) dx$. Alternatively, as $f(x) = \frac{dF(x)}{dx}$, we can express the expectation as $E[X] = \int_{0}^{w} x dF(x)$
4. A degenerate random variable: a random variable that has a single possible value. Naturally, the variance of a degenerate variable is 0.

Conditional expectation

Contrast the conditional expectation of $X$ given $X \leq x$ with the normal version:

Normal: $E[X] = \int_{0}^{w} x f(x) dx= \int_{0}^{w} x dF(x)$

Conditional: $E[X \leq x] = \frac{1}{F(x)}\int_{0}^{x} t f(t) dt$

Which can be viewed as a 2-step approach:

1. Change the top limit
2. Adjust the whole thing by multiplying $\frac{1}{F(x)}$

There is a graphical representation too that can be found in Krishna (2003)'s Book: "Auction Theory" in Appendix A:

$F(x) E[X | X \leq x] = \int_{0}^{x}tf(t)dt = xF(x) - \int_{0}^{x}F(t)dt$

The second equality is derived using integration by parts.

Intuition behind

Consider again the problem we encountered just now, $E[X | X \leq x]$ can be written as:

$E[X \leq x] = \int_{-\infty}^{\infty} t f(t | t \leq x) dt$

Notice that we can break the conditional distribution into: $E[X \leq x] = \int_{-\infty}^{\infty} t \frac{f(t)_{1(t < x)}}{F(x)} dt = \frac{1}{F(x)} \int_{-\infty}^{x} t f(t) dt$

The reason is that the indicator function is always $0$ for $t > x$, and always $1$ for $t \leq x$. Therefore, it is the same as truncating the top limit to $x$ precisely because any $t$ above $x$ will have $f(t)_{1(t < x)} = 0$.