$E(X \mid X > x)$ is increasing in $x$. Why? For two points $x < x'$ and a random variable $X$, we must have $E(X\mid X > x )\leq E(X\mid X > x' )$. This is "obviously" true because the center of the truncated distribution shifts to the right. How do I prove that?
I tried working with an iid copy $X^*$ of $X$ to show that the expectation of $X1(X>x)1(X^*>x')$ is smaller than the expectation of $X1(X>x')1(X^*>x)$ but I'm not having any luck with that.
All results I can find either focus on normality or assume densities.
 A: Let $Y$ be an iid copy of $X$.
Notice the following inequality holds $$(X-Y)(1_{X>x'}1_{Y>x}-1_{Y>x'}1_{X>x})\geq 0$$
and take expectations to find $$E(X1_{X>x'})P(Y>x)-E(X1_{X>x})P(Y>x')-E(Y1_{Y>x})P(X>x')+E(Y1_{Y>x'})P(X>x)\geq 0,$$
which rewrites
$$2E(X1_{X>x'})P(X>x)-2E(X1_{X>x})P(X>x')\geq 0,$$
thus
$$\frac {E(X1_{X>x'})}{P(X>x')}\geq \frac {E(X1_{X>x})}{P(X>x)}$$
A: Let $Y = X \mid X>x$. Then $Y \mid Y > x'$ is the same as $X \mid X > x'$, so it's enough to show that $\mathbb E[Y] \le \mathbb E[Y \mid Y > x']$: in other words, conditioning on $Y$ being high increases the expectation of $Y$.
For this, we have the law of total expectation:
$$
    \mathbb E[Y] = \mathbb E[Y \mid Y > x'] \Pr[Y > x'] + \mathbb E[Y \mid Y \le x'] \Pr[Y \le x'].
$$
In other words, $\mathbb E[Y]$ is a weighted average of $\mathbb E[Y \mid Y > x']$ and $\mathbb E[Y \mid Y \le x']$.
There are two cases:
Case 1. $\mathbb E[Y] \le x'$.  In this case, we have $\mathbb E[Y \mid Y > x'] \ge \mathbb E[Y] $ because because $Y \mid Y > x'$ is always bigger than $\mathbb E[Y]$.
Case 2. $\mathbb E[Y] > x'$. In this case, we always have $\mathbb E[Y \mid Y \le x'] \le \mathbb E[Y]$, because $Y \mid Y \le x'$ is always less than $\mathbb E[Y]$. To have the weighted average come out to $\mathbb E[Y]$, we must have $\mathbb E[Y \mid Y > x'] \ge \mathbb E[Y] $ to compensate.
In both cases, we get $\mathbb E[Y \mid Y > x'] \ge \mathbb E[Y] $.
A: For continuous random variables only:
Let $f(t) = E[X | X > t ] = \frac{ \int_t^\infty xf_X(x) dx}{\int_t^\infty f_X(x) dx}$
Define $g(t) = \int_t^\infty xf_X(x) dx$ and $h(t) = \int_t^\infty f_X(x) dx$ such that $f(t) = \frac{g(t)}{h(t)}$.
$g(t)$ and $h(t)$ are differentiable $\Rightarrow f(t)$ is differentiable by Quotient Rule.
$$
\begin{align*}
f'(t) &= \frac{-tf_X(t) \int_t^\infty f_X(x)dx + f_X(t)\int_t^\infty xf_X(x)dx}{h(t)^2} \\
\Rightarrow f'(t) h(t)^2 &= f_X(t) \bigg( \int_t^\infty (x - t)f_X(x)dx \bigg) \geq 0\\
\end{align*}
$$
Hence $f$ is an increasing function!
A: Non-only is the function non-decreasing in $x$, but its derivative is explicit.
Assume $X$ is non-negative for simplicity and let $F(x)=P(X<x)$ the cdf. As the CDF is monotone, the monotone differentiation theorem given in Theorem 53 of https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/ gives that $F(x)$ is differentiable almost everywhere.
The same goes for $G:x\mapsto \int_x^\infty P(X>t)dt$, it is monotone decreasing and thus differentiable everywhere with derivative equal to $P(X>x)$.
The identity holds $E[X|X>x] = x + (1-F(x))^{-1}\int_x^\infty P(X>t)dt$. Before giving a proof, let's explain why this shows monotonicity of $h(x) =  E[X|X>x]$. If
the cdf $F$ and $G$ are both diffentiable at $x$, then $h$ is also differentiable at $x$ as elementary addition/product/inverse of differentiable functions at by the product rule $h'(x) = 1 - 1 + F'(x)(1-F(x))^{-2} \int_x^\infty P(x>t)dt \ge 0$. Hence almost everywhere, the derivative of $h$ exists and is non-negative.
It remains to study the points at which $F$ is not differentiable: we can proceed ``guided'' by the differentiable case: for any $a>0$ since $(1-F(x+a))^{-1} \ge (1-F(x))^{-1}$
\begin{align}
E[X|X>x+a] - E[X|X>x]
&\ge (x+a) + (1-F(x))^{-1}\int_{x+a}^\infty P(X>t)dt
- x
- (1-F(x))^{-1}\int_{x}^\infty P(X>t)dt
\\&= a - (1-F(x))^{-1} \int_x^{x+a} P(X>t)dt
\\&\ge a - (x+a - x) P(X>x) \ge 0
\end{align}
thanks to $-P(X>t)\ge -P(X>x)$ for all $t\in[x,x+a]$ for the last inequality.
Why is the identity $h(x) = E[X|X>x] = x + \int_x^\infty P(X>t)dt$ true? It's a consequence of the well known identity $E[X]=\int_0^\infty P(X>t)dt$ for non-negative $X$ which follows from Fubini's theorem. Here,
\begin{align}
P(X>x)x + \int_x^\infty P(X>t)dt
&= P(X>x)\int_0^x 1 dt + \int_x^\infty P(X>t)dt
\\&= \int_0^\infty P(X> \max\{x,t\})dt 
\\&= E[I\{X>x\}X]
\end{align}
and the desired formula for $E[X|X>x]=E[I\{X>x\}X]/P(X>x)$ follows.
