$E(X \mid X > x)$ 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.

• What is the definition of $E(X\mid X > x )$ ? – Gabriel Romon Apr 3 at 14:30
• $E(X \mid X >x) = E(X1(X>x))/P(X > x)$ – Galton Apr 3 at 14:31
• @Galton Could you clarify the denominator? Is it $X_1 = X, if X > x ;\; 0 \;, o.w.$ and $E[X_1]$? – Kaind Apr 3 at 14:33
• The numerator is the expectation of a variable that equals $X$ if it is larger than $x$ and zero otherwise. The denominator is the probability that $X>x$ – Galton Apr 3 at 15:13

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

$$\frac {E(X1_{X>x'})}{P(X>x')}\geq \frac {E(X1_{X>x})}{P(X>x)}$$

• This is a very elegant solution. I had a hunch that working with an iid copy should do the trick but couldn't quite get it to work. – Galton Apr 3 at 18:30
• Nice idea taking an iid copy I would like to get better the intuition behind these proofs using probabilistic concepts. For the moment I was curious, it is so trivial that, in the general case, an iid copy of a r.v. always exists ? – Thomas Apr 4 at 19:46
• @Thomas see math.stackexchange.com/questions/250145/… – Gabriel Romon Apr 5 at 10:57

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 $$\mathbb E[Y \mid Y > x'] \ge \mathbb E[Y]$$.

• This solution very nicely captures the intuition that the mean has to shift to the right. – Galton Apr 3 at 18:32

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!

Non-only is the function non-decreasing in $$x$$, but its derivative is explicit.

Assume $$X$$ is non-negative for simplicity and let $$F(x)=P(X the cdf. As the CDF is monotone, the monotone differentiation theorem given in Theorem 53 of https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/ gives that $$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.