Conditional expectation bounded between an interval

Question

Let the random variable $X$ have density $f(x)>0$ for all $x \in [\underline{x}, \overline{x}]$. Consider the conditional expectation $\mathbb{E}\left(X| X \in [\underline{x}, \widehat{x}]\right)$ for some $\widehat{x} \le \overline{x}$. Then is it true that $$\underline{x}<\mathbb{E}\left(X| X \in [\underline{x}, \widehat{x}]\right) < \widehat{x}$$ My intuitive reasoning is that the answer is yes because we are effectively taking an "average" of $X$ knowing that $X$ is within a restricted interval and hence the average must be contained within that interval. Also, since the density mass is strictly positive, that means the inequality is strict. How can I prove this formally?

Answer 1

Here is a rigorous proof: let $A=\{\underline {x} \leq X \leq\widehat {x}\}$. We have to show that $\underline {x} <\frac 1 {P(A)} EXI_A < \widehat {x}$. To show that $\underline {x} <\frac 1 {P(A)} EXI_A$ note that $\underline {x} \leq \frac 1 {P(A)} EXI_A$ and if equality holds then $E(XI_A -\underline {x} P(A))=0$. Since $XI_A -\underline {x} P(A)$ is strictly positive on $A$ this is a contradiction. Similarly for the inequality on the other side.

Answer 2

Result is true.

Write your expectation as a (Riemann is good enough) integral; it is now an exercise in analysis- if you really wanted to go into the painful detail then pick your favourite definition of an integral to proceed.

The density function need not be strictly greater than 0 for strict inequality; it is sufficient that a density function exists (i.e. you have no atoms of probability)