How to calculate conditional expectation with inequality condition

Question

How to transition conditional expectation where the conditioned item is not an equation into a simpler form for calculation?

As a simple example, how to show that (from Introduction to Probability Models by S.M. Ross) $$E[R_1|R_1<R_2] = E[\min(R_1,R_2)]$$

where $R_1, R_2$ are continuous RVs following exponential distributions with rate $\lambda_1, \lambda_2$ respectively.

I did find: How to calculate conditional expectation $E[X|X \geq 0]$?, How to calculate conditional probability with inequality, and Conditional Expectation Multivariate Normal Distribution with inequality condition three questions, but it looks like they are either dealing with probabilities or not providing a concrete answer. I know there might not exist an explicit rule, but are there any general rules, or more examples or references?

Update: Please refer to the following image (Example 5.8 from the book). As is stated in @John Dawkins' answer, it holds when the two RVs follow a continuous distribution with finite mean.

Ross book Ex 5.8

Answer 1

Your claim is not true.

Consider independent $R_1, R_2 \sim Bern\frac{1}2$. We have $P(R_i = 0) = P(R_i =1) = \frac12$. Thus $ R_1 < R_2$ iff $R_1 = 0$ and $R_2 = 1$.

Thus $$E(R_1 | R_1 < R_2) = E(R_1 | ( \{R_1 = 0\} \cap \{ 1 = R_2\})) = 0$$ and $E\min(R_1, R_2) = P(R_1 = 1, R_2 = 1) > 0$.

Answer 2

I don't think the inequality is true in general.

The LHS is the same as $\mathbb E[\min(R_1,R_2)\mid R_1<R_2]$, which in general is not the same as $\mathbb E[\min(R_1,R_2)]$.

Consider, for instance, $R_1$ is uniform on $[0,1]$ and $R_2$ is almost surely $\frac{1}{2}$. The LHS is $\frac{1}{4}$ but the RHS is $\frac{1}{2}\left(\frac{1}{4}+\frac{1}{2}\right)=\frac{3}{8}$.

The equation you write will be true if, e.g. $\min(R_1,R_2)$ is independent of the event $R_1<R_2$. This is the case when, for example, $R_1$ and $R_2$ are independent exponential RVs.

Answer 3

The assertion is true if $R_1$ and $R_2$ are i.i.d. with continuous distribution (and finite mean). In this case $P[R_1<R_2] = 1/2$, and $$ \eqalign{ E[R_1\wedge R_2] &= E[R_1; R_1<R_2]+E[R_2; R_2<R_1]\cr &=2E[R_1; R_1<R_2] \qquad\hbox{(by symmetry)}\cr &=E[R_1; R_1<R_2]/P[R_1<R_2]\cr &=E[R_1\mid R_1<R_2]\cr } $$