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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

jasperhyp
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  • Do you suppose that $ER_i < \infty$ or smth. else? – Botnakov N. Jan 02 '21 at 14:19
  • Yes, in the book they follow simple Exponential distributions with parameters $\mu_i$. But that's merely a small example and I am actually asking about more general/complicated inequalities/distributions. @BotnakovN. – jasperhyp Jan 02 '21 at 14:26
  • $E(\xi | A) = \frac{E(\xi I_A)}{P(A)}$. If you know how to work with usual expectations, then there are no problems. Otherwise it's better to ask smth. about usual expectations. – Botnakov N. Jan 02 '21 at 14:35
  • As I mention in my answer, the result you quote happens to be true in the case of independent exponential RVs, which is the case in the text you are reading – jlammy Jan 02 '21 at 14:43
  • @jlammy Uh I got that now. Thanks for pointing it out. As I commented under your answer, despite using this kind of intrinsic properties (of RV themselves), are there some general rules/exmaples? – jasperhyp Jan 02 '21 at 14:47
  • @BotnakovN. Thanks for the point. I actually saw this transformation in another post. But it looks like this is more applicable in the case that a constant is in the inequality, but I guess it might not be that applicable to my context (despite the fact they could be dealt with w/ memoryless property), where two RVs are being compared...? – jasperhyp Jan 02 '21 at 14:50
  • @jasperhyp, by definition $E(\xi | A) = \frac{E\xi I_A}{P(A)}$, where $A$ is event with positive probability. In your context $P(A) = P(R_1 < R_2) > 0$, so it works. Is there any questions? – Botnakov N. Jan 02 '21 at 15:17

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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 $$ \eqalign{ E[R_1\wedge R_2] &= E[R_1; R_1<R_2]+E[R_2; R_2<R_1]\cr &=2E[R_1; R_1<R_2] \qquad\hbox{(by symmetry)}\cr &=E[R_1; R_1<R_2]/P[R_1<R_2]\cr &=E[R_1\mid R_1<R_2]\cr } $$

John Dawkins
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    Great answer! Could you clarify why the different means of the distributions $R_1$ and $R_2$ follow respectively do not affect $P[R_1<R_2]=\frac{1}{2}$? And does that hold if they follow different continuous distributions with different mean? Thanks! – jasperhyp Jan 02 '21 at 16:24
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    @jasperhyp John's answer states that your claim is true when $R_1$ and $R_2$ are independent and have the same distribution. So $R_1$ and $R_2$ have the same mean. – jlammy Jan 02 '21 at 18:14
  • @jlammy Thanks for the clarification. But in Ross's example, $R_1$ and $R_2$ are exponentially distributed with different means $\mu_i$. I guess in this case $P[R_1<R_2]=\frac{1}{2}$ might not hold? – jasperhyp Jan 03 '21 at 01:32
  • You are correct – jlammy Jan 03 '21 at 01:33
  • Thanks but so... How does Ross's example work in this condition? – jasperhyp Jan 03 '21 at 06:56
  • @jasperhyp the example in your text works because the variables are both exponential (albeit with different parameters). In this very special case, we don't need $R_1$ and $R_2$ to have the exact same distribution. – jlammy Jan 03 '21 at 13:29
  • @jlammy Thank you. I think I kind of understand why $\min (R_1,R_2)$ is independent of $R_1<R_2$ now. – jasperhyp Jan 03 '21 at 14:11
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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 $E\min(R_1, R_2) = P(R_1 = 1, R_2 = 1) > 0$.

Botnakov N.
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  • Copied from my reply to jilammy: Thanks for pointing out the mistake. It does not work for most rv's, and yes, in this specific example they are exponential rv's. Looks like they are using the memoryless property of exponential rv here. But are there some intuitions for more general cases (either distributions or form of inequality)? – jasperhyp Jan 02 '21 at 14:46
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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.

jlammy
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  • Thanks for pointing out the mistake. It does not work for most rv's, and yes, in this specific example they are exponential rv's. Looks like they are using the memoryless property of exponential rv here. But are there some intuitions for more general cases (either distributions or form of inequality)? – jasperhyp Jan 02 '21 at 14:45