Conditional expectation of exponential random variable

Click For Summary

Discussion Overview

The discussion centers around the conditional expectation of an exponential random variable, specifically the calculation of E{X|X>a} where a is a specified scale value. The conversation explores the mathematical derivation and understanding of this expectation in the context of probability theory.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks for help in proving that E{X|X>a} equals a + E{X}, expressing uncertainty about the validity of this claim.
  • Another participant suggests that the expectation can be computed using integration, specifically through the limits a to ∞ for both the numerator and denominator of the integral.
  • A follow-up question seeks clarification on the reasoning behind the integration method proposed.
  • A further response explains that the integral from 0 to ∞ represents the definition of E(X) for an exponential random variable, and changing the limits to (a,∞) reflects the condition that X is greater than a, with the denominator providing normalization.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants express varying levels of understanding and seek clarification on the reasoning behind the integration method used to derive the conditional expectation.

Contextual Notes

Participants do not fully resolve the assumptions involved in the integration process or the implications of changing the limits of integration. There is also no explicit agreement on the initial claim regarding E{X|X>a}.

new_for_ever
Messages
3
Reaction score
0
For an exponential random variable X with rate u What is E{X|X>a} where a is a scale value
from searching in internet I found that
E{X|X>a}=a+E{x}​
but I can not prove it
Help please
 
Physics news on Phys.org
It can be done by a straightforward integration.
∫xe-uxdx/∫e-uxdx with limits a,∞ for both the numerator and the denominator.
 
Thanks for reply, it is correct but why??
 
The integral in the numerator (properly normalized) from 0 to ∞ is the definition of E(X) for an exponentially distributed random variable. Changing the limits to (a,∞) means assuming X>a. The denominator supplies the normalization.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K