Finding expectation of peicewise mixed distribution density function

Click For Summary

Discussion Overview

The discussion revolves around finding the expected value of a piecewise mixed distribution density function, specifically in the context of its cumulative distribution function (CDF). Participants explore the implications of different configurations of the distribution, including cases with one or multiple pieces.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests that the expected value for a piecewise mixed distribution could be calculated as \(a + \text{integral(all the pieces)}\).
  • Another participant questions whether the expected value would simply be \(a + \text{integral of that piece}\) if there were only one piece.
  • A third participant proposes that if there is only one piece, the expected value would just be the integral of that piece, raising the question of how to handle multiple pieces.
  • A later reply advises differentiating each piece of the CDF to find the probability density function (PDF) and then using the definition of expectation. This response also mentions considerations for discrete distributions and non-integrable forms.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the calculation of expected values for piecewise distributions, with no consensus reached on the correct approach or formula.

Contextual Notes

Participants note that the expected value may depend on whether the distribution is discrete or continuous, and the discussion includes references to different mathematical forms such as Riemannian and Lebesgue.

torquerotates
Messages
207
Reaction score
0
If I am given the CDF of a piecewise mixed distribution density starting from a and ending at b, would the expected value just be a + integral(all the pieces) ?
 
Physics news on Phys.org
If there were only one "piece" would the expected value be a+ the integral of that piece?
 
I suppose not. It would just be the integral of that piece. But if it were two or more?
 
torquerotates said:
If I am given the CDF of a piecewise mixed distribution density starting from a and ending at b, would the expected value just be a + integral(all the pieces) ?

Hey torquerotates.

Why don't you just differentiate each piece of your CDF and then use the definition of expectation?

If you function is in discrete form or a form that is non-integrable (non-Riemannian) then you can either just look at the deltas if you have a discrete distribution for that piece, or if its in a Lebesgue form just use the properties of the characteristic function to get your PDf for that piece.
 

Similar threads

Undergrad Calculating the inverse of a function involving the error function
  • · Replies 3 ·
Replies
3
Views
2K
MHB Distribution and Density functions of maximum of random variables
  • · Replies 6 ·
Replies
6
Views
5K
  • Poll Poll
Graduate How to change the support of a probability density function?
  • · Replies 4 ·
Replies
4
Views
2K
MHB Cdf, expectation, and variance of a random continuous variable
  • · Replies 1 ·
Replies
1
Views
2K
Graduate Probability Density Function: Converting Experimental Observations to PDF
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Does the generalized gamma distribution have a mgf?
  • · Replies 1 ·
Replies
1
Views
3K
Graduate Constructing PDFs for Max Likelihood Density Estimation Problem
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Help me understand skewness in QQ-plots please
  • · Replies 1 ·
Replies
1
Views
2K
MHB Cumulative distribution function
  • · Replies 3 ·
Replies
3
Views
3K
High School Impossible to find the quantile of any equation
  • · Replies 2 ·
Replies
2
Views
1K