E[f(X)] - Expectation of function of rand. var.

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Discussion Overview

The discussion revolves around the expectation of a function of random variables, specifically how to compute the expectation of a variable Z defined piecewise based on disjoint conditions involving random variables X and Y. The scope includes theoretical aspects of probability and expectation in the context of random variables.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the expectation of Z can be expressed as E[Z]=E[X]*Pr(A)+E[Y]*Pr(B) under certain conditions.
  • Another participant counters that independence is required for this formulation and suggests using the indicator function to express Z as Z=X\mathbf{1}_A+Y\mathbf{1}_B, leading to the expectation E[Z]=E[X\mathbf{1}_A]+E[Y\mathbf{1}_B].
  • This second participant notes that to separate the expectations, independence between X and A, as well as between Y and B, is necessary.
  • A further contribution introduces an integral formulation for the expectation of a function of a random variable, suggesting that if a probability density function exists, the expectation can be computed using integrals.

Areas of Agreement / Disagreement

Participants express differing views on the conditions necessary for the expectation formulation, particularly regarding the need for independence. The discussion remains unresolved as no consensus is reached on the correct approach.

Contextual Notes

There are limitations regarding the assumptions about independence and the definitions of the random variables involved. The discussion also touches on the abstract versus practical computation of expectations.

Apteronotus
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Hi quick question:

Suppose you have a function of random variables given in the following way

Z=X if condition A
Z=Y if condition B

where both X and Y are random variables, and conditions A & B are disjoint.

Then would the expectation of Z be

E[Z]=E[X]*Pr(A)+E[Y]*Pr(B)?

Thanks in advance.
 
Last edited:
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No you need independence for that, what you really mean is
[tex]Z=X\mathbf{1}_A+Y\mathbf{1}_B[/tex]
where 1 is the indicator function. Now take the expectation
[tex]\mathbb{E}[Z]=\mathbb{E}[X\mathbf{1}_A]+\mathbb{E}[Y\mathbf{1}_B][/tex].

Now you know that [itex]\mathbb{E}[\mathbf{1}_A]=\mathbb{P}(A)[/itex], but to separate the expectations, you need independence between X and A, also between Y and B.
 
Thank you Focus for your reply. I see my error.
 
Last edited:
You can use
[tex]\mathbb{E}[f(X)]=\int_{\mathbb{F}}f(x)F(dx)[/tex]
where F is the law of X. This may be somewhat abstract so if you are working over the reals and have a pdf f_X then
[tex]\mathbb{E}[f(X)]=\int_{\mathbb{R}}f(x)f_X(x)dx[/tex].
 

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