Conditional expectation and variance

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Discussion Overview

The discussion revolves around the conditional expectation and variance of independent exponential random variables X and Y, specifically focusing on the random variable Z defined by the condition X < Y. Participants explore the computation of E(X|Z) and V(X|Z), including the necessary probability distributions and potential errors in calculations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes to find E(X|Z=z) using the integral of xf(x|z) but questions how to determine f(x|z).
  • Another participant clarifies that since Z is discrete, E[.|Z] and V[.|Z] are also discrete functions, leading to E[X|Z=1] = E[X|XY].
  • A third participant claims to have computed E[X|Z=1] = 1/9 and E[X|Z=0] = 8/9 by integrating and conditioning on Y, leading to a derived E(X) of 17/27, which they argue contradicts the expected value of X.
  • Another participant questions the computation of Pr{Z=0} and Pr{Z=1}, suggesting a need for clarification on these probabilities.

Areas of Agreement / Disagreement

Participants express differing views on the calculations of conditional expectations and variances, with at least one participant identifying a potential contradiction in the results. The discussion remains unresolved regarding the correctness of the computations and the implications of the derived values.

Contextual Notes

There are limitations in the discussion related to the assumptions made about the distributions and the calculations of probabilities associated with Z. The participants have not fully resolved the mathematical steps or the implications of their findings.

covariance64
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Let X, Y be independent exponential random variables with means 1 and 2 respectively.

Let
Z = 1, if X < Y
Z = 0, otherwise

Find E(X|Z) and V(X|Z).

We should first find E(X|Z=z)
E(X|Z=z) = integral (from 0 to inf) of xf(x|z).
However, how do we find f(x|z) ?
 
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Z is discrete. Since E[.|Z] and V[.|Z] are functions of Z, they too are discrete.

E[X|Z=1] = E[X|X<Y].
E[X|Z=0] = E[X|X>Y].

Similarly for V[X|Z].
 
I have found that

E[X|Z=1] = E[X|X<Y] = 1/9
E[X|Z=0] = E[X|X>Y] = 8/9
by integrating, and conditioning on the random variable Y.

So E(X) = E(E(X|Z)) = (1/9)(1/3) + (8/9)(2/3) = 17/27,

which contradicts the fact that E(X) = 1, for X exponential with mean 1.

I am wondering where is the error.
 
How are you computing Pr{Z=0} and Pr{Z=1}?
 

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