Probability Conditional Expectation

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Homework Help Overview

The discussion revolves around conditional expectation involving independent Poisson random variables, specifically focusing on the expression E[Y − X|X + Y = 10]. Participants are exploring the implications of independence and the relationships between the variables in the context of a probability problem encountered during a midterm exam.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the independence of X and Y and the implications of conditioning on the sum X + Y. There is an exploration of the probability mass function related to the conditional distribution of X given X + Y.

Discussion Status

The conversation is ongoing, with some participants expressing confusion and seeking alternative approaches. Hints have been provided that suggest a connection to familiar distributions, which some participants find helpful in guiding their reasoning.

Contextual Notes

Participants note the constraints of the problem as it relates to their midterm exam, highlighting the pressure to understand the material before the next class. There is an acknowledgment of the challenges faced in utilizing available resources for clarification.

ctownballer03
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Suppose X and Y are independent Poisson random variables with respective parameters λ and 2λ.
Find E[Y − X|X + Y = 10]3: I had my Applied Probability Midterm today and this question was on it. The class is only 14 people and no one I talked to did it correctly. The prof sent out an e-mail saying how no one did it correctly and we need to work on it and get it figured out and corrected by our next class and frankly I'm still super stuck (my professor is pretty useless, I can't utilize him as resource for anything in this class). Anyways, this is my attempt at doing this however I realize that I've made a mistake and even though X and Y are independent, Y=y and X+Y=10 are NOT independent events so the cancellation that I did is not a legal move.. Any other ideas of how to approach this problem, I feel like I'm back to square one and I'm not sure where to go. Thank you!

ElQtujT.jpg
 
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ctownballer03 said:
Suppose X and Y are independent Poisson random variables with respective parameters λ and 2λ.
Find E[Y − X|X + Y = 10]3: I had my Applied Probability Midterm today and this question was on it. The class is only 14 people and no one I talked to did it correctly. The prof sent out an e-mail saying how no one did it correctly and we need to work on it and get it figured out and corrected by our next class and frankly I'm still super stuck (my professor is pretty useless, I can't utilize him as resource for anything in this class). Anyways, this is my attempt at doing this however I realize that I've made a mistake and even though X and Y are independent, Y=y and X+Y=10 are NOT independent events so the cancellation that I did is not a legal move.. Any other ideas of how to approach this problem, I feel like I'm back to square one and I'm not sure where to go. Thank you!

ElQtujT.jpg

If ##X \sim \text{Po}(a)## and ##Y \sim \text{Po}(b)##, what is ##P(X = k | X + Y = n)##?

Hint: ## (X|X+Y=n)## is a familiar discrete random variable, whose values range from ##0## to ##n##.
 
Ray Vickson said:
If ##X \sim \text{Po}(a)## and ##Y \sim \text{Po}(b)##, what is ##P(X = k | X + Y = n)##?

Hint: ## (X|X+Y=n)## is a familiar discrete random variable, whose values range from ##0## to ##n##.

Things have just got interesting. I may have to post again later tonight if I get stuck but you have put me on the right track. So the PMF of X=k|X+Y=n where X+Y are poisson RVs is going to be a binomial.
 
Solved it now.

Thank you for the hint.
 
ctownballer03 said:
Solved it now.

Thank you for the hint.

Just as a matter of interest: for independent ##X_a \sim \text{Po}(a)## and ##X_b \sim \text{Po}(b)##, what did you obtain as the distribution of ##(X_a|X_a+X_b=n)##?
 

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