Probability Conditional Expectation

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The discussion revolves around a challenging problem involving independent Poisson random variables X and Y, specifically calculating E[Y − X|X + Y = 10]. Participants express frustration over their midterm performance, noting that no one in the small class managed to solve the problem correctly. Acknowledgment is made that while X and Y are independent, the events Y=y and X+Y=10 are not independent, complicating the calculations. A hint is provided that leads to the realization that the conditional distribution of X given X+Y=n follows a binomial distribution. Ultimately, the problem is clarified, and one participant successfully solves it after receiving guidance.
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)##?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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