Probability Theory - Expectation Problem

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Homework Help Overview

The discussion revolves around discrete random variables X and Y, which are defined to take positive integer values. The joint probability mass function is given as \( p_{X,Y}(x,y) = 2^{-x-y} \). Participants are tasked with determining the marginal probability mass functions \( p_X(x) \) and \( p_Y(y) \), assessing the independence of X and Y, and calculating their expectations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivation of the marginal probability mass functions and the independence of X and Y based on the joint probability mass function. There is uncertainty regarding the calculation of expectations, particularly concerning the range of values that X and Y can take. Some participants question whether it is appropriate to assume that X and Y can only take values 0 and 1, while others clarify that they are indeed positive integers.

Discussion Status

The conversation is ongoing, with participants exploring the implications of the positive integer constraint on the expectations of X and Y. Some guidance has been provided regarding the marginal probabilities and the nature of the random variables, but there remains a lack of consensus on how to proceed with the expectation calculations.

Contextual Notes

Participants note that the problem does not explicitly state the possible values for X and Y, leading to confusion about the expectations. The assumption that X and Y can take on infinite positive integers is acknowledged, which complicates the calculation of expectations.

rooski
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Homework Statement



Discrete random variables X and Y , whose values are positive integers, have the joint probability mass function , (, ) = 2−−. Determine the marginal probability mass functions () and (). Are X and Y independent? Determine [], [ ], and [ ].

The Attempt at a Solution



y=1ʃ∞ pX(x) = Σ pX,Y(x,y)
= y=1ʃ∞ Σ 2^(-x-y)
= y=1ʃ∞ 2^(-x) Σ 2^(-y)
= 2^-x

x=1ʃ∞ pY(y) = Σ pX,Y(x,y)
= x=1ʃ∞ Σ 2^(-x-y)
= x=1ʃ∞ 2^(-y) Σ 2^(-x)
= 2^-y

Since pX,Y(x, y) = pX(x) * pY(y), X and Y are independent of each other.

I am stuck figuring out the expectations. Are we to assume that x and y can only take the values 0 and 1? The expectation requires a weighted average of all the possible values of x and y but the problem does not tell us the possible values...
 
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rooski said:

Homework Statement



Discrete random variables X and Y , whose values are positive integers, have the joint probability mass function , (, ) = 2−−. Determine the marginal probability mass functions () and (). Are X and Y independent? Determine [], [ ], and [ ].

The Attempt at a Solution



y=1ʃ∞ pX(x) = Σ pX,Y(x,y)
= y=1ʃ∞ Σ 2^(-x-y)
= y=1ʃ∞ 2^(-x) Σ 2^(-y)
= 2^-x

x=1ʃ∞ pY(y) = Σ pX,Y(x,y)
= x=1ʃ∞ Σ 2^(-x-y)
= x=1ʃ∞ 2^(-y) Σ 2^(-x)
= 2^-y

Since pX,Y(x, y) = pX(x) * pY(y), X and Y are independent of each other.

I am stuck figuring out the expectations. Are we to assume that x and y can only take the values 0 and 1? The expectation requires a weighted average of all the possible values of x and y but the problem does not tell us the possible values...

Yes it does. Your first statement says X and Y take on positive integers.

You have figured out that pX(x) = P(X=x) = 2-x, right? And what is your formula for E(X) for a discrete probability function?
 
But if it tells me that it only takes on positive integers then technically it can take on infinite positive integers right? And from what i gather you need to know the range of x in order to calculate the expectations. :S
 
rooski said:
But if it tells me that it only takes on positive integers then technically it can take on infinite positive integers right? And from what i gather you need to know the range of x in order to calculate the expectations. :S

The random variable X can take on values 1,2,3,...with various probabilities. You have already figured out P(X = x) = 2-x

In other words
P(X = 1) = 1/2
P(X = 2) = 1/4
P(X = 3) = 1/8
and so on.

So I will ask you again, what is the formula for E(X) when you know its discrete probability function pX(x)?
 

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