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Probability Theory - Expectation Problem

  • Thread starter 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 eachother.

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...
 

Answers and Replies

  • #2
LCKurtz
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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 eachother.

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?
 
  • #3
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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
 
  • #4
LCKurtz
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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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