
#1
Dec110, 06:52 AM

P: 18

1. The problem statement, all variables and given/known data
The expectation value of the sum of two random variables is given as: [tex] \langle x + y \rangle = \langle x \rangle + \langel y \rangle[/tex] My textbook provides the following derivation of this relationship. Suppose that we have two random variables, x and y. Let [tex]p_{ij}[/tex] be the probability that our measurement returns [tex]x_{i}[/tex] for the value of x and [tex]y_{j}[/tex] for the value of y. Then the expectation value of the sum of [tex]x+y[/tex] is: [tex] \langle x + y \rangle = \sum\limits_{ij} p_{ij} (x_i + y_j) =\sum\limits_{ij} p_{ij} x_i + \sum\limits_{ij} p_{ij} x_j [/tex] Then I am given the following statement: But [tex]\sum\limits_j p_{ij} = p_i[/tex] is the probability that we measure [tex]x_i[/tex] regardless of what we measure for y, so it must be equal to [tex]p_i[/tex]. Similarly, [tex]\sum\limits_i p_{ij} = p_j[/tex], is the probability of measuing [tex]y_i[/tex] irrespective of what we get for [tex]x_i[/tex]. 2. Relevant equations The difficulty I have with this statement is that I do no see how [tex]\sum\limits_j p_{ij}[/tex] can be equal to [tex] p_i[/tex]. 3. The attempt at a solution Summing over j, we should have [tex](p_{i1} + p_{i2},+ .... p_{in})[/tex]. Now, is this equal to [tex]p_i[/tex]. And similarly how can [tex]\sum\limits_i p_{ij}[/tex] can be equal to [tex]p_j[/tex] I am hopefull that someone can clear this up for me. Thank you for your kind assitance. jg370[/quote] 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Dec110, 07:57 PM

HW Helper
P: 3,309

so you have 2 discrete random variables X & Y, with a joint distribution, pij
[tex]p_{ij} = P(X=x_i, Y = y_j)[/tex] the expectation is given by: [tex] <X+Y> = \sum_{ij} p_{ij} x_i y_j = [/tex] By definition, the marginal probabilities are [tex]P(X=x_i) = \sum_{j} p_{ij} = p_i [/tex] [tex]P(Y=y_j) = \sum_{i} p_{ij} = p_j [/tex] If the variables are independent then you have the further conidtion that [tex]p_{ij} = P(X=x_i, Y = y_j) = P(X=x_i)P(Y = y_j) = p_i p_j [/tex] 


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