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Covariance of random sum

  1. Dec 30, 2007 #1
    Suppose [tex]X_1,....,X_n[/tex] are independent and identically distributed random variables.
    Now suppose I picked [tex]m_1[/tex] random variables from the set [tex]X_1,....,X_n[/tex] and defined [tex]Y_1[/tex] as the sum of the [tex]m_1[/tex] variables, where [tex]m_1[/tex] is also a random variable.
    Now suppose I did this again and I picked [tex]m_2[/tex] random variables from the set [tex]X_1,....,X_n[/tex] and defined [tex]Y_2[/tex] as the sum of the [tex]m_2[/tex] variables, where [tex]m_2[/tex] again is a random variable.
    I also know the expected number of random variables from the set [tex]X_1,....,X_n[/tex], that are contained in both sums [tex]Y_1[/tex] and [tex]Y_2[/tex]. Call this number [tex]a[/tex].


    So I basically have two random sums, [tex]Y_1[/tex] and [tex]Y_2[/tex], and I want to find the covariance bwteen them, [tex]Cov(Y_1, Y_2)[/tex]. I came up with the following solution but it doesn't seem to work, so any pointers on what's wrong or how to go about doing it would be great.


    So I just simply used the definition of covariance of sums, ie. For sequences of random variables [tex]A_1,....,A_m[/tex] and [tex]B_1,....,B_n[/tex], we have [tex]Cov(\sum_{i=1}^{m}A_i,\sum_{j=1}^{n}B_j) = \sum_{i=1}^{m}\sum_{j=1}^{n}Cov(A_i,B_j)[/tex].
    So applying the above formula to my situation of [tex]Cov(Y_1, Y_2)[/tex], I have that because [tex]X_1,....,X_n[/tex] are independent, most of the terms in the double sum in the above formula will be zero, and will only be non-zero if [tex]X_i \equiv X_j[/tex], in which case [tex]Cov(X_i,X_j)[/tex] will be just [tex]Var(X_1)[/tex].
    Hence the [tex]Cov(Y_1, Y_2)[/tex], will be just [tex]a*Var(X_1)[/tex] ???

    There is something wrong in the logic above, as the formula [tex]a*Var(X_1)[/tex] doesn't seem to work, but I can't figure out where I am going wrong. Any help??
     
  2. jcsd
  3. Dec 30, 2007 #2

    EnumaElish

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    If all the X's are mutually independent then doesn't that allow you to make a statement about Cov(X1+X2, X3+X4+X5), for example?
     
  4. Dec 31, 2007 #3
    All the X's are mutually independent, so if I apply the definition [tex]Cov(\sum_{i=1}^{m}A_i,\sum_{j=1}^{n}B_j) = \sum_{i=1}^{m}\sum_{j=1}^{n}Cov(A_i,B_j)[/tex] to your example Cov(X1+X2, X3+X4+X5), then the answer is 0.

    But in my situation there is a certain amount of overlap, For example, suppose I have the set [tex]X_1,.....X_{20}[/tex], and [tex]m_1=5[/tex] and [tex]m_2=7[/tex], then I might have a situation where [tex]Y_1=X_1+X_3+X_6+X_8+X_9[/tex] and [tex]Y_2=X_1+X_2+X_3+X_6+X_{10}+X_{12}+X_{15}[/tex].

    So in this case if I apply the above covariance of sum definition then [tex]Cov(Y_1,Y_2)[/tex] will not be 0, as there will be 3 non-zero terms in the sum (ie. [tex]Cov(X_1,X_1), Cov(X_3,X_3), Cov(X_6,X_6))[/tex].
    So, as all X's are identically distributed, we get [tex]Cov(Y_1,Y_2)[/tex] = [tex]a*Var(X) = 3*Var(X)[/tex].

    Now this formula, [tex]a*Var(X)[/tex], works when [tex]m_1[/tex] and [tex]m_2[/tex] are not random variables, but when they are random variables it doesn't work anymore. When they are random variables, I know what the expected values of [tex]m_1[/tex] and [tex]m_2[/tex] are going to be, and also know what the expected number of overlapping elements will be, call this [tex]a[/tex].

    So from this information, anyone know how to get the expression for [tex]Cov(Y_1,Y_2)[/tex] when [tex]m_1[/tex] and [tex]m_2[/tex] are random variables??
     
    Last edited: Dec 31, 2007
  5. Jan 1, 2008 #4

    EnumaElish

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    To simplify, suppose you have X1, X2.

    Then m = 1 or 2, and n = 1 or 2.

    If m = 1 then Y1 is X1 or X2. If m = 2 then Y1 is X1+X2.

    Similarly if n = 1 then Y2 is X1 or X2. If n = 2 then Y2 is X1+X2.

    If you can make a table of these possible outcomes and assign a probability to each, you can calculate a probability-weighted average of the covariance formulas for each possible case.
     
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