Suppose [tex]X_1,....,X_n[/tex] are independent and identically distributed random variables.(adsbygoogle = window.adsbygoogle || []).push({});

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??

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# Covariance of random sum

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