Covariance between functions of 3 random variables

  • Thread starter EvenSteven
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  • #1
EvenSteven
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Find [tex]cov(Y,Z)[/tex] where [tex] Y = 2X_1 - 3X_2 + 4X_3[/tex] and [tex]Z = X_1 + 2X_2 - X_3[/tex]
Information given [tex]E(X_1) =4[/tex]
[tex]E(X_2) = 9 [/tex]
[tex]E(X_3) = 5[/tex]
[tex]E(Y) = -7[/tex]
[tex]E(Z) = 26[/tex]

I tried expanding cov(Y,Z) = E(YZ) - E(Y)E(Z) but can't figure out how to calculate E(YZ)
 

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  • #2
Ray Vickson
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Find [tex]cov(Y,Z)[/tex] where [tex] Y = 2X_1 - 3X_2 + 4X_3[/tex] and [tex]Z = X_1 + 2X_2 - X_3[/tex]
Information given [tex]E(X_1) =4[/tex]
[tex]E(X_2) = 9 [/tex]
[tex]E(X_3) = 5[/tex]
[tex]E(Y) = -7[/tex]
[tex]E(Z) = 26[/tex]

I tried expanding cov(Y,Z) = E(YZ) - E(Y)E(Z) but can't figure out how to calculate E(YZ)

You have an expression for Y and another expression for Z, both of them as linear combinations of the X_j, so you can expand out the product. However, you will need some information you have not written here: you need to know that variances of the X_j in order to complete the calculation. If you were not told those (or some equivalent information) there is no hope of obtaining numerical answers; you can still give a formula but it will contain some unevaluated input constants.

Note also that there is something wrong with the given information: from EX1 = 4, EX2 = 9 and EX3 = 5, it follow that EY = E(2X1 - 3X2 + 4X3) = 2(4) - 3(9) + 4(5) = +1, not the given value -7. You should also check the value of EZ.
 
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