If ๐X and ๐Y are independent random variables with ๐ธ[๐]=1E[X]=1, Var(๐)=2Var(X)=2, ๐ธ[๐]=3E[Y]=3, and Var(๐)Var(Y) = 4, then find ๐ธ[(2๐+3๐)2]E[(2X+3Y) 2 ].
Question
If ๐X and ๐Y are independent random variables with ๐ธ[๐]=1E[X]=1, Var(๐)=2Var(X)=2, ๐ธ[๐]=3E[Y]=3, and Var(๐)Var(Y) = 4, then find ๐ธ[(2๐+3๐)2]E[(2X+3Y) 2 ].
Solution
To find E[(2X+3Y)^2], we first need to expand the square. This gives us:
E[(2X+3Y)^2] = E[4X^2 + 12XY + 9Y^2]
Since X and Y are independent, E[XY] = E[X]E[Y]. We can substitute this into the equation:
E[(2X+3Y)^2] = E[4X^2] + E[12XY] + E[9Y^2] = 4E[X^2] + 12E[X]E[Y] + 9E[Y^2]
We know that E[X^2] = Var(X) + (E[X])^2 and E[Y^2] = Var(Y) + (E[Y])^2. Substituting these values in gives us:
E[(2X+3Y)^2] = 4(Var(X) + (E[X])^2) + 12E[X]E[Y] + 9(Var(Y) + (E[Y])^2) = 4(2 + 1^2) + 1213 + 9(4 + 3^2) = 43 + 36 + 913 = 12 + 36 + 117 = 165
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