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2#
发表于 2011-7-13 16:28
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Because Var(c*y)=c^2*Var(y) (where c is a constant and y is a random variable). Since weights are constants, the rule above applies. Let me prove the rule:
Obviously, E(c*y)=c*E(y). Now Var(c*y)=E[(c*y)^2]-(E(cy)]^2=E(c^2*y^2)-(c*E(y))^2=
c^2*(E(y^2)-(E(y))^2)=c^2*Var(y) |
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