Pdf that should be used to solve the question is a Gaussian mixture while
the one for Y = 0.25X1 + 0.75X2 is just normal. These are two different
things.
On 27 February 2012 23:13, Michail Tsagris <[log in to unmask]> wrote:
> Yes it is true, the two outputs are assumed independent. that is why.
> Then mean of Y is 0.25*m1+0.75*m2 and the variance of Y
> 0.0625*Var(x1)+0.5625*Var(X2)+zero covariance.
>
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--
Ufuk Mat
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