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I'm trying to simplify the following expectation so that I can later solve a maximization problem: $max_k E[(A - kB)^2]$, where $A$ ~ $N(0,\sigma^2_1)$, $B = A+ \epsilon$ and $\epsilon$ ~ $N(0,\sigma^2_2)$.

I thought I could linearize the expectation by expanding the square, and then plugging in $B$ in terms $A + \epsilon$, but each term I get depends on $A$ in this case, which has an expected value of 0, so I feel that I'm thinking about this the wrong way. What's the best way to simplify? Am I solving a conditional expectation here, since $B$ depends on $A$?

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Expanding the expectation yields:

$$f(k)=E[(A-kB)^2]=E[((1-k)A-k\epsilon)^2]=(1-k)^2E[A^2]+k^2E[\epsilon^2]-2k(1-k)E[A\epsilon]$$

Here, if $A$ and $\epsilon$ are independent, $E[A\epsilon]=E[A]E[\epsilon]=0$, and we'll have $f(k)=(1-k)^2\sigma_1^2+k^2\sigma_2^2$. We just take derivative wrt $k$ and have $$f'(k)=-2(1-k)\sigma_1^2+2k\sigma_2^2=0\rightarrow k=\frac{\sigma_1^2}{\sigma_2^2-\sigma_1^2}$$

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  • $\begingroup$ Ah I think the step I was missing was that the expected of $A^2$ and such is the variance. Thanks! $\endgroup$ – John Alperto Mar 14 at 12:25

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