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Could you give me some hints how to show that the function $$f(a):=\frac{(\mathbb E [a^t X_1]-\mathbb E [a^t X_2])^2}{\mathrm {Var} (a^t X_1-a^t X_2)}$$ is maximised by a vector $$a \propto C^{-1} (\mu_1-\mu_2) ?$$
$X_1 \sim N (\mu_1, C)$ and $X_2 \sim N(\mu_2, C)$ are two random vectros. All I managed to do was to calculate that $$ \frac{(\mathbb E [a^t X_1]-\mathbb E [a^t X_2])^2}{\mathrm {Var} (a^t X_1-a^t X_2)} = \frac{(a^t (\mu_1-\mu_2))^2}{a^t 2Ca}=\frac{a^t Aa}{a^t B a},$$ $$A:=(\mu_1-\mu_2)(\mu_1-\mu_2)^t, \quad B:=2C$$ and $$ [Df(a)]=\frac{2a^tA(a^tBa)-(a^t A a)2a^t B}{(a^t B a)^2}.$$ I don't how to proceed ;/

P Lrc
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  • The question seems clear (but way out of my expertise). But the title won't attract the people who can answer. Please provide a full title so your question will get looked at (and maybe answered) by those who have the requisite knowledge. – Harvey Motulsky Feb 13 '20 at 01:54
  • If this is a study exercise, consider adding the self-study tag. – Igor F. Feb 13 '20 at 08:10

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I would use the square-root $B^{1/2}$of a positive-definite matrix $B$ such that $$B^{1/2} B^{1/2} = B.$$ Using this notation, write $f(a)$ by $b=B^{-1/2}a$. Then, we have $$f(b) = \dfrac{b^{\rm T} \tilde{A}b}{b^{\rm T} b},$$ where $\tilde{A}=B^{1/2}AB^{1/2}$. Then, use the Cauchy-Schwarz to find the maximum at some $b_0$ and transform it back to $a_0=B^{1/2}b_0$.

inmybrain
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