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x,T are vectors with $cov(x_i,T_i)>0$. Without specifying f(x,T), is it possible to determine the sign of $\mathbb{E}[\frac{T'x}{T'T}]$?

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  • $\begingroup$ if $x,T$ are multi-dimensional (assuming because you take transpose of $T$), $\operatorname{cov}(x,T)$ won't be scalar. How do you define $>0$? $\endgroup$ – gunes Feb 15 '20 at 23:49
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    $\begingroup$ @gunes, you are right. But I just edited. x and T are single single valued. T'T is just the dot product of ot (or equivalently $\sum_i t_i^2$ $\endgroup$ – LucasMation Feb 16 '20 at 3:58
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Since it must hold for all $n$ (dimension of $X$ and $T$), consider $n=1$. The expectation reduces to $$\mathbb{E}\left[\frac{XT}{T^2}\right]=\mathbb{E}\left[\frac{X}{T}\right]$$

Some intuition: Consider zero-mean $X$, then this expression is equal to $\operatorname{cov}(X,1/T)$. Intuitively, if the correlation/covariance between $X,T$ is positive; e.g. they may increase or decrease together, I wouldn't expect positive correlation between $X,1/T$ in general (not saying it's impossible), because now while $X$ increases, $1/T$ decreases.

A counterexample: let $X=2T+1\rightarrow \operatorname{cov}(X,T)=2\operatorname{var}(T)>0$. For $E[X/T]=2+E[1/T]$ to be negative, we just need small enough $E[1/T]$, which can be set to any value arbitrarily, e.g. let $T$ be a binary RV having $-0.1$ and $-0.5$ values with equal probability. Then $E[1/T]=-(10+2)/2=-6$.

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