# If $cov(x_i,T_i)>0$ can I show $\mathbb{E}[\frac{T'x}{T'T}] > 0$?

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

• 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 $>0$? – gunes Feb 15 '20 at 23:49
• @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 $\sum_i t_i^2$ – LucasMation Feb 16 '20 at 3:58

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 $$E[1/T]=-(10+2)/2=-6$$.