Given are 6 independent standard normal distributed variables $x_i$. Simulation shows that the covariance of the transformed random variables $$f\sim a (x_5 - x_6) + b (x_4 - x_6)$$ $$g\sim (x_1 - x_3) (x_5 - x_6) - (x_2 - x_3) (x_4 - x_6)$$ $$\text{with } a,b\in \mathbb{R}^{-}$$ is close to 0. How to prove that $\text{cov}(f,g)=0$?
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From your problem set-up, $\mbox{Cov} \left[f,g\right]$ is the summation of two different covariance terms both of which evaluate to zero. 1) If $i \ne j \ne k$, then $\mbox{Cov}\left[x_i,x_jx_k\right] = \mbox{E} \left[x_ix_jx_k\right] - \mbox{E} \left[x_i\right]\mbox{E} \left[x_jx_k\right] = 0$ by independence and zero mean assumption. 2) If $i = j \ne k$, then $\mbox{Cov}\left[x_i,x_ix_k\right] = \mbox{E} \left[x_i^2x_k\right] - \mbox{E} \left[x_i\right]\mbox{E} \left[x_ix_k\right]= 0$ by independence and zero mean assumption. These terms were found by using the fact that the covariance of two sums is the sum of the covariances.
