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I looked at $cov(W_s^n-W_t^n, W_t^n) = \mathbb{E}(W_s^n-W_t^n)(W_t^n)-\mathbb{E}(W_s^n-W_t^n)\mathbb{E}(W_t^n)$, used Binomial theorem and Moments of Normal Distribution to simplify this, but still can't prove that Covariance is equal to zero (or the contrary).

user228494
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  • Interesting problem. It may help to first convert the process into a martingale by adding an appropriate deterministic term. For instance, by Ito's formula, $d(W^2_t)=2W_t dW_t + t dt$, so $W^2_t-t$ is a martingale. It may be easier to check whether you have independent increments for this process instead. – Ian Jun 03 '15 at 11:35
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    Using the moments of a standard normal distribution up to degree $6$, one gets, for every $s>t>0$, $$\mathrm{cov}(W_s^4-W_t^4,W_t^4)=72(s-t)t^3\ne0.$$ – Did Jun 03 '15 at 11:37

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