$X(t) = W^2(t) - W(t)$ where $W(t)$ is a Wiener process.
I'd like to find an expected value and a correlation function of it.
$M\{X(t)\} = M\{W^2(t)\} - M\{W(t)\} = t - 0 = t$
$K_X(t_1,t_2) = M\{[W^2(t_1)-W(t_1)][W^2(t_2)-W(t_2)]\} - M\{X(t_1)\}M\{X(t_2)\} = M\{W^2(t_1)W^2(t_2)\} - M\{W^2(t_1)W(t_2) \} - M\{W(t_1)W^2(t_2)\} + M\{W(t_1)W(t_2)\} - t_1t_2$
The last but one element is equal $M\{W(t_1)W(t_2)\}=min(t_1,t_2)$, but I have no idea how to find the others. Could you give me any tips, please?
Thank you!
Assuming that $t_1<t_2$ we have $M(W^{2}(t_1)W(t_2))=M(W^{2}(t_1)(W(t_2)-W(t_1))+M(W^{3}(t_1))=M(W^{3}(t_1))$ by independence of $W(t_2)-W(t_1)$ and $W(t_1)$. Now $M(W(t_1)^{3})=0$ because the distribution is symmetric. Hence $M(W^{2}(t_1)W(t_2))=0$. Now $M(W^{2}(t_2)W(t_1))=M(W(t_2)-W(t_1))^{2} W(t_1))=M((W(t_2)-W(t_1))^{2} M(W(t_1))=0$. [I dropped some terms in the first equality but those terms are $0$]. What remains is $M(W^{2}(t_1)W^{2}(t_2))$. This is $MW^{2}(t_1)[W(t_2)-W(t_1)]^{2}+2MW^{3}(t_1)W(t_2)-M(W(t_1))^{4}. $ Can you complete the calculation now?