G'day, today I have managed to derive $W^n_{t}$ in SDE form. However I am missing some critical rules to compute $W^1_{t}, W^2_{t}, W^3_{t}, W^{4}_{t}$.
SDE $dW^n_{t} = \frac{n(n-2)}{2} W^{n-2}_{t}dt+ nW^{n-1}_{t}dW_t$
a) n = 1 $W^1_{t} = \frac{1(1-1)}{2} W^{1-2}_{t}dt+ 1W^{1-1}_{t}dW_t = 0 W^{-1}_td_t + W^{0}_tdW_t = W_tdW_t = ???$
b) n = 2 $W^2_{t} = \frac{2(2-1)}{2} W^{2-2}_{t}dt+ 2W^{2-1}_{t}dW_t = 1 W^{0}_td_t + W^{1}_tdW_t = W_tdW_t + 2W_tdW_t= ???$
Can someone finish what I started?
Using Itô's formula one gets $$ d W^n_t = \frac{n(n-1)}{2} W^{n-2}_t dt + n W^{n-1}_t dW_t. $$ Now if $n=1$ we have the trivial identity $$ d W_t = \frac{1(1-1)}{2} W^{1-2}_t dt + 1 W^{1-1}_t dW_t = W^0_t dW_t = dW_t, $$ if $n = 2$ $$ d W^2_t = \frac{2(2-1)}{2} W^{2-2}_t dt + 2 W^{2-1}_t dW_t = dt + 2 W_t dW_t, $$ if $n = 3$ $$ d W^3_t = \frac{3(3-1)}{2} W^{3-2}_t dt + 3 W^{3-1}_t dW_t = 3 W_t dt + 3 W_t^2 dW_t, $$ and if $n=4$ $$ d W^4_t = \frac{4(4-1)}{2} W^{4-2}_t dt + 4 W^{4-1}_t dW_t = 6 W_t^2 dt + 4 W^3_t dW_t. $$