Suppose $dX_t = a(X_t) dt + b(X_t) dW_t$ and $Y_s=X_t$ where $s=t^2$. What SDE does $Y_s$ satisfy in the weak sense? Hint: calculate $E[ dY | \mathcal{F}_s]$ where $dY = Y_{s-ds} - Y_s$.
This is from an old course and can be found here as question 21.
Does anyone have ideas on how to approach this?
Let $h(t) = \int_0^t h'(s) ds$ be an absolutely continuous function with $h'>0$. Then $\{V_t = W_{h(t)},t\ge 0 \}$ has the same distribution as $\int_0^t \sqrt{h'(s)} dW_s$.
For $Y_s = X_{h(s)}$, write $$ dY_s = Y_{s+ds} - Y_s = X_{h(s) + h'(s) ds} - X_{h(s)} = a(X_{h(s)})h'(s)ds + b(X_{h(s)}) dV_s\\ = a(Y_s)h'(s) ds + b(Y_s) dV_s. $$ Hence, by the above remark, $Y$ is a weak solution to $$ dY_s = a(Y_s)h'(s) ds + b(Y_s)\sqrt{h'(s)} dW_s. $$