1

How can I solve $$dX_t=|X_t|^\alpha dW_t\ \ ?$$

What I tried is to set $X_t=u_tdt+v_tdW_t$. Then, by Itô formula $$df(X_t)=f'(X_t)u_tdt+f'(X_t)v_tdW_t+\frac{1}{2}f''(X_t)v_t^2dt.$$

So, we should have $$f'(X_t)u_t=\frac{1}{2}f''(X_t)v_t^2$$ and $$f'(X_t)v_t=|u_t+v_tdW_t|^\alpha .$$

but how can I continue ?

joshua
  • 1,001
  • The Ito formula for $Y=f(X)$ gives $$dY_t=\tfrac12f''(X_t)|X_t|^{2α},dt+f'(X_t)|X_t|^α,dW_t,$$ I don't see where your last formula comes from. – Lutz Lehmann May 01 '21 at 10:27
  • @LutzLehmann: Indeed. From your formula, do you see how to find $X_t$ ? Because I don't see how to conclude. – joshua May 01 '21 at 10:54
  • There are even less SDE that are symbolically solvable than ODE (conceptually, or for a given formula length). One could try for $f'(x)|x|^α=$const. as that would improve some numerical methods, Euler-Marayuma is then identical to the strong-order-1 Milstein method. – Lutz Lehmann May 01 '21 at 11:27

0 Answers0