Remove drift from exponential Weiner process I have the following problem: let $X_t$ solve $$ dX_t = b X_t \, dt + \sigma X_t \, dW_t$$ where $W_t$ is a Weiner process. Find $s(\cdot)$ such that $Y_t = s(X_t)$ is a martingale. * * * We can see by inspection that $X_t$ is an exponential Weiner process with drift $b - \sigma^2/2$, and the problem is to remove this drift somehow. There are lots of ways to do this if $s$ is a function of $t$ also, and of course there is the trivial solution $s = 0$, but I don't think either of those are what the question is looking for. My textbooks don't seem to have anything about finding such functions, so I assume there is some really simple answer that I'm just not seeing. Does anyone have any hints?

If you compute the derivative of $Y_t$ using Ito's lemma, you have

$Y'_t=s'(X_t) dX_t + 0.5 s''(X_t) = [s'(X_t) b X_t+ 0.5 s''(X_t) X_t^2\sigma^2] dt + s'(X_t) \sigma X_t dW_t$

$Y_t$ is a martingale if and only if its drift, $s'(X_t) b X_t+ 0.5 s''(X_t) X_t^2\sigma^2$, is $0$. Thus, unless $X_0=0$, $s(x)$ must be such that $$bs'(x) + 0.5 s''(x) x \sigma^2 = 0$$ Solving it yields $$s(x) = c_1 + c_2 x^{\left(\frac{2b}{\sigma^2}+1\right)}$$