Is there an inverse Lamperti transformation for diffusions?

The Lamperti transformation is commonly used to transform SDEs with state dependent coefficients into SDEs with constant diffusion.

For multidimensional processes there are some conditions on the drift that have to be valid before applying the transformation (e.g., drift should be the gradient of potential).

If we consider a general multidimensional SDE with constant drift $$dX_t = f(X_t) dt + \sigma dW_t,$$

can we get the other way around, i.e., can we convert it to a driftless SDE with state dependent coefficient? I.e., obtain an equation like this $$dY_t = \sigma(Y_t) dW_t$$,

or this $$dY_t =g(Y_t) dt+ \sigma(Y_t) dW_t$$. (here $$g()$$ shall conform to the conditions of the Lamperti transform).

If yes, what are the conditions for this?

• Uh, isn’t this doable by Girsanov’s theorem? Nov 20, 2022 at 15:38
• @NateRiver No the Girsanov requires the diffusion coefficient to remain unchanged. Here I want to change to the diffusion. Nov 20, 2022 at 15:56
• can you add more details? The Girsanov's applies to coefficients that depends on both X,t $$dX_t=\mu(X_t,t)dt+\sigma(X_t,t)dW_t, dY_t=(\mu(Y_t,t)+\nu(Y_t,t))dt+\sigma(Y_t,t)dW_t$$. See wiki article en.wikipedia.org/wiki/… Nov 21, 2022 at 16:53
• @ThomasKojar the two SDEs you mention have the same diffusion function. I would like to transform from $dX_t = f(X_t) dt + \sigma dW_t$ to $dY_t = \sigma(Y_t) dW_t$, i.e., I want to change the diffusion coefficient, if possible. Nov 21, 2022 at 20:15
• do you mean different sigma? see here for mutually singular example math.stackexchange.com/questions/79875/… Nov 21, 2022 at 21:34

Girsanov's applies to state dependent states. So it actually gives simpler answer if initially the volatility didn't depend on the state.

But for changing to a different volatility, the result is false because it changes the quadratic function and thus creates a mutually singular process, see here:

Change of time or change of measure

where they consider the event $$A=\{\omega:[\omega]_T=a\sigma^2T\}$$ which is $$1$$ for one process and $$0$$ for the other.

Now in particular for the Lamperti transform. Following the results 2.1.5/2.1.6 from "The Lamperti Transform" by de Boer,

first using Ito to compute for general map $$\psi$$.

then for this specific $$\psi$$ below, they turn the volatility coefficient to be one.

So since this mapping is 1-1 we can reverse the above procedure: $$\psi$$ is invertible and so starting from concrete $$Z_{t}$$, we can define $$X_{t}:=\psi^{-1}(Z_{t})$$. This will give the state-dependent diffusion from 2.1.5.

• When you employ a Lamperti transform what property of the stochastic process do you assume that remains the same? clarusft.com/lamperti-transform Nov 21, 2022 at 21:53
• See here mathoverflow.net/a/69299/483817 Nov 21, 2022 at 21:55
• in this note, they have a nice explanation too: google.com/… Nov 21, 2022 at 22:02
• so since that mapping psi is 1-1 in Theorem 2.1.6, you could try to consider the inverse $X_t=\psi^{-1}(Z_{t})$ i.e. starting from some concrete Zt and then obtaining Xt, but then you will naturally pick up a drift term. Nov 21, 2022 at 22:04