If $$dX_t = \mu(t,X_t)dt + \sigma(X_t)dW_t$$ with $\sigma$ positive, show there exists a function $f$ such that $$d\left(f(X_t)\right) = v(t,X_t)dt + V dW_t$$ where $V$ is constant. How unique is $f$?

The solution by Mark Joshi says: The volatility of $f(X_t)$ will, from Ito's lemma, be $$f'(X_t)\sigma(X_t)$$ So we need to solve $$f'(X_t) = \sigma^{-1}(X_t)A$$ for some constant $A$. We deduce $$f(X) = C + A\int_{0}^{X}\sigma(S)^{-1}dS$$ with $A$ and $C$ arbitrary constants.

Is this a complete solution? I don't see the associated steps in showing if $f$ is unique. Also, I do not understand why when he asserts from Ito's lemma that the volatility part of $f(X_t)$ leads to solving $f'(X_t) = \sigma^{-1}(X_t)A$.


1 Answer 1


It is a complete solution.

Bearing in mind the SDE verified by $(X_t)_{t \geq 0}$, applying Itô's lemma to compute the (stochastic) differential of $f(X_t)$ yields \begin{align} df(X_t) &= \underbrace{\frac{\partial f}{\partial t}}_{0} dt + \frac{\partial f}{\partial X}(X_t) dX_t + \frac{1}{2} \frac{\partial^2 f}{\partial X^2}(X_t) d\langle X \rangle_t \\ &= f'(X_t) dX_t + \frac{1}{2}f''(X_t)\sigma^2(X_t) dt \\ &= \left( f'(X_t)\mu(t,X_t) + \frac{1}{2}f''(X_t)\sigma^2(X_t) \right) dt +f'(X_t) \sigma(X_t) dW_t \\ &:= v(t,X_t) dt + V dW_t \end{align} such that by identifiying the diffusion terms we indeed need to solve $$ f'(X_t)\sigma(X_t) = V $$ Solving this simple ODE gives $$ f(X) = C + V \int_0^X \sigma^{-1}(x) dx,\,\,\, \forall C \in \Bbb{R} $$ which shows the function $f$ is not unique since defined up to a scalar constant (obviously if $V$ is not fixed then it is also an arbitrary parameter).


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