# Equivalence of formulations of Radon Nikodym derivative

Let $$N$$ be a numeraire associated with the probability measure $$Q^N$$ and $$U$$ be a numeraire associated with the probability measure $$Q^U$$, both of which are equivalent to the physical probability measure $$Q^0$$. The Radon-Nikodym derivative defining the measure $$Q^U$$ is given by: $$\frac{dQ^U}{dQ^N} = \frac{U_t N_0}{U_0 N_t}$$ Now consider a scalar diffusion process $$X$$ , whose dynamics under $$Q^N$$ and $$Q^U$$ are given respectively by: $$dX_t = \mu_t^N(X_t) dt + \sigma_t dW_t^N, Q^N$$ $$dX_t = \mu_t^U(X_t) dt + \sigma_t dW_t^U, Q^U$$ We can apply Girsanov's theorem to deduce the Radon-Nikodym derivative between $$Q^N$$ and $$Q^U$$ from the dynamics of $$X$$ under the two measures: \begin{align} &\frac{dQ^U}{dQ^N} \\ &= \exp \left( - \frac{1}{2} \int_0^t \lvert \frac{\mu_s^U(X_s) - \mu_s^N(X_s)}{\sigma_s(X_s)} \rvert^2 ds - \int_0^t \frac{\mu_s^U(X_s) - \mu_s^N(X_s)}{\sigma_s(X_s)} dW_s^N \right) \end{align} Can someone help me show how the two formulations are equivalent? I assume that this is something standard so some simple reference would also be appreciated.

• Wouldn't proving that these are equivalent be equivalent to proving Girsanov's theorem? Commented Apr 19 at 11:33