Deriving the Change-of-Numeraire Radon–Nikodym Derivative

# Radon Nikodym derivative when changing numeraires


# Radon Nikodym derivative when changing numeraires












I note from Wikipedia that if $Q$ and $Q^N$ are two measures corresponding to numeraires $M$ and $N$, then the Radon Nikodym derivative is given by: $$\frac{dQ^N}{dQ} = \frac{M(0)}{M(T)}\frac{N(T)}{N(0)}.$$

However I do not understand how this formula comes from the traditional definition of a Radon-Nikodym derivative, which is a random variable such that the following holds for all RV $Z$: $E_N(Z)=E_M\left(\frac{dQ^N}{dQ}Z\right)$

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/53776

For any price process $Z$, given that $N$ and $M$ are numeraire processes, \begin{align*} E_N\left(\frac{Z_T}{N_T} \right) = E_M\left(\frac{Z_T}{M_T}\frac{M_0}{N_0} \right), \end{align*} as they both equal to $Z_0/N_0$. Note also that \begin{align*} E_N\left(\frac{Z_T}{N_T} \right) = E_M\left(\frac{dQ_N}{dQ_M}\frac{Z_T}{N_T} \right). \end{align*} Then \begin{align*} E_M\left(\frac{dQ_N}{dQ_M}\frac{Z_T}{N_T} \right) &= E_M\left(\frac{Z_T}{M_T}\frac{M_0}{N_0} \right)\\ &=E_M\left(\frac{N_T}{M_T}\frac{M_0}{N_0}\frac{Z_T}{N_T}\right). \end{align*} Since $\frac{Z_T}{N_T}$ can be arbitrary, we conclude that \begin{align*} \frac{dQ_N}{dQ_M} = \frac{N_T}{M_T}\frac{M_0}{N_0}. \end{align*}

## Answer by siou0107 (score 7)

https://quant.stackexchange.com/a/53669

Any nonnegative random variable $Z$ with expectation 1 is a Radon-Nikodym derivative: $$ \mathbb{E}^{\mathbb{P}} \left(Z\right) = \mathbb{E}^{\mathbb{P}} \left(\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}\right) = \mathbb{E}^{\mathbb{Q}} \left(1\right) = \int{\mathrm{d}\mathbb{Q}} = 1 $$ $$ \mathbb{Q} \left(A\right) = \mathbb{E}^\mathbb{P} \left(Z 1_A\right) \in \left[0, 1\right] $$ If $Z$ is positive, the probability measure $\mathbb{Q}$ that it defines is equivalent to the original probability measure $\mathbb{P}$.

Now, by definition of a numeraire, under its associated probability measure, all asset prices expressed as units of the numeraire are martingales. For $\mathbb{Q}$ with numeraire $M$ and $N$ a positive asset price process, $$ \mathbb{E}^{\mathbb{Q}} \left(\frac{N_T}{M_T}\right) = \frac{N_0}{M_0} \Rightarrow \mathbb{E}^{\mathbb{Q}} \left(\frac{M_0}{M_T}\frac{N_T}{N_0}\right) = 1 $$ Since a numeraire is always chosen to be a strictly positive asset price process, the random variable $\frac{M_0}{M_T}\frac{N_T}{N_0}$ defines the Radon-Nikodym derivative of measure $\mathbb{Q}^N$ with respect to $\mathbb{Q}$. If $X$ is an (arbitrary) asset price process, $$ \mathbb{E}^{\mathbb{Q}^N} \left(X_T \frac{N_0}{N_T}\right) = \mathbb{E}^{\mathbb{Q}} \left(X_T \frac{N_0}{N_T}\frac{M_0}{M_T}\frac{N_T}{N_0}\right) = \mathbb{E}^{\mathbb{Q}} \left(X_T\frac{M_0}{M_T}\right) = X_0 $$ That shows that $N$ is indeed the numeraire for measure $\mathbb{Q}^N$.