Deriving the General Change-of-Numeraire Measure

# Change of Numeraire formula


# Change of Numeraire formula












The general change of Numeraire formula gives the following Radon-Nikodym derivative:

$$ \frac{dN_2}{dN_1}(t)|\mathcal{F}_{t_0}=\frac{N_1(t_0)N_2(t)}{N_1(t)N_2(t_0)} $$

I am able to derive this Radon-Nikodym for specific examples, such as changing from the risk-neutral measure $Q$ to the T-Forward Measure associated with a zero-coupon bond $P(t_0,t)$: in this case, we have under $Q$:

$$ \frac{S_0}{N_Q(t_0)=1}=\mathbb{E}^Q\left[\frac{S_t}{N_Q(t)=e^{rt}}|\mathcal{F}_{t_0}\right] $$

So that:

$$ (i) S_0 = \mathbb{E}^Q\left[S_t\frac{N_Q(t_0)=1}{N_Q(t)=e^{rt}}|\mathcal{F}_{t_0}\right] $$

Under the T-forward Bond numeraire:

$$ \frac{S_0}{N_{P}(t_0)=P(t_0,t)}=\mathbb{E}^{P_t}\left[\frac{S_t}{N_P(t)=1}|\mathcal{F}_{t_0}\right]$$

So that:

$$(ii) S_0 = P(t_0,t)\mathbb{E}^{P_t}\left[\frac{S_t}{N_P(t)=1}|\mathcal{F}_{t_0}\right]$$

Equating (i) to (ii) we get:

$$\mathbb{E}^Q\left[S_t\frac{N_Q(t_0)}{N_Q(t)}|\mathcal{F}_{t_0}\right]=N_P(t_0)\mathbb{E}^{P_t}\left[\frac{S_t}{N_P(t)}|\mathcal{F}_{t_0}\right]$$

Since $N_P(t)$ at time $t$ is by definition constant (equal to one), it is easy to take it out of the expectation and group all the Numeraire terms on the LHS, so that:

$$ \mathbb{E}^Q\left[S_t\frac{N_Q(t_0)N_P(t)}{N_Q(t)N_P(t_0)}|\mathcal{F}_{t_0}\right]=\mathbb{E}^{P_t}\left[S_t|\mathcal{F}_{t_0}\right] $$

And the result follows be inspection.

Note: in general, the numeraire $N_2(t)$ would not be a constant at time $t$, as is the case for the numeraire associated with the T-forward maturing bond. So it would not be possible to take $N_2(t)$ out of the expectation $\mathbb{E}_{t_0}^{N_2}[]$ as in the case above. It would therefore not be so straight forward to group all the numeraire terms and deduce the Radon-Nikodym derivative by inspection.

Question: How can the change of Numeraire Radon-Nikodym formula be derived or proved in the general case? (not thinking about specific numeraires as in the case above).

## Answer by Daneel Olivaw (score 5, accepted)

https://quant.stackexchange.com/a/55497

We work on a probability space $(\Omega,\mathcal{N},\mathfrak{F})$ with filtration $(\mathfrak{F}_t)_{0\leq t\leq T}$ and $\mathfrak{F}_T:=\mathfrak{F}$. Let $\xi$ be a $\mathfrak{F}_T$-mesurable contingent claim, and $N_t$ and $M_t$ two assets with positive prices. We assume the process $M_t/N_t$ is a martingale under the probability measure $\mathcal{N}$. Define for $0\leq t\leq T$ the process: $$Z_t:=E^\mathcal{N}\left(\left.\frac{N_0M_T}{N_TM_0}\right|\mathfrak{F}_t\right) = \frac{N_0M_t}{N_tM_0}$$ We notice that $E^\mathcal{N}(Z_t)=1$ for all $t$ per the martingale property. Therefore the random variable $Z:=Z_T$ is a valid Radon-Nikodym derivative and $Z_t$ its associated process: $$Z_t=\left.\frac{d\mathcal{M}}{d\mathcal{N}}\right|_{\mathfrak{F}_t}$$ We can define a new probability measure $\mathcal{M}$ as follows: $$\mathcal{M}(E):=\int_EZ(\omega)d\mathcal{N}(\omega)=E^\mathcal{N}(1_{E}Z)$$ Now define the $\mathfrak{F}_T$-measurable random variable: $$Y:=\frac{\xi}{M_T}$$ Per Lemma 5.2.2 in Shreve (2004): $$\begin{align} \tag{1} E^\mathcal{M}\left(\left.Y\right|\mathfrak{F}_t\right) & = \frac{1}{Z_t}E^\mathcal{N}\left(\left.YZ_T\right|\mathfrak{F}_t\right) \\[6pt] & = \frac{1}{E^\mathcal{N}\left(\left.\frac{M_T}{N_T}\right|\mathfrak{F}_t\right)} E^\mathcal{N}\left(\left.\frac{\xi}{N_T}\right|\mathfrak{F}_t\right) \end{align}$$ That is: $$\begin{align} M_tE^\mathcal{M}\left(\left.\frac{\xi}{M_T}\right|\mathfrak{F}_t\right) =N_tE^\mathcal{N}\left(\left.\frac{\xi}{N_T}\right|\mathfrak{F}_t\right) \end{align}$$ As an addenda, notice that we can make the following construct based on Equation $(1)$: $$ \left.\frac{d\mathcal{M}}{d\mathcal{N}}\right| _{\mathfrak{F}_t}^{\mathfrak{F}_T} := \frac{N_tM_T}{N_TM_t} = \frac{N_tM_0}{N_0M_t}\frac{N_0M_T}{N_TM_0} = \frac{Z_T}{Z_t} $$