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:


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).

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    $\begingroup$ Under the measure associated to a given numéraire, the value of any self-financing strategy should be a martingale, when it's expressed in numéraire units by absence of arbitrage. So basically, $N_Q(t)$ and $N_P(t)$ can be any positive-valued traded assets in the equations you wrote, they don't need to be specified. For the rest you can also check: quant.stackexchange.com/questions/53581/… $\endgroup$ – Quantuple Jul 7 at 10:10
  • $\begingroup$ @Quantuple: thank you. The thing is that for a general Numeraire, $N_2(t)$ might not be a constant (or indeed "1" as in the case of the T-forward measure), and then the math gets more difficult: I found it no longer easy to derive the formula when the second Numeraire doesn't equal a constant at time $t$. $\endgroup$ – Jan Stuller Jul 7 at 10:51
  • $\begingroup$ OK I understand :) Note that if you really want to keep the "conditional on" on the LHS I would rather write $$ \left. \frac{d \Bbb{N}}{d \Bbb{M}} \right\vert_{\mathcal{F}_t} = \frac{M_0 N_t}{M_t M_0} $$ not $\mathcal{F}_{t_0}$. $\endgroup$ – Quantuple Jul 7 at 11:01
  • $\begingroup$ @Quantuple: why? We are conditioning on the information we know "today", which is equivalent to $t_0$. If we condition on filtration as of time $t$, then $\mathbb{E}[N_t|\mathcal{F_t}]$ doesn't make sense, because $N_t|\mathcal{F_t}$ is a constant, not a random variable. $\endgroup$ – Jan Stuller Jul 7 at 11:07
  • $\begingroup$ just in the sense that you have $$ \Bbb{E}^{\Bbb{P}}_0 [ X_t ] = \Bbb{E}^{\Bbb{Q}}_0 \left[ X_t \left. \frac{d\Bbb{P}}{d\Bbb{Q}} \right\vert_{\mathcal{F}_t} \right] $$ so to stress the fact that both $X_t$ and the RN derivative at $t$ should be $\mathcal{F}_t$-measurable processes. $\endgroup$ – Quantuple Jul 7 at 11:15

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 $(1)$: $$ \left.\frac{d\mathcal{M}}{d\mathcal{N}}\right| _{\mathfrak{F}_t}^{\mathfrak{F}_T} := \frac{N_tM_T}{N_TM_t} = \frac{N_tM_0}{N_0M_t}\frac{N_0M_T}{N_TM_0} = \frac{Z_T}{Z_t} $$

| improve this answer | |
  • $\begingroup$ Thank you. I knew that Daneel Olivaw would be the most likely person to answer this question :). Epic answer, as always. $\endgroup$ – Jan Stuller Jul 7 at 10:48
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    $\begingroup$ Thank you @JanStuller, you're welcome. I've cleaned up a bit the answer. $\endgroup$ – Daneel Olivaw Jul 8 at 8:30

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