Deriving the Change-of-Numéraire Conditional Expectation Identity

# Numéraire -- couldn't understand the wiki explanation


# Numéraire -- couldn't understand the wiki explanation












I'm trying to understand Numéraire concept so am reading the wiki page:

I couldn't understand the last formula's 2nd equation:

$$ E_{Q}\left[\left.\frac{M(0)}{M(T)}\frac{N(T)}{N(0)}\frac{S(T)}{N(T)}\right| \mathcal{F}(t)\right]/ E_Q\left[\left.\frac{M(0)}{M(T)}\frac{N(T)}{N(0)}\right| \mathcal{F}(t)\right] = \frac{M(t)}{N(t)}E_{Q}\left[\left.\frac{S(T)}{M(T)}\right| \mathcal{F}(t)\right] $$

Why so? Which part from the left hand side is mapped into $\frac{M(t)}{N(t)}$ and which part mapped to $E_{Q}\left[\left.\frac{S(T)}{M(T)}\right| \mathcal{F}(t)\right] $?

Just for reference, below is copied from the wiki page.

-- begin of wiki >>

In a financial market with traded securities, one may use a change of numéraire to price assets. For instance, if $M(t)=exp(∫_0^t r(s)ds)$ is the price at time $t$ of $\$1$ that was invested in the money market at time $0$, then all assets (say $S(t)$), priced in terms of the money market, are martingales with respect to the risk-neutral measure, (say $Q$). That is $$\frac{S(t)}{M(t)}=E_Q\left[\left.\frac{S_T}{M_T} \right| F_t\right], ∀t≤T$$

Now, suppose that $N(t)>0$ is another strictly positive traded asset (and hence a martingale when priced in terms of the money market). Then, we can define a new probability measure $Q^N$ by the Radon–Nikodym derivative $$\frac{d Q^N}{dQ}=\frac{N_T/N_0}{M_T/M_0}$$

Then, by using the abstract Bayes' Rule it can be shown that $S(t)$ is a martingale under $Q^N$ when priced in terms of the new numéraire, $N(t)$:

$$ E_{Q^N}\left[\left.\frac{S(T)}{N(T)}\right| \mathcal{F}(t)\right] $$ $$= E_{Q}\left[\left.\frac{M(0)}{M(T)}\frac{N(T)}{N(0)}\frac{S(T)}{N(T)}\right| \mathcal{F}(t)\right]/ E_Q\left[\left.\frac{M(0)}{M(T)}\frac{N(T)}{N(0)}\right| \mathcal{F}(t)\right] $$ $$ = \frac{M(t)}{N(t)}E_{Q}\left[\left.\frac{S(T)}{M(T)}\right| \mathcal{F}(t)\right] = \frac{M(t)}{N(t)}\frac{S(t)}{M(t)} = \frac{S(t)}{N(t)} $$

<< end of wiki--

## Answer by Probilitator (score 8, accepted)

https://quant.stackexchange.com/a/10776

If you are interested in the proof of the Baye's Rule for conditional expectations you can find it here

The sake of completeness:

> The Baye's rule for conditional expectations states $$ E^Q[X|\mathcal{F}]E^P[f|\mathcal{F}]=E^P[Xf|\mathcal{F}] $$ With $f=dQ/dP$ - thus being the Radon-Nikodyn derivative and $X$ being some random variable and $\mathcal{F}$ being some sigma-algebrad.

Now we need to apply that rule to the change of numeraire context. From the Change of Numeraire Theorem we not that $dQ^N/dQ^M$ is given by $$ f=\frac{dQ^N}{dQ^M}=\frac{M(0)N(T)}{M(T)N(0)} $$

In the next step we insert this $f$ into above theorem and also subtitute $X$ for $S(T)/N(T)$

$$ E^{Q^N}\left[\frac{S(T)}{N(T)}|\mathcal{F}_t\right]E^{Q^M}\left[\frac{M(0)N(T)}{M(T)N(0)}|\mathcal{F}_t\right]=E^{Q^M}\left[\frac{S(T)}{N(T)}\frac{M(0)N(T)}{M(T)N(0)}|\mathcal{F}_t\right] $$

$\frac{S(T)}{N(T)}\frac{M(0)N(T)}{M(T)N(0)}$ simplifies to $\frac{S(T)}{M(T)}\frac{M(0)}{N(0)}$.

Now perhaps the crucial step. $N(t)$ is a numeraire and thus a tradeable asset. $M(t)$ is also a numeraire and $Q^M$ is its equivalent measure. Thus $N(t)/M(t)$ is a martingale under $Q^M$. This leads to

$$E^{Q^M}\left[\frac{M(0)N(T)}{M(T)N(0)}|\mathcal{F}_t\right]=\frac{M(0)N(t)}{M(t)N(0)}$$

Deviding by this fraction results in

$$ E^{Q^N}\left[\frac{S(T)}{N(T)}|\mathcal{F}_t\right]=\frac{N(0)M(t)}{N(t)M(0)}E^{Q^M}\left[\frac{S(T)}{M(T)}\frac{M(0)}{N(0)}|\mathcal{F}_t\right]= \frac{N(0)M(t)}{N(t)M(0)}\frac{M(0)}{N(0)}E^{Q^M}\left[\frac{S(T)}{M(T)}|\mathcal{F}_t\right]$$

This leads to

$$E^{Q^N}\left[\frac{S(T)}{N(T)}|\mathcal{F}_t\right]=\frac{M(t)}{N(t)}E^{Q^M}\left[\frac{S(T)}{M(T)}|\mathcal{F}_t\right]$$

Now we know that $S(t)/M(t)$ is a martingale under $Q^M$. Thus the desired result follows.

$$E^{Q^N}\left[\frac{S(T)}{N(T)}|\mathcal{F}_t\right]=\frac{M(t)S(t)}{N(t)M(t)}$$