Given, a numéraire $(N(t))_{0\leq t \leq T}$ and an index $(X(t))_{0\leq t\leq T}$ that is a $\mathbb Q^{N}$-martingale, we consider the natural payoff $V_{N}(T)$, where it pays

$$V_{N}(T):=X(T)N(T) \; \; \text{in }T,$$

i.e. it pays the index $X(T)$ in units of $N(T)$.

Now let us consider the payoff $V_{M}(T)$, where

$$ V_{M}(T):=X(T)M(T)\; \; \text{in }T.$$

Question: It is stated that the value of $V_{M}(T)$ equals the value of the instrument that pays a "new" index $\frac{\tilde{X}(0)}{X(0)}X(T)$ in units of $N(T)$, where $$\tilde{X}(0):=X(0)+\frac{N(0)}{M(0)}\mathbb E^{\mathbb Q^{N}}\left[\int_{0}^{T}d\frac{V_{N}(t)}{N(t)}\cdot d\frac{M(t)}{N(t)}\right]$$


I know how to arrive at $\tilde{X}(0)$ when defining $\tilde{X}(0)$ such that $$N(0)\cdot \mathbb E ^{\mathbb Q^{N}}\left[\frac{V_{M}(T)}{N(T)}\right]=V_{M}(0)=:\tilde{X}(0)\cdot M(0)$$

I just really do not understand the statement on the values of $V_{M}(T)$ and $\frac{\tilde{X}(0)}{X(0)}X(T)\cdot N(T)$ being equal.

In my attempt, the value of the "new" index is:

$N(0)\mathbb E^{\mathbb Q^{N}}\left[\frac{\frac{\tilde{X}(0)}{X(0)}X(T)\cdot N(T)}{N(T)}\right]=\tilde{X}(0)N(0)$ which of course does not necessarily equal $\tilde{X}(0)\cdot M(0)$

I think I may be missing something rather fundamental here, any ideas? Or is this simply a typo?


1 Answer 1


One way to attack this problem is obviously by invoking Girsanov theorem. Let's try to reach the same conclusion without it.

The first contingent claim delivers a payout $V^N(T) = X(T) N(T)$. Assuming that $(X(t))_{0 < t \leq T}$ is a $\Bbb{N}$-martingale, under the measure associated to the numéraire $N(t)$ we then get: $$ V^N(0) = N(0) \Bbb{E}_0^\Bbb{N} \left[ X(T) \right] = N(0) X(0) $$

The second contingent claim delivers a payout $V^M(T) = X(T) M(T)$. Under the measure associated to numéraire $M(t)$ we get: $$ V^M(0) = M(0) \Bbb{E}_0^\Bbb{M} \left[ X(T) \right] = M(0) \tilde{X}(0) \ne M(0) X(0) $$ since $(X(t))_{0 < t \leq T}$ is not a $\Bbb{M}$-martingale a priori but we defined $$ \tilde{X}_0 := \Bbb{E}_0^\Bbb{M} \left[ X(T) \right] $$

One can then write \begin{align} \tilde{X}_0 &= \Bbb{E}_0^\Bbb{M} \left[ X(T) \right] \\ &= \Bbb{E}_0^\Bbb{N} \left[ X(T) \frac{M(T)}{N(T)} \frac{N(0)}{M(0)} \right] \\ &= \frac{N(0)}{M(0)} \Bbb{E}_0^\Bbb{N} \left[ \frac{V^N(T)}{N(T)} \frac{M(T)}{N(T)} \right] \\ &= \frac{N(0)}{M(0)} \left( \Bbb{E}_0^\Bbb{N} \left[ \frac{V^N(T)}{N(T)} \right] \Bbb{E}_0^\Bbb{N} \left[ \frac{M(T)}{N(T)} \right] + \text{cov}\left( \frac{V^N(T)}{N(T)} , \frac{M(T)}{N(T)} \right)\right) \\ &=\frac{N(0)}{M(0)} \left( X(0) \frac{M(0)}{N(0)} + \int_0^T \Bbb{E}_0^\Bbb{N} \left[ d\left\langle \frac{V^N}{N}, \frac{M}{N} \right\rangle_t \right] \right) \\ &= X(0) + \frac{N(0)}{M(0)} \Bbb{E}_0^\Bbb{N} \left[ \int_0^T d\left\langle \frac{V^N}{N}, \frac{M}{N} \right\rangle_t \right] \end{align} hence the corresponding convexity adjustment. In the above, we have respectively used the following identities to move from one line to the other

  • definition of (change of) numéraire
  • definition of $V^N$
  • definition of (terminal) covariance between 2 random variables
  • Martingale property of $V^N(t)/N(t)$ and $M(t)/N(t)$ under $\Bbb{N}$ (both $V^N(t)$ and $M(t)$ represent the $t$-values of a self-financing traded strategy within our model economy, as such they their prices are martingales when expressed in $N_t$ units) along with Itô isommetry to tie terminal covariance to quadratic covariation
  • Linearity of expectation operator.

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