4
$\begingroup$

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 $$\tilde{X}(0):=X(0)+\frac{N(0)}{M(0)}\mathbb E^{\mathbb Q^{N}}\left[\int_{0}^{T}d\frac{V_{N}(t)}{N(t)}\cdot d\frac{M(t)}{N(t)}\right]$$

Comment:

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?

$\endgroup$

1 Answer 1

2
+50
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.