# How am I supposed to understand the following statement on the convexity adjusted rate

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?

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.