Why Numeraire Changes Simplify LIBOR Market Model Pricing

# Libor Market Model: numeraire change


# Libor Market Model: numeraire change












I am currently studying the Libor forward market model, and although I get the mechanics behind the main arguments, I still do not have an intuitive idea of what's exactly the objective behind changing the numeraire to obtain the desired dynamics.

Why do we need to do this so that the dynamics of the forward rate make sense? Or, if we put it from a different angle: if this numeraire change would not "exist", what could we not achieve exactly in this model?

## Answer by roym00 (score 5, accepted)

https://quant.stackexchange.com/a/15954

EDIT: I changed the answer to have it more on topic.

### Summary

It boils down to Mark Joshi's answer. I wanted to add something more.

### Answer

A probability measure $Q1$ and a numeraire $N1(t)$ are associated if all prices expressed relative to $N1$ are martingales under $Q1$:

$$\frac{price(t)}{N1(t)} = \mathbb{E}^{Q1} \left[ \left. \frac{price(T)}{N1(T)} \, \right| \, F_t\right] \quad \Rightarrow \quad price(t) = \mathbb{E}^{Q1} \left[ \left. \frac{N1(t) \cdot price(T)}{N1(T)} \, \right| \, F_t \right] $$

Having another probability measure $Q2$ with associated numeraire $N2(t)$, then you can change the numeraire:

$$ \begin{array}{ccl} price(t) & = & \mathbb{E}^{Q1} \left[ \left. \frac{N1(t) \cdot price(T)}{N1(T)} \, \right| \, F_t \right] = N1(t) \cdot \mathbb{E}^{Q1} \left[ \left. \frac{price(T)}{N1(T)} \, \right| \, F_t \right] \\ & = & \mathbb{E}^{Q2} \left[ \left. \frac{N2(t) \cdot price(T)}{N2(T)} \, \right| \, F_t \right] = N2(t) \cdot \mathbb{E}^{Q2} \left[ \left. \frac{price(T)}{N2(T)} \, \right| \, F_t \right] \end{array} $$

Given information at time $t$, you are able to take out both numeraires at the same time since they are known.

If you set

- $Q1$ equal to the risk-neutral probability measure $QRN$ and $N1(t)$ equal to the risk-free asset, i.e. $N1(t) = B(t) = e^{\int_0^t r(s) ds}$

- $Q2$ equal to the T-forward probability measure $QT$ and $N2(t)$ equal to the zero-coupon bond price with maturity $T$ computed at $t$, i.e. $N2(t) = P(t,T)$

then you have the following:

$$ \begin{array}{ccl} price(t) & = & \mathbb{E}^{QRN} \left[ \left. \frac{B(t) \cdot price(T)}{B(T)} \, \right| \, F_t \right] = B(t) \cdot \mathbb{E}^{QRN} \left[ \left. \frac{price(T)}{B(T)} \, \right| \, F_t \right] \\ & = & \mathbb{E}^{QT} \left[ \left. \frac{P(t, T) \cdot price(T)}{P(T, T)} \, \right| \, F_t \right] = P(t, T) \cdot \mathbb{E}^{QT} \left[ \left. \frac{price(T)}{P(T, T)} \, \right| \, F_t \right] \\ \end{array} $$

The problem under the risk-neutral measure is that $B(T)$ and $price(T)$ are not independent, especially for interest rate derivatives. However, $P(T,T)$ is perfectly known: it's just $1$.

$$ price(t) = P(t, T) \cdot \mathbb{E}^{QT} \left[ \left. \frac{price(T)}{P(T, T)} \, \right| \, F_t \right] = P(t, T) \cdot \mathbb{E}^{QT} \left[ \left. price(T) \, \right| \, F_t \right] $$

So with the change of numeraire you simplified a lot the computation, this is what you achieved.

In general, the change of numeraire is useful mainly for two reasons ($X_t$ is a process and $N_t$ is the numeraire):

- $X_t = \frac{tradable asset}{N_t}$ is a martingale under the new measure, i.e. you were not able to prove directly $X_t$'s martingality but you are able under the new measure with the new numeraire.

- $\frac{X_t}{N_t}$ becomes a very easy quantity to compute; this is the case of the T-forward measure in the LIBOR market model, since the expectation of the discounted payoff in the risk-neutral world becomes a deterministic discount factor times the the expectation today of the payoff.

I add that a theorem guarantees that all results are the equal: if there exists a probability measure associated to a numeraire and all assets expressed under this measure are martingales, then any other (traded) asset can be used as numeraire and any numeraire choice will lead to the same price.

## Answer by Mark Joshi (score 3)

https://quant.stackexchange.com/a/15556

Different measures have different properties. Using a particular measure may make it easy to derive an analytic formula since a rate is driftless. When performing Monte Carlo, the sign of the drifts changes with measure which affects convergence. There is also the problem in the terminal measure that the numeraire can get very small and so some paths can lead to very large values.

Essentially, it comes down to convenience. I have extensive discussion of these points in "more mathematical finance."