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Why Bond Price Ratios Are Martingales Under the Forward Measure

Article Quant Q&A · Author: R. Rayl
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Summary

The document asks why a zero-coupon bond price divided by a bond maturing at T is a martingale under the T-forward measure. The key principle is that this measure uses the T-maturity bond as numeraire, so prices of tradable assets expressed in units of that bond are martingales, subject to the usual assumptions behind the pricing framework.

One response sketches a change-of-measure argument using a second forward measure and a Radon–Nikodym density; another states the numeraire principle directly. The attempted conditional-expectation argument in the question does not establish the result as written, since it effectively assumes the martingale property it aims to prove. The discussion is brief and provides no model-specific derivation or treatment of integrability and measure definitions.

Key ideas

  • The T-forward measure takes the zero-coupon bond maturing at T as numeraire.
  • Tradable asset prices divided by that numeraire are martingales under the corresponding measure, under standard pricing assumptions.
  • A change-of-measure density can relate forward measures with different maturities.
  • The original conditional-expectation argument is circular unless the relevant martingale property is independently established.

Tags

Fixed incomeStatistics
Full text
# Show that a zero-coupon bond discounted by a bond with mautrity $T$ is a martingale under the $T$-Forward measure


# Show that a zero-coupon bond discounted by a bond with mautrity $T$ is a martingale under the $T$-Forward measure












Here's the exact question:

Show that for any $s>0$, $\frac{P(t,s)}{P(t,T)}$ is a $Q^T$-martingale.

Here's my attempt:

Let $t^\prime < t$. First consider the case $s>T$. \begin{aligned} \mathbb{E}_{Q^T}\Big[\frac{P(t,s)}{P(t,T)} \lvert \mathcal{F}_{t^\prime}\Big] &= \mathbb{E}_{Q^T}\Big[P(T,s) \lvert \mathcal{F}_{t^\prime}\Big] \\ &= \mathbb{E}_{Q^T}\Big[\frac{P(t^\prime,s)}{P(t^\prime,T)} \lvert \mathcal{F}_{t^\prime}\Big] \\ &= \frac{P(t^\prime,s)}{P(t^\prime,T)} \end{aligned} And then you can use a similar argument for when $T > s$. But this argument has to be wrong surely, as this is not specific to $Q_T$. Could anyone help and point out where I've gone wrong?

## Answer by NN2 (score 0)

https://quant.stackexchange.com/a/51733

Suppose that $T<S$. Arcording to Girsanov, we have $$\frac{dQ^T}{dQ^S}|_{F_t'}= \frac{P(T,T)/P(t',S)}{P(T,S)/P(t',S)} =\frac{1}{P(T,S)}\frac{P(t',S)}{P(t',T)}$$ So $$dQ^T|_{F_t'} =\frac{1}{P(T,S)}\frac{P(t',S)}{P(t',T)} dQ^S|_{F_t'}$$ $$E_{Q^T} (\frac{P(t,S)}{P(t,T)}|F_{t'}) =E_{Q^S} (\frac{P(t,S)}{P(t,T)} *\frac{1}{P(T,S)}\frac{P(t',S)}{P(t',T)}|F_{t'}) =\frac{P(t',S)}{P(t',T)} E_{Q^S} (\frac{P(t,S)}{P(t,T)} *\frac{1}{P(T,S)}|F_{t'}) = \frac{P(t',S)}{P(t',T)}$$ We use similar demonstration for the case $T>S$. We can conclude that $\frac{P(t,S)}{P(t,T)}$ is a $Q^T$ martingale.

## Answer by Andrea (score 0)

https://quant.stackexchange.com/a/81248

It is true by definition of the $T$-forward measure.

Which is the measure that makes every tradable asset discounted by the $T$-bond a martingale.

For the uniqueness of one of the comments, not all $T$-measures are different, since in a model with deterministic interest rates (make $r=0$, it is even easier), they are all the same.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.