Skip to content
All library documents

Risk-Neutral Bond Pricing and the Money Market Account

Article Quant Q&A · Author: Inawashiro Hiromichi
Read the original

Summary

This exchange explains a common mistake in risk-neutral bond valuation: treating the money market account, or its reciprocal, as a martingale under the risk-neutral measure. The bond price is expressed as the conditional expectation of the accumulated short-rate discount factor. The question then incorrectly applies a martingale argument to conclude that this expectation equals one, which would imply every bond trades at par regardless of maturity or interest rates.

The answers clarify that risk-neutral pricing concerns the bond price discounted by the money market account; it does not imply the account itself is a martingale. One reply points to constant interest rates as a simple counterexample. Another derives the bond price from the martingale property of the discounted bond and notes that its exact value depends on the dynamics of the short rate. The discussion is conceptual rather than a full model-based calculation, and one displayed expression in a reply appears to omit an exponential around the integrated rate.

Key ideas

  • Risk-neutral bond prices are conditional expectations of discount factors based on the short rate.
  • The money market account is not generally a martingale under the risk-neutral measure.
  • The martingale condition applies to appropriately discounted asset prices.
  • A bond's exact price cannot be determined without specifying the interest-rate dynamics.

Tags

Fixed incomeDerivatives pricingStatistics
Full text
# Bond price under the risk-neutral measure


# Bond price under the risk-neutral measure












Could you point out where I am making mistake in the process below?

It follows from the term structure equation and the Feynman-Kac theorem that the bond price is given by

$ p(t,T) = E_t^Q\left[ \exp\left( -\int_t^T r(u) du \right) \right], $

where $E_t^Q$ denotes the expectation at time $t$ under the risk neutral measure $Q$.

Let the money market account be

$ B(t) = \exp\left( \int_0^t r(u) du \right), $

and the bond price expression above is written as

$ p(t,T) = E_t^Q\left[\frac{B(t)}{B(T)} \right]. $

Since the numeraire of $Q$ is $B$, it follows from the martingale property that

$ E_t^Q\left[ \frac{B(t)}{B(T)} \right] = E_t^Q\left[ \frac{B(t)}{B(t)} \right] = 1. $

Thus, $p(t,T)=1$.

## Answer by masi (score 1)

https://quant.stackexchange.com/a/60138

No, $B$ is not a $Q$ martingale, neither is $1/B$, which you have assumed in your calculation (try using constant $r$, for an example of why this can be the case). The measure $Q$ is a risk neutral measure if the stock price processes that are discounted by $B$ are martingales.

## Answer by p.vitzliputzli (score 0)

https://quant.stackexchange.com/a/60168

I think you have your argument slightly mixed up. Assuming the existence of a risk-neutral measure $Q$, you know from the risk-neutral pricing formula that the discounted bond price is a martingale, i.e. ($D(t)$ being the discount factor) $$D(t)p(t,T) = E_t^Q[D(T)p(T,T)] = E_t^Q[D(T)].$$ With $D(t) = \mathrm{exp}(-\int_0^t r(u) du)$, we immediately get $$p(t,T) = \frac{1}{D(t)}E_t^Q[D(T)] = E_t^Q[-\int_t^T r(u) du].$$ Therefore, the equation for the bond price includes already the numeraire. Using it again wouldn't make sense. Also, without further information about $r(t)$, it is not possible to derive an exact solution for the bond price.

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.