In Shreve II, on p. 265 he states the Hull-White interest rate model as $$ dR(u) = \left( a(u) - b(u)R(u)\right) dt + \sigma(u)d\tilde{W}(u), $$ and then mentions "...$\tilde{W}(u)$ is a Brownian motion under a risk-neutral measure $\tilde{\mathbb{P}}$." However, when he defines a risk-neutral measure on p. 228, he states that $\tilde{\mathbb{P}}$ is a measure under which the discounted stock price is a martingale.

This definition doesn't really apply here, so what is meant by a "risk-neutral measure" when modelling interest rates? Also, why do interest rate models always seem to be stated under these risk-neutral probabilities?


2 Answers 2


It is a very interesting question. There is a brief explanation in the book Martingale methods in financial modelling. Basically, it says that, the interest short rate $r_t$ can be modeled in any martingale measure $Q$, however, as long as the zero-coupon bond price $P(t, T)$ is defined by \begin{align*} P(t, T) = E^{Q}\Big(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t\Big) \end{align*} then the discounted bond price $$\frac{P(t, T)}{B(t)},$$ is a $Q-$martingale, and is arbitrage free. Here $B(t)= e^{\int_0^tr_sds}$ is the money market account value. This provides us the freedom to choose the martingale measure, and people always assume that the interest rate model is defined under the risk-neutral probability measure.

  • $\begingroup$ Okay, thanks for that. Discounted bond price = $e^{-\int_t^T r_s ds} B(t,T) $? Discounted by what? Actually, if $B(t,T) = E_{P^*}\left(e^{-\int_t^T r_s ds}|\mathcal{F}_t\right)$, isn't $B(t,T)$ itself already a $P^*$-martingale? $\endgroup$
    – bcf
    Commented Jun 5, 2015 at 14:03
  • $\begingroup$ @bcf, I added more details. The discounting is relative to the money market account, or saving account, value. $\endgroup$
    – Gordon
    Commented Jun 5, 2015 at 14:11
  • $\begingroup$ @bcf, note that $B(t, T)$ itself is not a martingale. $\endgroup$
    – Gordon
    Commented Jun 5, 2015 at 14:18
  • $\begingroup$ I've actually confused myself again, in a chicken vs. egg mystery. Are we using the FTAP to define $B(t,T)$ as a conditional expectation? As in, this is really the price of a self-financing portfolio replicating the payoff $B(T,T) = 1$? Don't we usually go the other way in derivatives pricing? I.e., we first find the measure for which the discounted (bond? or rate?) process is a martingale, then invoke the FTAP to get the above formula? Thanks again. $\endgroup$
    – bcf
    Commented Jun 9, 2015 at 19:25
  • $\begingroup$ Good answer. What about a corporate (defaultable) bond? $\endgroup$
    – tosik
    Commented Feb 28, 2019 at 17:24

The definition of a risk-neutral probability measure depends on the model. The (one factor) Interest Rate Model in Shreve II consists of a single zero-coupon bond $B(t,T)$ with maturity $T$ and of a money market account. So we want discounted bond price to be a martingale under risk-neutral probability measure. We define it as usual (i.e. Shreve II, 5.2.2.):

Assume that the interest rate $R(t)$ and the bond $B(t,T)$ processes satisfy their respective stochastic differential equations under the actual probability: $$ dR(t) = \xi(t,R(t))dt + \phi(t, R(t))dW(t)$$ $$ dB(t,T) = \mu(t,T)B(t,T)dt + \sigma(t,T)B(t,T)dW(t)$$ where $W(t)$ is a Brownian motion.

The discount process $D(t) = e^{-\int_0^t R(s)ds}$ so as usual $ dD(t) = -R(t)D(t)dt$

We want the discounted bond price to be a martingale: $$ d(D(t)B(t,T)) = D(dB(t,T) - R(t)B(t,T)dt) = D(t)B(t,T)\sigma(t,T)\Big(\frac{\mu(t,T) -R(t)}{\sigma(t,T)}dt + dW(t)\Big) = D(t)B(t,T)\sigma(t,T)\Big(\theta(t)dt + dW(t)\Big)$$

where we defined the market price of risk $\theta(t) = \frac{\mu(t,T) -R(t)}{\sigma(t,T)}$.

We introduce risk-neutral probability measure $\tilde{\mathbb{P}}$ using Girsanov's theorem as usual.

The above considerations do not depend on the form of SDE for the interest rate process $R(t)$ so it is ok to start right from the riks-neutral probability measure as it is done in Shreve's book.


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