15
$\begingroup$

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?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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
    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
    Jun 5, 2015 at 14:11
  • $\begingroup$ @bcf, note that $B(t, T)$ itself is not a martingale. $\endgroup$
    – Gordon
    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
    Jun 9, 2015 at 19:25
  • $\begingroup$ Good answer. What about a corporate (defaultable) bond? $\endgroup$
    – tosik
    Feb 28, 2019 at 17:24
2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.