Is the interest rate allowed to be truly stochastic in the binomial pricing model and in continuous models so that we are still able to switch to the risk-neutral measure?

Shreve mentions multiple times that the interest rate has to be adapted to the stock price process's filtration. My understanding is as follows: let a binomial pricing model of two periods be given. Let us index prices with $i$: $S_i$ for the stock price in $i^{th}$ period. Also let us index prices with $u$ and $d$ for when the previous move in the price was up or down. In other words, $S_{1}^{u}$ stands for the stock price in the first period, whereby it has just moved up ($S_0*u$). Also let the interest rate be fully random and not adapted to the filtration of the discrete stock price process. Index the interest rates similarly but let $r^A$ stand for outcome $A$ and $r^B$ stand for outcome $B$. In other words, $r$ is a discrete R.V. and has two possible outcomes. As always, suppose $d \leq (1+r) \leq u$ to prevent arbitrage. To hedge a short position in that option, we have to hedge eight states of nature:

  1. $r_1^A, S_2^{uu}$
  2. $r_1^A, S_2^{ud}$
  3. $r_1^B, S_2^{uu}$
  4. $r_1^B, S_2^{ud}$
  5. $r_1^A, S_2^{du}$
  6. $r_1^A, S_2^{dd}$
  7. $r_1^B, S_2^{du}$
  8. $r_1^B, S_2^{dd}$

Also, because we can readjust hedge only based on the information (filtration) available to us at time $t$, let us denote $M$ as the initial investment into the money market account and let $\delta_i(S_i^{w}, r_i^{w})$ be the hedging position in stocks at time $i$, depending on the outcomes $w$. Say, $\delta_1(S_1^u, r_1^A)$ would the position in stocks that we take in period 1 when the stock has moved up and the interest rate has ended up $A$. Then there are six such hedging positions based on filtration possible:

  1. $\delta_0$
  2. $M$
  3. $\delta_1(S_1^u, r_1^A)$
  4. $\delta_1(S_1^d, r_1^A)$
  5. $\delta_1(S_1^u, r_1^B)$
  6. $\delta_1(S_1^d, r_1^B)$

Eight equations in six unknowns - unhedgeable. We need one more leverage factor to hedge this. Is this correct?

A colleague quant has told me that the interest rate process does not have to adapted to the filtration of the stock process and that it's only a technical assumption. Well yes but then we cannot switch to the risk-neutral measure because we cannot hedge, and the risk-neutral measure is all about hedging... Is this the correct understanding of mine?


1 Answer 1


I think both are right in different ways! Risk neutral measure is the measure associated with the bank account, and the standard definition of bank account is indeed an adapted process. And technical conditions are indeed there for a reason.

But then you can have, say the T-measure, which uses zero coupon, whose return is stochastic in each step, except in the last step from $T-1$ to $T$, because the payoff at T is known. And the bank account return is nothing but the return of the bond that matures at next time step.

  • $\begingroup$ thank you for your answer. what is exactly the T-measure? could you pls provide textbook references or formal definition? could not google it... but still, what I don't understand is how one can hedge with unadapted interest rate. Essentially, we can 'imagine' that discounted stock process is a martingale (risk-neutral measure) when we can construct a perfect hedge... for this the risk-free rate must be either constant or adapted to filtration of the stock process. If this is not the case, there is no perfect hedge, and risk-neutral measure is not justified?... $\endgroup$ Commented Aug 4, 2019 at 11:26
  • $\begingroup$ Please see page 38 (section 2.5) of Brigo's interest rate models theory and practice $\endgroup$ Commented Aug 4, 2019 at 12:54
  • $\begingroup$ thank you, @MagicIsInTheChain, I check it out the next days $\endgroup$ Commented Aug 4, 2019 at 15:25

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