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Why Zero-Coupon Bond Prices Are Stochastic Under Stochastic Rates

Article Quant Q&A · Author: apelle

Summary

The document explains why a zero-coupon bond price can be stochastic even when it is defined as a risk-neutral conditional expectation. The key distinction is that conditioning on information available at time t produces a value measurable with respect to that information, not necessarily a deterministic constant. As the market learns more about future interest rates, the expected discounted payoff can change.

The discussion connects this point to the risk-neutral dynamics of discounted asset prices and the resulting stochastic differential equation for the bond. It uses an example of updating an expected future stock price as new price information arrives. The example is intuitive rather than a derivation, and the text does not work through the bond dynamics or specify assumptions on the rate process. Its central lesson is that a conditional expectation is generally random when the conditioning information varies across states.

Key ideas

  • A conditional expectation given current information is measurable with respect to that information and can remain random.
  • A zero-coupon bond price can vary with the observed state of stochastic interest rates.
  • Risk-neutral bond pricing takes the conditional expectation of future discounting given current information.
  • The document uses an updating expectation example to explain why conditioning does not make a value deterministic.

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Full text
# Stochastic representation of a zero-coupon bond


# Stochastic representation of a zero-coupon bond












In Chapter 9 of Shreve's book Stochastic Calculus for Finance II, the main theorem is the 9.2.1. Defining the discounting process $D(t)=\mathrm{e}^{-\int_0^t du r(u)}$ and $r(u)$ the, possibly stochastic, interest rate process; the theorem states that any positive price process $N$, under the risk neutral measure with Brownian motion $\tilde{W}_t$, obeys the following equation: $$(i) d(D(t)N(t))=D(t)N(t)\nu(t)\cdot d\tilde{W}_t, $$ where $\nu(t)$ is defined using the Martingale Representation Theorem. Moreover, massaging the equation we can rewrite equation (i) as: $$(ii) dN(t)=r(t)N(t)dt+N(t)\nu(t)\cdot d\tilde{W}_t.$$ In the same chapter in section 9.4 he defines the zero-coupon bond as $$ (iii) B(t,T)=\tilde{E}[D(T)/D(t)|\mathcal{F}_t]=\tilde{E}[\mathrm{e}^{-\int_t^T du r(u)}|\mathcal{F}_t], $$ where $B(T,T)=1$ and the expectation is over the risk neutral measure. Clearly, since $B$ is defined as an expectation value it is a non-random function. Then, in the following, in order to define the T-forward measure, the author uses theorem 9.2.1 to write equation (i) for the zero-coupon bond as: $$(iv)d(D(t)B(t,T))=D(t)B(t,T)\sigma(t,T)\cdot d\tilde{W}_t$$ Now since the theorem applies also equation (ii) it is supposed to make sense and then: $$(v)dB(t,T)=r(t)B(t,T)dt+B(t,T)\sigma(t,T)\cdot d\tilde{W}_t.$$ I.e. the zero-coupon bond price is described by a stochastic Ito equation!

Now my problem is that since equation (iii) defines $B$ as a deterministic function,is it possible to write equation (v) for $B$? It makes sense that equation (iv) is an Ito equation since $D(t)$ is, in principle, a random function, but what is the meaning of equation (v) then? A possible justification could be constructed considering that, actually, in equation (v) the stochastic interest rate $r$ and the volatility process $\sigma$ given by the Martingale Representation theorem are connected to each other, and then it would be possible (in theory) to "simplify" equation (v) and obtain a deterministic differential equation describing $B$ as a function of $t$. Is this the correct way to look at equation (v)?

Or maybe I am wrong and $B(t,T)$ is a stochastic function. Then why an expectation value is not deterministic function?

## Answer by Rylan (score 2, accepted)

https://quant.stackexchange.com/a/78079

$B(t, T)$ (and in general conditional expectations of random variables, under the usual conditions) are not deterministic but $F(t)$ measurable. $r(t)$ here is also $F(t)$ measurable.

One intuitive way of explaining conditional expectations being $F(t)$ measurable but not deterministic: suppose a stock is worth \$100 on Jan 1 2024. On Jan 1 2024, we may expect it to be worth $105 on Jul 1 2024, then worth \$110 on Jan 1 2025. However, if at Jul 1 2024 it's reached a value of \$150, then we will likely adjust our expectation of the stock's price on Jan 1 2025 to be higher than \$110.

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.