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Price European Calls with Deterministic Time-Varying Interest Rates

Article Quant Q&A · Author: Tyler D

Summary

This discussion considers European call pricing when the risk-free rate varies deterministically over time. It starts from a stock following geometric Brownian motion under the risk-neutral measure and distinguishes the bank-account-discounted stock from the payoff at maturity. The pricing expression discounts the payoff using the growth of the risk-free account; the payoff itself is expressed in terms of the actual terminal stock price, not the discounted stock.

The answer also presents a numéraire-change interpretation. Using a zero-coupon bond as numéraire makes the bond-forward stock price a martingale under the corresponding forward measure. It decomposes the call price into a stock component valued under the stock measure and a strike component valued under the bond measure, each involving an exercise probability. The derivation assumes deterministic rates for the simplification equating the standard risk-neutral and bond-forward measures; the broader measure-change framework is stated separately, and no specific volatility distribution is needed for the decomposition.

Key ideas

  • The call payoff is based on the actual terminal stock price and is discounted by the risk-free account.
  • With deterministic rates, discounting the payoff is equivalent to using the zero-coupon bond price.
  • Under the bond-forward measure, the stock price divided by the bond price is a martingale.
  • A call price can be decomposed into two exercise-probability terms under stock and bond measures.
  • The probability decomposition does not require a particular stock-price distribution.

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Full text
# Martingale pricing with time-dependent risk-free rate


# Martingale pricing with time-dependent risk-free rate












I want to find the price of a European call-option under the assumption that the risk-free rate $r$ is time-dependent, i.e.

$$ d\beta = r(t)\beta dt \leftrightarrow \beta(T) = e^{\int_0^T r(u)du} $$

I want to express the price in terms of zero-coupon bond prices $P$, where we know that $P(t;T) = \beta(t)/\beta(T)$.

I just learned about margingale pricing, but here is my strategy: My starting point is GBM under the risk-neutral measure $Q$: $dS = rSdt + \sigma S dW^Q$. Introducing $\hat S = S/\beta(t)$ and using Ito's product rule we find

$$ d\hat S = \sigma \hat S dW^Q $$

In other words, the discounted GBM is a martingale under $Q$, which in turn means that we can find the call price as

$$ C(t) = \frac{\beta(t)}{\beta(T)}\mathbb E^Q[(S(T)-K)^+] $$

Question: In the expectation value for calculating $C(t)$, do I need $S(T)$ or $\hat S(T)$?

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/61159

### Numéraire Change

The time-$t$ price of a zero-coupon bond maturing at time $T$ is $$P(t,T)=\mathbb{E}^\mathbb{Q}_t\left[\exp\left(-\int_t^T r_s\text{d}s\right)\right].$$

Let $\mathbb{Q}$ be our standard risk-neutral probability measure which uses a locally risk-free bank account, $\text dB_t=r_tB_t\text dt$, as numéraire. From Geman et al. (1995), we know \begin{align} \frac{\text d\mathbb Q^T}{\text d\mathbb Q}\Bigg|_{\mathcal{F}_t}=\frac{P(T,T)}{P(t,T)}\frac{B_t}{B_T}=\frac{1}{P(t,T)}\frac{B_t}{B_T}. \end{align} Then, the forward price $\frac{S_t}{P(t,T)}$ is a $\mathbb{Q}^T$-martingale, i.e. \begin{align*} S_t = P(t,T)\mathbb{E}^{\mathbb{Q}^T}_t[S_T]. \end{align*} When interest rates are deterministic, $\mathbb{Q}=\mathbb{Q}^T$ and, as always, $$ S_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t[S_T].$$

For an equivalent probability measure which uses the (reinvested) stock as numéraire, we get \begin{align} \frac{\text d\mathbb{Q}^S}{\text d\mathbb Q}\Bigg|_{\mathcal{F}_t}=\frac{S_Te^{qT}}{S_te^{qt}}\frac{B_t}{B_T}. \end{align}

### Option Pricing

The initial value of a call option is thus \begin{align*} \text{Call}(S_0;K,T)&=\mathbb{E}^\mathbb{Q}_0\left[\frac{B_0}{B_T}\max\{S_T-K,0\}\right] \\ &= \mathbb{E}^\mathbb{Q}_0\left[\frac{B_0}{B_T}S_T\mathrm{1}_{\{S_T\geq K\}}\right]-K\mathbb{E}^\mathbb{Q}_0\left[\frac{B_0}{B_T}\mathrm{1}_{\{S_T\geq K\}}\right] \\ &= S_0e^{-qT}\mathbb{E}^{\mathbb{Q}^S}_0\left[\mathrm{1}_{\{S_T\geq K\}}\right]-KP(0,T)\mathbb{E}^{\mathbb{Q}^T}_0\left[\mathrm{1}_{\{S_T\geq K\}}\right] \\ &= S_0e^{-qT}\mathbb{Q}^S\left[\left\{S_T\geq K\right\}\right]-KP(0,T)\mathbb{Q}^T\left[\left\{S_T\geq K\right\}\right]. \end{align*}

This is Theorem 2 in Geman et al. (1995) and beautifully decomposes option prices into two exercise probabilities. Note that we made no assumptions about the distribution of the stock price. Assuming constant interest rates and normally distributed stock returns nests the Black and Scholes (1973) formula.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.