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 $\hat S(T)$?


1 Answer 1


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.


Your Answer

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

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