Black-Scholes under stochastic interest rates

Question

I'm trying to implement the Black-Scholes formula to price a call option under stochastic interest rates. Following the book of McLeish (2005), the formula is given by (assuming interest rates are nonrandom, i.e. known):

$E[exp\{-\int_0^Tr_t dt\}(S_T-k)^+]$

=$E[(S_0 exp\{N(-0.5\sigma^2T,\sigma^2T)\}-exp\{-\int_0^Tr_tdt\}K)^+]$

=$BS(S_0,k,\bar{r},T,\sigma)$

where $\bar{r}=\frac{1}{T}\int_0^Tr_tdt$ is the average interest rate over the life of the option .

If interest rates are random, "we could still use the Black-Scholes formula by first conditioning on the interest rates, so that

$E[e^{-\bar{r}T}(S_T-K)^+|r_s, 0<s<T]= BS(S_0,K,\bar{r},T,\sigma)$

and then computing the unconditional expected value of this by simulating values of $\bar{r}$ and averaging".

I'm not sure how can I calculate $\bar{r}$ given a simulated sample paths.

Answer 1

We assume that the short interest rate $r_t$ follows the Hull-White model, that is, the short rate $r$ and the stock price $S$ satisfies a system of SDEs of the form \begin{align*} dr_t &= (\theta_t -a\, r_t)dt + \sigma_0 dW_t^1,\\ dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big)\Big], \end{align*} where $a$, $\sigma_0$, $\sigma$, and $\rho$ are constants, and $\{W_t^1, t\ge 0\}$ and $\{W_t^2, t\ge 0\}$ are two independent standard Brownian motions.

Note that, \begin{align*} &\ E\bigg(\exp\Big(-\int_0^T r_t dt \Big) (S_T-K)^+\bigg) \\ =& \ E\bigg(e^{-\bar{r}T} \Big(S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T - \sigma \big(\rho W_T^1 + \sqrt{1-\rho^2}W_T^2\big)} -K\Big)^+ \bigg)\\ =& \ E\Bigg(E\bigg(e^{-\bar{r}T} \Big[S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T + \sigma \big(\rho W_T^1 + \sqrt{1-\rho^2}W_T^2\big)} -K\Big]^+ \Bigg\vert r_s, 0<s \leq T\bigg)\Bigg)\\ =& \ E\Big(F(S_0,K,\bar{r},T,\sigma, W_T^1) \Big\vert r_s, 0<s \leq T\Big), \end{align*} for a certain function $F$. Note the random variable $W_T^1$ in the formula.

If $\rho=0$, that is, $S$ and $r$ are independent, then \begin{align*} &\ E\bigg(\exp\Big(-\int_0^T r_t dt \Big) (S_T-K)^+\bigg) \\ =& \ E\Bigg(E\bigg(e^{-\bar{r}T} \Big(S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T + \sigma W_T^2} -K\Big)^+ \bigg\vert r_s, 0<s \leq T\bigg)\Bigg)\\ =&\ E\Big(BS(S_0,K,\bar{r},T,\sigma) \Big\vert r_s, 0<s \leq T \Big). \end{align*} That is, the formula provided in the question holds if the stock price and the interest rate are independent. In this case, $\bar{r}$ can be approximated by a Riemann sum.

EDIT

Here, we provide an analytical valuation formula for the above vanilla European option. From this question, the zero-coupon bond price is given by \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \Big\vert \mathcal{F}_t \right)\\ &=\exp\left(-B(t, T) r_t - \int_t^T \theta(s) B(s, T) ds + \frac{1}{2}\int_t^T \sigma_0^2 B(s, T)^2 ds\right), \end{align*} where \begin{align*} B(t, T) = \frac{1}{a}\Big(1-e^{-a(T-t)} \Big). \end{align*} Then \begin{align*} d\ln P(t, T) &=-e^{-a(T-t)}r_tdt -B(t, T)dr_t + \theta(t)B(t, T)dt - \frac{1}{2} \sigma_0^2 B(t, T)^2 dt\\ &=\left(r_t-\frac{1}{2} \sigma_0^2 B(t, T)^2\right) dt - \sigma_0 B(t, T)dW_t,\tag{1} \end{align*} or \begin{align*} \frac{dP(t, T)}{P(t, T)} = r_t dt - \sigma_0 B(t, T)dW_t. \end{align*}

Let $Q$ denote the risk-neutral measure and $Q^T$ denote the $T$-forward measure. Moreover, let $B_t = e^{\int_0^t r_s ds}$ be the money market account value. From $(1)$, \begin{align*} \frac{dQ^{T}}{dQ}\Bigg|_t &= \frac{P(t, T)B_0}{P(0, T)B_t}\ \ (\text{with } B_0=1) \\ &=\exp\left(-\frac{1}{2}\int_0^t \sigma_0^2 B(s, T)^2 ds - \int_0^t \sigma_0 B(s, T) dW_s\right). \end{align*} Then by the Girsanov theorem, under $Q^T$, the process $\{(\widehat{W}_t^1, \widehat{W}_t^2), t \ge 0 \}$, where \begin{align*} \widehat{W}_t^1 &= W_t^1 + \int_0^t \sigma_0 B(s, T) ds,\\ \widehat{W}_t^2 &= W_t^2, \end{align*} is a standard two-dimensional Brownian motion. Moreover, under $Q^T$, \begin{align*} \frac{dP(t, T)}{P(t, T)} &= r_t dt - \sigma_0 B(t, T)dW_t^1 \\ &=\big(r_t +\sigma_0^2 B(t, T)^2\big)dt - \sigma_0 B(t, T)d\widehat{W}_t^1 \\ \frac{dS_t}{S_t} &= r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big) \\ &=\big(r_t- \rho\sigma_0\sigma B(t, T)\big) dt + \sigma \Big(\rho d\widehat{W}_t^1 + \sqrt{1-\rho^2} d\widehat{W}_t^2\Big).\tag{2} \end{align*}

Note that, the forward price $F(t, T)$ has the form \begin{align*} F(t, T) &= E_{Q^T}(S_T \mid \mathcal{F}_t)\\ &=\frac{S_t}{P(t, T)}. \end{align*} which is a martingale under the $T$-forward measure $Q^T$ and satisfies an SDE of the form \begin{align*} dF(t, T) &= \frac{dS_t}{P(t, T)} -\frac{S_t}{P(t, T)^2}dP(t, T) \\ &\qquad - \frac{d\langle S_t, P(t, T)\rangle}{P(t, T)^2} + \frac{S_t}{P(t, T)^3}d\langle P(t, T), P(t, T)\rangle\\ &= F(t, T)\left[\sigma \Big(\rho d\widehat{W}_t^1 + \sqrt{1-\rho^2} d\widehat{W}_t^2\Big) + \sigma_0 B(t, T)d\widehat{W}_t^1 \right]\\ &= F(t, T) \left[ \big(\sigma\rho + \sigma_0 B(t, T)\big) d\widehat{W}_t^1 + \sigma \sqrt{1-\rho^2} d\widehat{W}_t^2 \right]. \end{align*} Let $\hat{\sigma}$ be a quantity defined by \begin{align*} T\hat{\sigma}^2 &= \int_0^T\Big[\big(\sigma\rho + \sigma_0 B(s, T)\big)^2 + \sigma^2\big(1-\rho^2\big) \Big] ds\\ &=\int_0^T\Big[\sigma^2 + 2\rho\sigma\sigma_0 B(s, T) + \sigma_0^2 B^2(s, T)\Big] ds\\ &=\sigma^2T + \frac{2\rho\sigma\sigma_0}{a}\Big[T-\frac{1}{a}\big(1-e^{-aT}\big)\Big] + \frac{\sigma_0^2}{a^2}\Big[T+\frac{1}{2a}\big(1-e^{-2aT} \big) - \frac{2}{a}\big(1-e^{-aT} \big) \Big]\\ &=\sigma^2T + \frac{2\rho\sigma\sigma_0}{a}\Big[T-\frac{1}{a}\big(1-e^{-aT}\big)\Big] + \frac{\sigma_0^2}{a^2}\Big[T-\frac{1}{2a}e^{-2aT}+\frac{2}{a}e^{-aT} -\frac{3}{2a} \Big]. \end{align*} Then \begin{align*} F(T, T) = F(0, T)\exp\left(-\frac{1}{2}\hat{\sigma}^2T + \hat{\sigma}\sqrt{T} Z \right), \end{align*} where $Z$ is a standard normal random variable. Consequently, \begin{align*} E_Q\left(\frac{(S_T-K)^+}{B_T}\right) &= E_Q\left(\frac{(F(T, T)-K)^+}{B_T}\right)\\ &=E_{Q^T}\left(\frac{(F(T, T)-K)^+}{B_T} \frac{dQ}{dQ^T}\bigg|_T \right)\\ &=P(0, T)E_{Q^T}\left((F(T, T)-K)^+\right)\\ &=P(0, T)\big[F(0, T)N(d_1) - KN(d_2) \big], \end{align*} where $d_1 = \frac{\ln F(0, T)/K + \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}$ and $d_2 = d_1 - \hat{\sigma} \sqrt{T}$.

Answer 2

As Gordon explained very clearly, if you assume your IR model is normal, you have closed form formulas.

The important thing here is that the Forward with maturity T is lognormal under the $T$-forward measure.

Why is that? Why do we care?

As soon as you have stochastic interest rates, you should basically forget about the risk neutral measure and think in terms of forward measures instead. The change of measure formula is: $$ V_t = \mathbb{E}^{\mathbb{Q}^{RN}}_t[e^{-\int_t^T r_u\,du} V_T] = Z_{t,T}\mathbb{E}^{\mathbb{Q}^T}_t[V_T] $$ where $$ Z_{t,T} = \mathbb{E}^{\mathbb{Q}^{RN}}_t[e^{-\int_t^T r_u\,du} ] $$ is the ZCB price, i.e. the value of receiving 1 unit of currency at time $T$, as seen from time $t$ (I write $\mathbb{E}_t$ for the conditional expectation wrt the filtration representing the information available at time $t$).

The ZCB price is typically known/implied from liquid rates instruments at time $t$. So the above formula factors out the stochasticity of interest rates. For non-path-dependent products, this means that we can forget about the risk-neutral measure altogether. The only thing that matters is the distribution of the terminal cash-flow $V_T$ under the $T$-forward measure $\mathbb{Q}^T$ associated with the numeraire $Z_{t,T}$. Most people without a rates background feel uncomfortable with this measure at first. Why introduce this fictional measure when we have the risk neutral one?

Well, first, the so-called risk-neutral measure is just as fictional. It is purely a mathematical construct whose existence is derived, under some strong assumptions, from the only measure that matters: the historical measure $\mathbb{P}$.

Moreover, this is how the market participants actually think! Indeed, in option markets, participants quote implied volatilities. If $C_t(T,K)$ is the value of a call with maturity $T$ and strike $K$ at time $t$, the corresponding BS implied volatility is $$ C_t(T,K) = Z_{t,T}BS\left(t,F_{t,T};T,K;\Sigma_{BS}\right) $$
where $$ BS(t,F;T,K;\sigma) = FN\left( -\frac{\log(K/F)}{\sigma\sqrt{T-t}} + \frac{1}{2}\sigma\sqrt{T-t} \right) - KN\left(-\frac{\log(K/F)}{\sigma\sqrt{T-t}} - \frac{1}{2}\sigma\sqrt{T-t} \right) $$ In order to agree on the current price, participants need to agree on the vol and on $Z_{t,T}$. But, in practice, market participants do not need to agree on the fair price. What is required is for each counterparty to estimate that the trade is beneficial to them. If you have a better estimate of $Z_{t,T}$ then you can arbitrage the other counterparty. This is exactly what happened after the 2008 crisis when some were still using USD Libor rates as "risk-free" discount rates when others were discounting at OIS rates (the interest rate on collateral).

Writing $F_{t,T} = S_t/Z_{t,T}$, the implied volatility can be seen as a function $\Sigma_{BS}(t,S,Z;T,K)$ where the variables after the semi-colon are fixed (they refer to the maturity and strike in the option contract) while those before that will evolve stochastically with $t$. The dependency wrt to the strike is the well-known volatility smile. The dependency wrt to the spot $S$ is known as the volatility backbone. The dependency wrt to $t$ is essentially what people call Theta (or at least its volatility component). The dependency wrt $Z$ corresponds to the IR risk. This risk is negligible in short dated options but not in long-dated ones.

Answer 3

In order to define option price we should follow Black Scholes construction to construct riskless portfolio at t then to state that instantaneous rate of return of this portfolio equal risk free rate r ( t ) where r is a random on [ t , t + dt ] interval. We actually then arrive at the problem which could not be embedded in BS pricing world.