Black–Scholes Pricing with Time-Varying Rates and Volatility
Summary
The note extends European call pricing to deterministic, time-varying interest rates and volatility under a risk-neutral model. It starts from the asset’s lognormal solution and defines the forward price using the accumulated short rate. The terminal log price has variance equal to the integral of instantaneous variance over the option’s life. That integrated variance can be represented by an equivalent constant volatility for the maturity, allowing the familiar Black–Scholes terms to be adapted using the forward price and the strike. The excerpt gives the distributional argument and the adjusted d terms, but it ends before presenting the complete call-price expression or deriving the replicating strategy. It therefore supports the pricing transformation, while the strategy question raised by the author remains unanswered in the supplied text. The result assumes deterministic continuous inputs and risk-neutral dynamics.
Key ideas
- The forward price reflects the accumulated time-varying risk-free rate.
- Terminal log-price variance is the integral of instantaneous variance over the option’s life.
- An equivalent maturity-specific volatility is the square root of average variance over the life.
- The excerpt does not finish the call formula or establish the replicating strategy.
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Full text
# derivation of general black-scholes formula
# derivation of general black-scholes formula
I would like to find a derivation for the Black-Scholes fomrula in the general case (i.e., where the volatility function $\sigma : [0,T] \to \mathbb{R}^+$ and the investment rate $r: [0,T] \to \mathbb{R}^+$ are continuous deterministic functions of $t$ and its associated replicating strategy.
I know that for constant $\sigma, r$, the Black-Schole's formula is $C_t = c(S_t, T-t) = S_tN(d_1(S_t, T-t)) - Ke^{-r(T-t)}N(d_2(S_t, T-t))$ where $N$ is the standard Gaussian CDF and $d_{1,2}(S_t, T-t) = \frac{\ln(S_t/K) + (r\pm \sigma^2)t}{\sigma\sqrt{T-t}}$ and the replicating strategy, $\phi$, satisfies $\phi_t^1 = \partial_{s}c(S_t, T-t), \phi_1^2 = e^{-rt}(c(S_t, T-t) - \phi_t^1 S_t)$
Now for he general case. We know that under the risk-neutral measure $\mathbb{P}^*$, $S$ satisfies $dS_t = r(t)S_t dt + \sigma(t) S_t dW_t^*$, where $W_t^*$ is Brownian motion under $\mathbb{P}^*$, and this has the unique solution $$S_t = S_0 \exp \left(\int_0^t \sigma(u) dW_u^* + \int_0^t (r(u) - \frac{1}{2}\sigma^2(u)) du \right)$$
Now, using the risk neutral formula, we know that the price of the European call option satisfies $$C_t = e^{-\int_t^Tr(u)du}E_{\mathbb{P}^*}((S_T - K)^+ | \mathcal{F}_t)$$ How do I use this combined with my formula for $S_t$ to obtain the generalised formula? Also, would I be right in saying that the generalised replicating strategy is the same?
## Answer by Quantuple (score 4, accepted)
https://quant.stackexchange.com/a/40202
\begin{align} S_t &= S_0 \exp \left(\int_0^t \sigma(u) dW_u^* + \int_0^t \left(r(u) - \frac{1}{2}\sigma^2(u)\right) du \right) \\ &= F(0,t) \mathcal{E}\left( \int_0^t \sigma(u) dW_u^* \right) \end{align} If we let $$\mathcal{E}(X_t) := \exp\left(X_t - \frac{1}{2}\langle X \rangle_t\right) $$ the Doléans Dade exponential of a process $(X_t)_{t \geq 0}$ and $$F(0,t) := \Bbb{E}_0^*\left[S_t \right] = S_0 \exp \left(\int_0^t r(u) du\right)$$ represent the underlying forward price.
In any case, you observe that $\forall T \geq 0$, $S_T$ is lognormally distributed since $$ \ln(S_T) \sim N \left( \ln \left(F(0,T)\right) - \frac{1}{2}\int_0^T \sigma^2(u) du, \int_0^T \sigma^2(u) du \right) $$ which you should compare to the traditional Black-Scholes result $$ \ln(S_T) \sim N \left( \ln \left(F(0,T)\right) - \frac{1}{2}\sigma_{BS}^2 T, \sigma_{BS}^2 T\right) $$ meaning you can treat this new case as the standard one with an equivalent "BS" volatility given by $$\hat{\sigma}_T := \sqrt{ \frac{1}{T} \int_0^T \sigma^2(u) du }$$
Repeating the usual steps you then find that standard $d_\pm$ terms involved in the original BS formula need to be replaced by their time-dependent counterparts $$ d_\pm(T) = \frac{ \ln\left(\frac{F(0,T)}{K}\right) \pm \frac{1}{2}\hat{\sigma}_T^2 T}{\hat{\sigma}_T \sqrt{T}} $$ whereShown 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.