Pricing a Time Integral of a Stock Under Black–Scholes
Summary
The document considers the time-t price of a claim paying the integral of a stock price over a fixed horizon under the Black–Scholes model. The questioner expands the risk-neutral expectation using the stock’s geometric Brownian motion and asks whether the resulting expression is correct and can be simplified. A correction in the discussion fixes the normal moment: the expectation of the exponential Brownian increment depends linearly on elapsed time in its exponent, rather than on elapsed time squared.
The accepted response uses the risk-neutral expected stock price, which equals its forward value in the no-dividend model, then integrates that expectation and discounts the payoff. For a valuation starting at time zero, it gives a closed-form expression and notes a first-order approximation for small rate-times-horizon. The response says the argument extends to conditioning at a later time, though it does not write out that full formula. The result relies on the stated Black–Scholes assumptions and no dividends; other carry terms or payoff definitions would change the calculation.
Key ideas
- Under risk-neutral pricing, the expected stock price in the no-dividend Black–Scholes model grows at the risk-free rate.
- The expectation of the exponential Brownian increment uses elapsed time, not elapsed time squared, in its variance term.
- Linearity of expectation allows the expected stock value to be integrated over the payoff horizon.
- Discounting that integral gives a closed-form value, with a simpler first-order approximation when rate times horizon is small.
- The calculation assumes the stated no-dividend model and requires adjustment for different carry assumptions.
Tags
Full text
# Black Scholes price of exotic claim
# Black Scholes price of exotic claim
Given a time horizon N, I want to know the time-$t$ Black-Scholes fair price of $$\int_0^T S_u du$$ where $S_u$ denotes the time-$u$ stock price. I have used the formula I have been given as follows: $$\begin{align*} &e^{-r(T-t)}E^Q\left(\int_0^T S_u du\mid \mathcal{F}_t \right)\\=&e^{-r(T-t)}\int_0^t S_u du+e^{-r(T-t)}E^Q\left(\int_t^T S_u du\mid \mathcal{F}_t\right)\\=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}E^Q\left(\int_t^T \frac{S_u}{S_t} du\mid \mathcal{F}_t\right)\\=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}E^Q\left(\int_t^T e^{(r-\sigma^2/2)(u-t)}e^{\sigma(W_u-W_t)} du\mid \mathcal{F}_t\right)\\=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}E^Q\left(\int_t^T e^{(r-\sigma^2/2)(u-t)}e^{\sigma(W_u-W_t)} du\right) \end{align*}$$ We remove the conditioning since $W_u-W_t$ is independent of the sigma algebra. Now we swap the integrals and use the mgf of a normal distribution. $$\begin{align*}=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}\left(\int_t^T e^{(r-\sigma^2/2)(u-t)}e^{\sigma^2(u-t)^2/2} du\right) \end{align*}$$
I was wondering if firstly, this is correct and secondly, if this was as simple as we could go for the pricing or if we could simplify this expression.
EDIT: Thanks to @LucaMac for pointing out $E(e^{\sigma(W_u-W_t)})=e^{\sigma^2(u-t)/2}$
## Answer by Soumirai (score 3, accepted)
https://quant.stackexchange.com/a/60142
That looks correct, but a bit complicated. We know that under Black-Scholes with no dividends, $E^Q(S_t) = Forward = S_0 e^{rt}$
$e^{-rT}E^Q(\int_0^TS_tdt) = e^{-rT}\int_0^TE^Q(S_t)dt \\ = e^{-rT}\int_0^T S_0 e^{rt} dt = S_0 e^{-rT}\int_0^T e^{rt} dt \\ = S_0 e^{-rT} \frac{1}{r}(e^{rT} - 1) = S_0\frac{1-e^{-rT}}{r}$
It is straightforward to generalize to the case where the filtration is not $F_0$ but $F_t$.
At order 1 in $rT$:
$S_0\frac{1-e^{-rT}}{r} \approx S_0\frac{1-(1-rT)}{r} = S_0T$, which is what you expect to have approximately. Because you integrate $E^Q(S_t)$ from 0 to T $\approx S_0e^{r\frac{T}{2}}$. So $e^{-rT}\int_0^T S_0e^{r\frac{T}{2}} dt = S_0e^{-r\frac{T}{2}}T$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.