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 $E(e^{\sigma(W_u-W_t)})=e^{\sigma^2(u-t)/2}$

  • 1
    $\begingroup$ Almost correct, $\mathbb E[e^{\sigma (W_u-W_t)}] = e^{\frac12\sigma^2(u-t)}$ and not what you wrote, so that the dependency on the volatility $\sigma$ disappears $\endgroup$
    – LucaMac
    Dec 24, 2020 at 22:39
  • $\begingroup$ Ahhhh yes, Thank you so much $\endgroup$ Dec 25, 2020 at 8:25

1 Answer 1


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 $e^{-rT}\int_0^T S_0e^{r\frac{T}{2}} dt = S_0e^{-r\frac{T}{2}}T$


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