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

• 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 Dec 24, 2020 at 22:39
• Ahhhh yes, Thank you so much Dec 25, 2020 at 8:25

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