Pricing an Integral-of-Spot Claim Under Risk-Neutral Valuation

# Time-zero price of two specific contingent claims


# Time-zero price of two specific contingent claims












I am unsure how to start with the following problem.

I have two contingent claims where contingent claim (1) pays $\int_0^T S_u du$ and contingent claim (2) pays $(\log S_T)^2$ at time $T$

Now I would like to use the Black-Scholes model to get their time-zero prices

Using the BS formula $C(S_0,K,\sigma,r,T)=S_0\Phi(d_1)-Ke^{-rT}\Phi(d_2)$ with $d_1=d_2+\sigma\sqrt{T}, d_2=\frac{\log{K/S_0}-(r-\frac{1}{2} \sigma^2)T}{\sigma\sqrt{T}}$

where I can I include the expressions from above?

## Answer by math (score 3, accepted)

https://quant.stackexchange.com/a/10051

Assuming the filtration is generated by Brownian Motion, you know that the price of a contingent claim is just the expectation under the risk neutral measure $Q$. Hence for the first one

$$E_Q[\int_0^TS_udu]$$

where $S$ has the dynamic: $S_t=S_0e^{\sigma W^Q_t-(\frac{1}{2}\sigma^2-r)t}$, where $W^Q_t=W_t+\frac{\mu-r}{\sigma}$ is the Girsanov chagned Brownian Motion. Hence $S_t$ has a lognormal distribution under $Q$. Therefore $S_t>0$ and you can use Fubinis Theorem to interchange the order of integration:

$$e^{-rT}\int_0^T S_0E_Q[S_u]du=S_0\int_0^Te^{-\frac{1}{2}\sigma^2u+\frac{1}{2}\sigma^2u+ru}du=\frac{1}{r}e^{-rT}S_0(e^{rT}-1)=\frac{1}{r}S_0(1-e^{-rT})$$

You can approach the second one in the exact same way.

## Answer by wsw (score 2)

https://quant.stackexchange.com/a/10052

I think we can compute the price of the first contingent claim at time 0 without using any models (e.g., Black-Scholes). For the first claim, $$ V_0 = e^{-r T} \mathbb{E}^Q \left[ \int_0^T S_u \; du\middle \vert \cal{F}_0\right] = e^{-r T} \int_0^T \mathbb{E}^Q \left[ S_u \middle \vert \cal{F}_0\right] du \; . $$ Since the price of a forward that matures at time $T$ is $\mathbb{E}^Q \left[ S_T \middle \vert \cal{F}_0\right] = S_0 e^{rT}$, I get $$ V_0 = \frac{S_0}{r} \left( 1-e^{-rT}\right) \; . $$ I want to stress again that this result is model-free.