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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?

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2 Answers

up vote 3 down vote accepted

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.

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Thanks for your answer, in case of (b), I have $E_Q[(\log S_t)^2]=S_0*E_Q[(\sigma W_T^Q-\frac{1}{2}\sigma^2 T)^2]=S_0*(\sigma W_T^Q-\frac{1}{2}\sigma^2 T)^2$, but this cant be right? –  TI Jones Jan 25 at 23:38
    
I think your expression for $S_t =S_0 e^{\sigma W^Q_t-\frac{1}{2}\sigma^2t}$ is wrong. Under the risk-neutral measure $Q$, $S_t= S_0 e^{(r-\sigma^2/2)t + \sigma W_t}$. –  William S. Wong Jan 26 at 1:41
    
@WilliamS.Wong I should have mentioned that I used discounted quantities. Your suggestions is for undiscounted ones. –  user8 Jan 26 at 9:26
    
But how can I get rid of the $W_T$ term? –  TI Jones Jan 26 at 13:16
    
@TIJones I think you don't understand what $W_T$ is. It is a random variable. Hence you can't it pull it outside the expectation. $X:=(\sigma W^Q_T-\frac{1}{2}\sigma^2T)\sim \mathcal{N}(-\frac{1}{2}\sigma^2T,\sigma^2T)$. Look at the second moment of a normal r.v. –  user8 Jan 26 at 16:13
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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.

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