Pricing a contract that pays G at T>0

Question

The problem is as follows:

Consider the generalized Black-Scholes model with a risky asset and assume that this model is complete. Calculate the price $\Pi_0[G]$ at $t=0$ of a contract that, at a time $T>0$, pays

a) $G=1$

b) $G=S_T$ . In this case, assume the rate of instant dividend payments $\delta_t$ is deterministic.

NOTE: According to my professor, a complete model means that if there exists a measure $\mathbb{Q}$, then there are no arbitrage opportunities.

My attempted solution is:

For a replicating portfolio, its value at time $t$ is given by $$V_t=\mathbb{E}^{\mathbb{Q}}\left[\frac{G_T}{B_t}\right]$$ where $B_t^{-1}=exp\{-\int_0^{T}r\cdot ds\}$. I assume $r$ is constant. If there are no arbitrage opportunities, then $V_0^{\alpha, \beta}=\Pi_0[G]$.

Since $\mathbb{E}^{\mathbb{Q}}\left[\frac{G_T}{B_t}\right]$ is $\mathbb{Q}$-martingale, then at time $t=0$

\begin{equation}\Pi_0[G]=\mathbb{E}^{\mathbb{Q}}\left[\frac{G_T}{B_t}\right]=e^{-rT}\mathbb{E}^{\mathbb{Q}}[G_T] \end{equation}

• This is where I don't know what to do next. I don't know the probability distribution of the contract. I know that it pays $G=1$

For part b) $\mathbf{G_T=S_T}$

We have:

$e^{-rT}\mathbb{E}^{\mathbb{Q}}[G_T] = e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T] = e^{-rT}\int_0^{\infty}S_0e^{(r-\delta_t-\frac{\sigma^2}{2})T}\phi_{\mu, Var}(x)dx$

But yet again, I'm not sure about what the next step would be... I assume that $G$ has a normal distibution with mean $\mu$ and some variance, which I wrote as $\phi_{\mu, Var}(x)$.

I'd appreciate any feedback on my reasoning, or any bibligraphy that could help me with this excercise. I'm sorry if something is not clear, english is not my first language.

Answer 1

If the filtration $\mathcal{F}(t)$ is generated by the underlying Wiener process (or Brownian Motion), then all $\mathcal{F}(T)$-measurable payoff functions can be hedged and the market is said to be complete.

What your professor has said is not correct (or your understanding, anyway!). It is not enough for any measure $\mathbb{Q}$ to exist for there to be no arbitrage opportunities - it needs to be a risk neutral measure (or martingale measure).

You have made a couple of mistakes:

• In the definition of the money market account $B$, you should have that $B(t)=e^{\int_{0}^{t} r(s) ds}$.
• The replicating value (also called the no-arbitrage value, the risk neutral value, etc) should be $V(t)=\mathbb{E}^\mathbb{Q}\left[ \frac{B(t)}{B(T)} V(T) \vert \mathcal{F}(t) \right]$.

In the case that $G(T)\equiv G=1$, this is like getting one dollar in $T$ years discounted back to today, so the present value should be $e^{-\int_{t}^{T} r(s) ds}\cdot1 =e^{-\int_{t}^{T} r(s) ds} $ at time $t$. Let us see it formally:

$V(t)=\mathbb{E}^\mathbb{Q}\left[ \frac{B(t)}{B(T)} G \vert \mathcal{F}(t) \right] = \mathbb{E}^\mathbb{Q}\left[ \frac{B(t)}{B(T)} \cdot 1 \vert \mathcal{F}(t) \right] = \frac{B(t)}{B(T)} $

To price the second payoff, we must assume the underlying asset $S$ follows some SDE, say

$ dS_t=(r_t-\delta_t) S_t dt+\sigma_t S_t dW_t. $

This has the solution

$ S_T=S_te^{\int_{t}^{T}(r_s-\delta_s) ds} \cdot \frac{z_T}{z_t}, $

where the stochastic exponential process

$z_t=e^{\int_{0}^{t} \sigma_s dW_s - \frac{1}{2} \int_{0}^{t} \sigma_s^2 ds}$

is a martingale under the risk neutral measure.

Then we get that

$V(t)=\mathbb{E}^\mathbb{Q}\left[ \frac{B(t)}{B(T)} S_T \vert \mathcal{F}(t) \right] = S_t e^{-\int_{t}^{T} \delta_s ds} \mathbb{E}^\mathbb{Q}\left[ \frac{z_T}{z_t} \vert \mathcal{F}(t) \right] = S_t e^{-\int_{t}^{T} \delta_s ds}. $