Compute $E^{\mathbb{Q}} \left[ \exp(S_T) | \mathcal{F}_t \right] $ with exponential Vasicek Model (1978)

Question

Please show me how to solve (3) with computation processes. (1) and (2) were initial questions which I solved. I show the answers (1) and (2) below.

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(1) Solve the S.D.E. \begin{eqnarray} dS_t=\alpha S_t dt + \sigma dB_t \end{eqnarray} in terms of $\alpha, \sigma >0$, and the initial condition $S_0$.

(2) For which values $\alpha_M$ of $\alpha$ is the discount price process $\tilde{S}_t = e^{-rt} S_t$, $t \in [0, T]$, a martingale under $\mathbb{P}$?

(3) Compute the arbitrage price \begin{eqnarray} C(t, S_t)=e^{-r(T-t)} E^{\mathbb{Q}} \left[ \exp(S_T) | \mathcal{F}_t \right] \end{eqnarray} at time $t \in [0, T]$ of the contingent claim $\exp(S_T)$, with $\alpha=\alpha_M$.

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(1) My Answer

• Let $f(t, x)=e^{-\alpha t} x$, use Itô's formula. \begin{eqnarray} d e^{-\alpha t} S_t &=& -\alpha e^{-\alpha t} S_t dt + e^{-\alpha t} d S_t + \frac{1}{2} 0 d[S_t] \\ &=& e^{-\alpha t} \left( -\alpha S_t dt +\alpha S_t dt + \sigma dB_t \right) \\ &=& \sigma e^{-\alpha t} dB_t \\ \int^t_0 d e^{-\alpha s} S_s &=& \int^t_0 \sigma e^{-\alpha s} dB_s \\ %e^{-\alpha t} S_t -S_0 &=& \int^t_0 \sigma e^{-\alpha s} dB_s \\ S_t &=& e^{\alpha t} \left( S_0 + \sigma \int^t_0 e^{-\alpha s} dB_s \right) \end{eqnarray}

$\square$

(2) My Answer

• The following formula shows that the risk-free interest rate $ r $ is the discount rate for the question, and is $ \alpha_M = \alpha = r $. \begin{eqnarray} \tilde{S}_t = e^{-rt} S_t = e^{-\alpha t} S_t ＝S_0 + \sigma \int^t_0 e^{-\alpha s} dB_s \end{eqnarray}

$\square$

(3) Thank you for your help in advance.

Answer 1

First, express $S_T$ in terms of $S_t$, which can be obtained from your solution in (1)

\begin{eqnarray} S_T &=& e^{\alpha T} \left( S_0 + \sigma \int^t_0 e^{-\alpha s} dB_s + \sigma \int^T_t e^{-\alpha s} dB_s\right) \\ &=&e^{\alpha (T-t)}S_t + \sigma \int^T_t e^{\alpha (T-s)} dB_s \end{eqnarray}

Next, evaluate the expectation, \begin{eqnarray} E_t\left[\exp{(S_T)}\right] &=& \exp\left(e^{\alpha (T-t)}S_t\right) E_t\left[ \exp\left(\sigma \int^T_t e^{\alpha (T-s)} dB_s\right)\right] \\ &=& \exp\left(e^{\alpha (T-t)}S_t\right)\exp\left(\frac{1}{2}\sigma^2 \int^T_t e^{2\alpha (T-s)} ds\right) \\ &=& \exp\left[e^{\alpha (T-t)}S_t + \frac{\sigma^2}{4a}(e^{2a(T-t)}-1) \right] \end{eqnarray}

Finally, the answer to (3) is

$$C(t,S_t) = \exp\left[-r(T-t) + \frac{\sigma^2}{4a}(e^{2a(T-t)}-1) +e^{\alpha (T-t)}S_t \right] $$