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I recently learnt about the Girsanov-Cameron-Martin theorem, which basically says, that if $(\tilde{B}(t),t\in[0,T])$ is some Brownian motion with a (possibly stochastic) drift $\theta(t)$ defined on $(\Omega,\mathcal{F},\mathbb{P})$, then we can construct a new probability measure $\mathbb{\tilde{P}}$, that is, assign new weights to the sample paths $\tilde{B}(\cdot,\omega)$) by mutiplication with the factor:

$$M(T) = \exp\left[-\int_{0}^{T}\theta(t)d\tilde{B}(t)-\frac{1}{2}\int_{0}^{T}\theta^2(t)dt\right]$$

so that:

$$\mathbb{\tilde{P}}(\mathcal{E})=\mathbb{\tilde{E}}[1_\mathcal{E}]=\mathbb{E}[M(T) 1_\mathcal{E}]$$

for any event $\mathcal{E} \in \mathcal{F}$.

The result is that, $\tilde{B}(t)$ is a standard brownian motion under $\mathbb{\tilde{P}}$.

Moreover, we can write $d\tilde{B}(t) = dB(t) + \theta(t)dt$. The expectation under $\mathbb{\tilde{P}}$ could be written as:

$$\mathbb{\tilde{E}}[V(T)] = \int_{\Omega}V(T)d\mathbb{\tilde{P}}=\int_{\Omega}V(T)\left(\frac{d\mathbb{\tilde{P}(\omega)}}{d\mathbb{P}(\omega)}\right)d\mathbb{P}(\omega)=\int_{\Omega}V(T)M(T)d\mathbb{P}(\omega)=\mathbb{E}[M(T)V(T)]$$

That means, the risk-neutral pricing formula becomes:

\begin{align} V(0) &= \mathbb{\tilde{E}}\left[\exp\left(-\int_{0}^{T}r(t)dt\right)V(T)\right]\\ &= \mathbb{E}\left[\exp\left(-\int_{0}^{T}r(t)dt\right)M(T)V(T)\right]\\ &= \mathbb{E}\left[\exp\left(-\int_{0}^{T}(r(t)+\frac{1}{2}\theta^2(t))dt-\int_{0}^T \theta(t)dB(t)\right)V(T)\right] \end{align}

Is the above pricing formula correct?


Edit: $\theta(t)$ represents the market price of holding the risky asset. For a stock, this is $\theta(t)=(\mu(t)-r(t))/\sigma(t)$, $\mu(t)$ is the expected return on the stock.

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    $\begingroup$ The $M(T)$ in your first formula should be an exponential. $\endgroup$
    – Kurt G.
    Nov 2, 2023 at 8:29
  • $\begingroup$ Corrected the typo. $\endgroup$
    – Quasar
    Nov 2, 2023 at 19:16

1 Answer 1

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Assuming that $V(t)$ is the price process of some (perhaps implicitly) traded claim, it is correct. A typical interpretation would be the following: Given that the $T$-claim $\mathcal{X}$ is replicable in the market $(S^0, S^1)$ (with $S^0$ locally riskless, since you mentioned one risky asset), by some self-financing portfolio $h = (h^0, h^1)$, it has a corresponding price process, given by $$ \Pi_t(\mathcal{X}) = V^h_t := h^0_t S^0_t + h^1_t S^1_t. $$ Then indeed your formula would hold for $V(T) = V^h_T = \mathcal{X}$, where the last equality is by replication. The reason for this is that $V^h_t / S^0_t$ is a martingale under $\tilde{\mathbb{P}}$, like $S^1_t / S^0_t$.

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