Risk-Neutral Derivative Pricing Through a Change of Measure
Summary
The document asks whether a derivative’s discounted expected payoff under a risk-neutral measure can be rewritten as an expectation under the original probability measure using the Girsanov density. It outlines how a Brownian motion with drift can be transformed into standard Brownian motion under a new measure, and how the likelihood adjustment weights the payoff in the original measure.
The answer affirms the pricing expression when the claim is traded or replicable in a market with a locally riskless asset and a risky asset. In that setting, a self-financing replicating portfolio gives the claim’s price process, and its value relative to the riskless asset is a martingale under the risk-neutral measure. The conclusion is conditional: the discussion does not establish pricing for arbitrary nonreplicable claims or detail the technical assumptions needed for the measure change, such as integrability conditions.
Key ideas
- Girsanov’s theorem changes probability weights so a Brownian motion with drift becomes standard Brownian motion under a new measure.
- A Radon–Nikodym density can express expectations under the new measure as weighted expectations under the original measure.
- The discounted expected payoff formula applies to a claim whose price process is supported by replication in the described market.
- The claim’s value relative to the locally riskless asset is a martingale under the risk-neutral measure.
- The argument relies on market and measure-change assumptions that the discussion does not spell out fully.
Tags
Full text
# Derivative pricing under $\mathbb{P}$
# Derivative pricing under $\mathbb{P}$
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.
## Answer by Viktor Nilsson (score 2, accepted)
https://quant.stackexchange.com/a/77230
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 $S^1_t / S^0_t$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.