When Risk-Neutral Expectations Depend on the Full Path
Summary
The document explains when the risk-neutral pricing expectation conditional on the current asset price is equivalent to conditioning on the full information set available at time t. For a European claim whose payoff depends only on the terminal asset price, geometric Brownian motion has independent future increments, so the conditional payoff expectation can be expressed as a function of time and the current price. Observing the earlier price path adds no relevant information under these assumptions.
The answer contrasts this with a barrier claim whose payoff also depends on the running maximum. In that case, the current price alone does not reveal whether the barrier was previously crossed, so the conditional expectation must retain path information, represented here by the running maximum. The equivalence therefore relies on the payoff and process assumptions; other path-dependent claims may require additional state variables. The document does not address more general models or filtrations.
Key ideas
- For a terminal-price-only payoff under the stated diffusion, the future price depends on the current price and independent future Brownian increments.
- The conditional expectation given the full filtration then reduces to one given the current asset price.
- A barrier payoff may depend on whether the running maximum crossed a threshold before time t.
- For such a claim, the current price is insufficient, and the running maximum is an additional state variable.
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# Is this the same as the conditial expectation in risk neutral pricing formula?
# Is this the same as the conditial expectation in risk neutral pricing formula?
Let the dynamic of underlying asset $S_t$ under objective probability measure $\mathbb{P}$ be as follow $$dS = \mu Sdt + \sigma S dW_t^{\mathbb{P}}.$$ We now define another probability measure $\mathbb{Q}$ under which the $S$-process has a different probability distribution. This is done by defining the $\mathbb{Q}$-dynamics of $S_t$ as $$dS_t = r S_t dt + \sigma S_t dW_t^{\mathbb{Q}}$$
We know the arbitrage free price of the claim $f(S_T)$, at time $t$ is given by $$\Pi(t) = e^{-r(T-t)}\mathbb{E}^Q_{t,s}\left[ f(S_T)\right]$$. The subscripts $t,s$ indicate that we take the expectation given the initial condition that $S(t) = s$. Is this theorem equivalent to writing $E^Q \left(f(S_T) | \mathcal{F}_t\right)$, that is, the expectation "given what we know at time $t$", which is another way that I've seen the risk neutral formula written down?
The reason for my confusion is that in the former, we only assume that we know the price at time $t$ exactly, but in the latter with the conditional expectation, we have access to the entire path information contained in $\mathcal{F}_t$, right? So how can they be equivalent, when the latter is a narrower form of expectation?
## Answer by LocalVolatility (score 1)
https://quant.stackexchange.com/a/31545
In the case you mentioned, the terminal payoff of the contingent claim only depends on $S_T$. You have
\begin{equation} S_T = S_t \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) (T - t) + \sigma \left( W_T - W_t \right) \right\}. \end{equation}
I assume that $\mathcal{F}_t$ is the natural filtration generated by the processes $S$ or equivalently $W$. In that case, $S_t$ is $\mathcal{F}_t$-measurable and the increment $W_T - W_t$ is independent of $\mathcal{F}_t$. Consequently,
\begin{equation} \mathbb{E} \left[ \left. f \left( S_T \right) \right| \mathcal{F}_t \right] = \mathbb{E} \left[ \left. f \left( S_T \right) \right| \sigma \left( S_t \right) \right] = g \left( t, S_t \right). \end{equation}
I.e. the additional information from observing the full path of $S$ over the interval $[0, t]$ is not relevant for computing the conditional expectation.
You could construct path-dependent contingent claims, where this does not hold. Consider e.g. a knock-out barrier option with terminal payoff
\begin{equation} f \left( S_T, M_T \right) = h \left( S_T \right) \mathrm{1} \left\{ M_T < B \right\}, \end{equation}
where $M_t = \max_{u \in [0, t]} S_u$ is the running maximum of the process $S$. In this case,
\begin{equation} \mathbb{E} \left[ \left. f \left( S_T \right) \right| \mathcal{F}_t \right] \neq \mathbb{E} \left[ \left. f \left( S_T \right) \right| \sigma \left( S_t \right) \right] \end{equation}
since by just knowing $S_t < B$, we don't know if also $M_t < B$. However,
\begin{equation} \mathbb{E} \left[ \left. f \left( S_T \right) \right| \mathcal{F}_t \right] = \mathbb{E} \left[ \left. f \left( S_T \right) \right| \sigma \left( S_t, M_t \right) \right] = g \left( t, S_t, M_t \right). \end{equation}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.