# 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?

• What is the evolution of Stock price? – user16651 Dec 16 '16 at 19:06

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}