Markov Property and Conditional Expectations for Diffusions
Summary
The document asks when the conditional expectation of a terminal payoff given the full Brownian filtration can be expressed using only the current state of a diffusion. Its proposed argument relies on future Brownian increments being independent of the past and on the process evolving from its current state. The displayed answer instead applies Itô’s lemma to a sufficiently smooth solution of the backward Kolmogorov equation with terminal value equal to the payoff. If the resulting stochastic integral is a true martingale, the conditional expectation equals that solution evaluated at the current state, which is measurable with respect to the state alone.
The argument illustrates the Markov property underlying diffusion pricing. Its PDE proof needs existence and regularity of the solution, suitable conditions on the coefficients, and integrability or boundedness sufficient for the martingale step. The document notes that existence of the required PDE solution is not established. The original increment argument also needs care: Brownian motion at time t is not independent of the filtration at t; future increments are.
Key ideas
- For a Markov diffusion, the conditional distribution of a future state given the past depends on the current state.
- A backward PDE solution can represent the conditional expectation of a terminal payoff.
- Itô’s lemma turns the PDE solution along the diffusion into a local martingale.
- Additional regularity and integrability conditions are needed to ensure the local martingale is a true martingale.
- Brownian increments after the conditioning time are independent of the past, while Brownian motion at that time is not.
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Full text
# Prove that $E[g(X_T)|\mathscr F_t] = E[g(X_T)|X_t]$
# Prove that $E[g(X_T)|\mathscr F_t] = E[g(X_T)|X_t]$
Let $T > 0$. Let $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)$ be a filtered probability space where $\mathscr F_t = \sigma(W_u, u \in [0,t])$ where $W_t$ is standard Brownian motion.
Let the stochastic process $X=(X_t)_{t \in [0,T]}$ solve the SDE
$$dX_t = \beta(t,X_t)dt + \sigma(t, X_t)dW_t$$
with initial condition $X_t = x$ where $x \in \mathbb R$
Prove that
$$E[g(X_T)|\mathscr F_t] = E[g(X_T)|X_t]$$
where $g$ is a Borel-measurable function and $E[|g(X_T)||X_t=x] < \infty$
What I tried: $\forall t \in [0,T]$.
$$X_t = X_0 + \int_0^t \beta du + \int_0^t \sigma dW_t$$
Choose $t=T$ to get:
$$X_T = X_0 + \int_0^T \beta du + \int_0^T \sigma dW_t$$
$$\to X_T = X_t + \int_t^T \beta du + \int_t^T \sigma dW_t$$
Define another Borel-measurable function $h(x,y)$ s.t.
$$h(X_t, W_t) := g(X_t + \int_t^T \beta du + \int_t^T \sigma dW_t)$$
$$\to g(X_T) = h(X_t, W_t)$$
$\because X_t \in m\mathscr F_t$ and $W_t$ is independent of $\mathscr F_t$, we have
$$E[h(X_t, W_t)|\mathscr F_t] = E[h(x, W_t)]|_{x=X_t} \tag{*}$$
Also, $\because X_t \in m\mathscr F_t$, $W_t$ is independent of $X_t$.
Thus,
$\because X_t \in mX_t$ and $W_t$ is independent of $X_t$, we have
$$E[h(X_t, W_t)|X_t] = E[h(x, W_t)]|_{x=X_t} \tag{**}$$
Combining $(*)$ and $(**)$ gives us what we want. QED
Is that right? Any other assumptions to make such as continuity, integrability or boundedness?
## Answer by Danny Pak-Keung Chan (score 3)
https://quant.stackexchange.com/a/34687
This is a corollary of Feynman-Kac theorem. For self-containedness, I re-produce the proof as follows.
Assume that there exists a $C^{1,2}$-function $F=F(t,x)$ defined on $[0,T]\times\mathbb{R}$ that satisfies the PDE on the interior $$ F_{t}+\beta F_{x}+\frac{1}{2}\sigma^{2}F_{xx}=0, $$ and the boundary condition: $F(T,x)=g(x)$. Consider the process $\left(F(t,X_{t})\right)_{t}$. By Ito's lemma, \begin{eqnarray*} & & dF(t,X_{t})\\ & = & F_{t}dt+F_{x}dX_{t}+\frac{1}{2}F_{xx}dX_{t}dX_{t}\\ & = & \left\{ F_{t}+\beta F_{x}+\frac{1}{2}\sigma^{2}F_{xx}\right\} dt+\sigma F_{x}dW_{t}\\ & = & \sigma F_{x}dW_{t}. \end{eqnarray*} (In the above, $\beta$, $\sigma$, and all partial derivatives of $F$ are evaluated at $(t,X_{t})$ (i.e., $\beta$ denotes $\beta(t,X_{t})$, etc...)) By assuming enough boundedness about the process $\left(\sigma(t,X_{t})F_{x}(t,X_{t})\right)_{t}$, $\left(F(t,X_{t})\right)_t$ is a martingale (rather than just a local martingale). Therefore, \begin{eqnarray*} F(t,X_{t}) & = & E\left[F(T,X_{T})\mid\mathcal{F}_{t}\right]\\ & = & E\left[g(X_{T})\mid\mathcal{F}_{t}\right]\mbox{ a.s.} \end{eqnarray*} Observe that the left hand side is $\sigma(X_{t})$-measurable, so does the right hand side, the result follows. For clarity, I work out the details as follows: Clearly $\sigma(X_{t})\subseteq\mathcal{F}_{t}$. Therefore, by tower property of conditional expectation, we have \begin{eqnarray*} & & E\left[g(X_{T})\mid X_{t}\right]\\ & = & E\left[E\left[g(X_{T})\mid\mathcal{F}_{t}\right]\mid X_{t}\right]\\ & = & E\left[F(t,X_{t})\mid X_{t}\right]\\ & = & F(t,X_{t})\\ & = & E\left[g(X_{T})\mid\mathcal{F}_{t}\right]. \end{eqnarray*}
Remarks: I do not have enough knowledge about PDE, so I am not sure that the function $F$ defined in above really exists.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.