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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?

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    $\begingroup$ The proof is pretty technical. But, if you really want to know, please consult Section 6 in Chapter 5 of the book Stochastic Integration and Differential Equations, the second edition, by P.E. Protter. $\endgroup$ – Gordon Dec 14 '15 at 14:32
  • $\begingroup$ @Gordon You mean mine is wrong? :( $\endgroup$ – BCLC Dec 31 '15 at 5:19
  • $\begingroup$ I do not understand why you can have $ g(X_T) = h(X_t, W_t)$, and why $W_t$ is independent of $\mathscr F_t$. The book above is a good source for this problem. $\endgroup$ – Gordon Dec 31 '15 at 15:48
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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.

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  • $\begingroup$ When are you using the Feyman-Kac theorem @Danny Pak-Keung Chan? The final steps for proving the equality are based on measurability arguments, but once you have proved $F(t,X_t)=E[g(X_T)|\mathcal{F}_t]$, couldn't you simply invoke Feyman-Kac theorem to state that $F(t,X_t)=E[g(X_T)|X_t]$ given all the conditions are fulfilled? Or are you explicitly avoiding Feyman-Kac to ensure self-containdness of the proof? $\endgroup$ – Daneel Olivaw Jun 16 '17 at 9:50
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    $\begingroup$ Actually, I reproduced the proof of Feynman-Kac Theorem. However, for self-containedness, I included everything. Please bear in mind that my proof is still not completely rigorous (notice the wording "assume", which I don't know how to justify). $\endgroup$ – Danny Pak-Keung Chan Jun 16 '17 at 13:46

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