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I am solving the (boundary?) value problem (from Bjork I think, see below)


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By Feynman-Kac, any solution has the form of a conditional expectation

$$F(t,x) = E[\psi(X_T)|X_t = x]$$

where

  1. $$\psi(x) = x^2$$

  2. $\{W_t\}_{t \in [0,T]}$ is standard Brownian motion in the filtered probability space $(\Omega, \mathfrak F, \{\mathfrak F_t\}_{t \in [0,t]}, \mathbb P)$ where $\mathfrak F_t = \mathfrak F_t^W$, the natural filtration of standard Brownian motion.

  3. $\{X_t\}_{t \in [0,T]}$ is any (I guess the only different thing is $X_0$, and that doesn't matter?) stochastic process in the same probability space satisfying either of the two SDEs

$$dX_t = \pm \sigma dW_t$$

and thus

$$X_T = X_t \pm \sigma (W_T -W_t)$$


So, I try to evaluate $$E[\psi(X_T)|X_t] \tag{*}$$ and then replace $X_t$ with $x$ as follows:

$$(*) = [(X_t \pm \sigma (W_T -W_t))^2|X_t]$$ $$ = E[X_t^2 \pm 2\sigma (W_T -W_t)X_t + (\sigma (W_T -W_t))^2|X_t]$$ $$ = E[X_t^2|X_t] \pm 2\sigma E[(W_T -W_t)X_t|X_t] + E[(\sigma (W_T -W_t))^2|X_t]$$ $$ = X_t^2 \pm 2\sigma X_t E[(W_T -W_t)|X_t] + E[(\sigma (W_T -W_t))^2|X_t]$$ $$ = X_t^2 \pm 2\sigma X_t E[W_T -W_t|X_t] + (\sigma^2)E[(W_T -W_t)^2|X_t]$$

$$ = X_t^2 \pm 2\sigma X_t E[W_T -W_t|\mathscr F_t] + (\sigma^2)E[(W_T -W_t)^2|\mathscr F_t] \tag{**}$$

$$ = X_t^2 \pm 2\sigma X_t E[W_T -W_t] + (\sigma^2)E[(W_T -W_t)^2]$$

$$ = X_t^2 \pm 2\sigma X_t (0) + \sigma^2(T-t)$$

$$ = X_t^2 + (\sigma^2)(T-t)$$

$$\to E[\psi(X_T)|X_t] = X_t^2 + \sigma^2(T-t)$$

$$\to E[\psi(X_T)|X_t=x] = x^2 + \sigma^2(T-t)$$

$$\to F(t,x) = x^2 + \sigma^2(T-t)$$


About the use of the Markov property in $(**)$,

I know that $W_T - W_t$ is independent of $\mathfrak{F_t}$, but I think that

$$X_t \ \in \ m \mathfrak F_t \ \to \ W_T - W_t \ \text{is independent of} \ X_t \tag{3}$$

If $(3)$ is wrong, why?

If $(3)$ is right, does this mean we don't need to use the Markov property? Everything after $(2)$ holds if we replace $\mathscr F_t$ with $X_t$? Why/Why not?


The problem seems to be taken from Bjork's Arbitrage Theory in Continuous Time. I got the problem from my class notes. Neither Bjork nor Wikipedia seems to use the Markov property


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    $\begingroup$ What do you mean by $dX_t = \pm \sigma dW_t$? $\endgroup$ – saz Jan 29 '15 at 10:35
  • $\begingroup$ @saz There are 2 possible SDEs. Same solution in either case. $\endgroup$ – BCLC Dec 9 '15 at 21:48
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Why do you think the expression implies such a lack of independence? Intuitively, the Markov property allows expectations conditional on randomness at a single point in time, $t$, to be replaced by expectations conditional on all of the randomness at points in time up to and including time $t$. Thats why $\mathfrak{F}_t$, which includes all the sigma-algebras up to index $t$, becomes relevant.

In fact, by proceeding with the computation and using the independence of $W_T-W_t$ and $\mathfrak{F}_t$ so that,

$$ \mathbb{E}\left[\left(W_T-W_t\right)\,\bigg\vert\,\mathfrak{F}_t\right]{}={}\mathbb{E}\bigg[\left(W_T-W_t\right)\bigg]{}={}0\,, $$ we have

$$ \begin{eqnarray*} \mathbb{E}\left[\left(X_t{}\pm{}\sigma\left(W_T-W_t\right)\right)^2\,\vert\,\mathfrak{F}_t\right]&{}={}&\mathbb{E}\left[X^2_t{}+{}\sigma^2\left(W_T-W_t\right)^2\pm2X_t\left(W_T-W_t\right)\,\vert\,\mathfrak{F}_t\right]\newline &{}={}&\mathbb{E}\left[X^2_t\,\vert\,\mathfrak{F}_t\right]{}+{}\sigma^2\mathbb{E}\left[\left(W_T-W_t\right)^2\,\vert\,\mathfrak{F}_t\right]\newline &&{}\pm2{}\mathbb{E}\left[X_t\left(W_T-W_t\right)\,\bigg\vert\,\mathfrak{F}_t\right]\newline &{}={}&X_t^2{}+{}\sigma^2\left(T-t\right){}{}\pm2X_t{}\mathbb{E}\left[\left(W_T-W_t\right)\,\bigg\vert\,\mathfrak{F}_t\right]\newline &{}={}&X_t^2{}+{}\sigma^2\left(T-t\right)\,, \end{eqnarray*} $$

which is required to hold at each point of realisation for the stochastic process up to, and including time $T$. Consequently, $$ F(t,x){}={}x^2{}+{}\sigma^2\left(T-t\right) $$

solves the given PDE.

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  • $\begingroup$ What? I know how to compute the expectation. I'm just wondering why those equations won't hold if you replace F_t with X_t. My answer to your 1st question is that in this case, I think, Markov property is not needed, but it might be for different PDE (see penultimate paragraph in OP pls) $\endgroup$ – BCLC Jan 30 '15 at 18:30
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    $\begingroup$ @BCLC, I see. You are wondering why $\sigma(X_t)$ isn't sufficient for the computation, instead of the apparent "overkill" of $\mathfrak{F}_t$? $\endgroup$ – ki3i Jan 30 '15 at 18:52
  • $\begingroup$ @BCLC, ...or asked another way, you are wondering why the problem can't be solved using the random variables $\mathbb{E}[ X_T^2\vert X_t]$, each of which are $\sigma(X_t)$-measurable respectively, instead of the martingale $\mathbb{E}[ X_T^2\vert \mathfrak{F}_t]$, which is adapted to the filtration $\{\mathfrak{F}_t\}_{t\ge 0}$? One reason is the stochastic integrals that justify the solution (e.g. showing rigorously that the solution satisfies the PDE) are defined in terms of martingales. More intuitively, as stated in my answer, the history of the process is required to satisfy the PDE. $\endgroup$ – ki3i Jan 30 '15 at 20:22
  • $\begingroup$ I don't follow at all. :( Again, is the incremement not independent of X_t? To your last statement, why? $\endgroup$ – BCLC Jan 31 '15 at 2:18
  • $\begingroup$ @BCLC, Yes, the increment is independent of $X_t$,...but that is not enough to solve the PDE using a martingale approach since, in general, $\mathbb{E}\left[X^2_T\vert X_t \right]$ is not a martingale adapted to the relevant filtration. It is, however, related to $\mathbb{E}\left[X^2_T\vert \mathfrak{F}_t \right]$ which is a martingale, and the Markov-property allows us to cast our argument in terms of this martingale. And, why is the process history required to satisfy the PDE? Because, we are trying to solve this particular PDE with the process; not just any unconstrained process will do. $\endgroup$ – ki3i Feb 1 '15 at 15:52
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In applying the Feynman-Kac formula, there is no need to use the Markov property.

In proving the Feynman-Kac formula, the Markov property is needed.

Showing that $G(t,x)$ indeed satisfies the PDE requires showing that $G(t,X_t)$ is a martingale which relies on $X_t$ having the Markov property, which it has because it is a solution of an SDE.

Or something like that.


From Shreve's Stochastic Calculus for Finance:


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