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I am looking to compute the following using Ito's formula.

$$u(t,\beta_t) = \mathbb{E}(\int_t^T\beta_s^2ds|\beta_t)$$

Knowing the properties of brownian motion, it is rather easy to show that the above is equivalent to $\frac{1}{2}(T^2-t^2)$; however, i'm looking to apply Ito's formula to come up with a similar result. Given that $u$ is a martingale, it follows from Ito's formula that $u$ satisfies the homogenous heat equation:

$$u_t = \frac{1}{2}u_{xx}$$ Though I am struggling to see how the solution aligns with what I found using the easier approach.

Side note:

My boundary conditions: $$u(T,x) = 0$$ $$u(0,0) = \mathbb{E}(\int_0^Tds) = T $$ Though I could be off here, as the expectation is confusing me

Edit:

My approach to finding $\frac{1}{2}(T^2-t^2)$ through knowledge of B.M.:

(1) By the tower property, using the fact that $\beta_t\in F_t$ $$u(t, \beta_t) = \mathbb{E}(\mathbb{E}(\int_t^T\beta_s^2ds|F_t)|\beta_t)$$

(2)Then given the integral is not within $F_t$, we have $$u(t,\beta_t) = \mathbb{E}(\mathbb{E}(\int_t^T\beta_s^2ds)|\beta_t)$$

(3)

$$u(t,\beta_t) = \mathbb{E}((\int_t^T\mathbb{E}(\beta_s^2)ds|\beta_t)$$

(4) Lastly,

$$u(t,\beta_t) = \mathbb{E}(T-t|\beta_t) = \frac{1}{2}(T^2-t^2)$$ (trivially)

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  • $\begingroup$ Going from (3) to (4) looks incorrect to me. Doesn’t it depend on $\beta_t$? And also have you integrated with respect to s properly ? $\endgroup$
    – dm63
    Jul 27 at 11:34
  • $\begingroup$ You wrote "given that u is a martingale". Are you sure? $\endgroup$
    – user34971
    Jul 27 at 12:13
  • $\begingroup$ @dm63 Yes made a mistake here - just fixed $\endgroup$ Jul 27 at 12:15

1 Answer 1

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I am not sure anymore what exactly your question is, but the correspondence between PDEs and probability / SDEs is given by Feynman-Kac. See for instance here.

Thus, using Feynman-Kac, the PDE satisfied by $u(t,\beta_t)$ is $$ \left\{ \partial_t + \frac12 \partial^2_{\beta_t\beta_t} \right\} u(t,\beta_t) = - \beta_t^2 $$ with terminal condition $$ u(T,\beta_T) = 0 $$ with the understanding that $\beta$ is standard Brownian motion.

Notice also that $u$ is not a martingale (hence my comment above).

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  • $\begingroup$ This was the correct approach - thank you $\endgroup$ Jul 28 at 15:04

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