# Conditional Expectation of Integral of Squared Brownian Motion - PDE Approach

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)

• 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 ?
– dm63
Jul 27, 2022 at 11:34
• You wrote "given that u is a martingale". Are you sure?
– user34971
Jul 27, 2022 at 12:13
• @dm63 Yes made a mistake here - just fixed Jul 27, 2022 at 12:15

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).