Using Feynman–Kac for the Expected Integral of Squared Brownian Motion
Summary
The document asks how to derive the conditional expectation of the future integral of squared standard Brownian motion using Itô’s formula and a partial differential equation. The key correction is that this conditional expectation is not itself a martingale. Because the integral accumulates the running quantity β², the relevant Feynman–Kac equation has a source term: the time derivative plus half the second spatial derivative of the value function equals minus x², with zero terminal value.
For Brownian motion at value x and time t, this yields a conditional expectation of (T−t)x² plus one half of (T−t)². The discussion highlights why the proposed homogeneous heat equation and boundary condition at the origin do not represent the problem: the conditional value depends on the current Brownian level, and its accumulated running cost must appear in the PDE. The original derivation also mishandles conditioning and integration, so its claimed expression is not valid for general current states.
Key ideas
- The conditional expected future integral depends on both the current time and Brownian level.
- Feynman–Kac gives an inhomogeneous backward PDE when the expectation includes a running quantity.
- The value function has terminal condition zero because no future integral remains at maturity.
- The conditional expectation is not a martingale because the accumulated integral has a nonzero running contribution.
- Conditioning must preserve the current Brownian state rather than replace future squared values by unconditional expectations.
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# Conditional Expectation of Integral of Squared Brownian Motion - PDE Approach
# 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)
## Answer by user34971 (score 1, accepted)
https://quant.stackexchange.com/a/71742
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).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.