Proof verification: conditional distribution of integral of brownian motion

I am looking to compute the conditional distribution of $$S_T=\int_0^T W_t dt$$ given $$W_T=x$$. Thanks to this question and using the fact that $$d(tW_t)=W_tdt + tdW_t$$ by Itô's formula, we get \begin{align*} W_t dt &= d(tW_t) - t dW_t\\ S_T=\int_0^T W_t dt & = T W_T - \int_0^T t dW_t \end{align*} and hence $$\mathbb{E}[S_T|W_T = x]= Tx$$ because the second term is the stochastic integral of a deterministic process, which is normally distributed with mean 0. Furthermore, \begin{alignat*}{2} \text{Var}(S_T|W_T=x) & = \text{Var}\left(Tx - \int_0^T t dW_t\right)\\ &= \text{Var}\left(\int_0^T t dW_t\right)\\ & = \mathbb{E}\left[\left(\int_0^T t dW_t\right)^2\right] &&\text{ null mean}\\ &= \int_0^T t^2 dt &&\text{ isometry}\\ & = \frac{T^3}{3} \end{alignat*} Is this correct?

• "because the second term is the stochastic integral of a deterministic process": I'm concerned. It's the stochastic integral with respect to $W_t$, which conditioned on $W_T = x$ is no longer a Brownian motion; not even a martingale, so I don't see how you can conclude its conditional expectation is zero. Commented Apr 13, 2022 at 5:53
• For a more vivid example, obviously $E[\int_0^T dW_t] = E[W_T] = 0$ unconditionally, but $E[\int_0^T dW_t \mid W_T = x] = x \ne 0$. Commented Apr 13, 2022 at 5:55
• I would be inclined to solve this by noting that $(S_T, W_T)$ are jointly normal, and computing their (unconditional) covariance matrix. Fubini's theorem helps. Then it is easy to find the conditional distribution of one coordinate of a jointly normal vector given another. Commented Apr 13, 2022 at 5:58

First, by witnessing the random vector $$(S_T, W_T)$$ as the $$L^2$$-limit of Gaussian vectors, we note that $$(S_T, W_T)$$ itself is Gaussian. As you have observed, integration by parts gives us $$S_T = TW_T - \int_0^T s dW_s$$ From here, we get $$E(S_T) = 0$$ and \begin{align*} E(S_T^2) &= T^2 E(W_T^2) - 2T E \left(W_T \int_0^T s dW_s\right) + E \left[\left(\int_0^T s dW_s\right)^2\right] \\ &= T^3 - 2T E \int_0^T s ds + E \int_0^T s^2 ds \\ &= \frac{T^3}{3} \end{align*} where we use the Itô isometry to compute the second and third terms. Likewise, we have $$E(W_T) = 0$$, $$E(W_T^2)=T$$, and, \begin{align*} E(S_T W_T) &= E \left(TW_T^2 - W_T \int_0^T sdW_s\right) =\frac{T^2}{2} \end{align*}
Thus, $$(S_T, W_T) \sim N (0, \Sigma)$$ where \begin{align*} \Sigma &= \begin{pmatrix} \frac{T^3}{3} & \frac{T^2}{2} \\ \frac{T^2}{2} & T \end{pmatrix} \end{align*}
It follows for the formula for bivariate normal distributions that $$S_T | (W_T = x) \sim N \left( \frac{T}{2}x, \frac{T^3}{12} \right)$$
• Nice answer. One can also compute the expectations without integrating by parts, using just Fubini's theorem. We have $S_T^2 = \left(\int_0^T W_t\,dt\right)^2 = \int_0^T \int_0^T W_t W_s\,ds\,dt$. So by Fubini $E[S_T^2] = \int_0^T \int_0^T E[W_t W_s]\,ds\,dt = \int_0^T \int_0^T t \wedge s\,ds\,dt$. Commented Apr 17, 2022 at 4:38
• Similarly, $E[S_T W_T] = E\left[\int_0^T W_s W_T\,ds\right]$. Since $E[W_s W_T] = s$ for $s \le T$, we get $E[S_T W_T] = \int_0^T s\,ds$. Commented Apr 17, 2022 at 4:39
• Can you explain why $\mathbb{E}\left[W_T\int_0^T s dW_s\right] =\int_0^T sds$? Commented Apr 17, 2022 at 16:04
• @SimonCello94 For processes $X, Y$ and martingales $M,N$, one has $\mathbb{E}\left[ \int_0^T X_s dM_s \int_0^T Y_s dN_s \right] = \mathbb{E} \left[ \int_0^T X_s Y_s d \langle M , N \rangle \right]$. This can be proven by the polarisation formula. Now, write $W_T = \int_0^T dW_s$ and apply the formula. Commented Apr 17, 2022 at 16:11