Bound on the conditional expectation of a stochastic integral given W_T

Question

Suppose that $f \colon \mathbb R \to \mathbb R$ is a bounded function and let $X$ denote the stochastic integral $$ X = \int_0^T f(W_t) \, d W_t. $$ Is it possible in general to bound conditional expectations of the form: $$ \mathbb E [X | W_T = x] \, ? $$

Answer 1

Define $F(t) =\int_0^tf(s)ds$ and apply the Ito's lemma to the function $F(W_t)$: $$dF(W_t)=f(W_t)dW_t + \frac{1}{2}f'(W_t)dt$$ Then $$\begin{align} &F(W_T) =\int_0^Tf(W_t)dW_t + \frac{1}{2}\int_0^Tf'(W_t)dt\\ &\implies \int_0^Tf(W_t)dW_t = F(W_T)-\frac{1}{2}\int_0^Tf'(W_t)dt \\ &\implies \mathbb{E} \left(X|W_T \right) = F(W_T)-\frac{1}{2} \mathbb{E} \left(\left.\int_0^Tf'(W_t)dt \right|W_T\right) \tag{1} \end{align}$$ From the property of Brownian Bridge, we have: $$\begin{align} W_t|W_T &= \frac{t}{T} W_T + \underbrace{\left(W_t -\frac{t}{T} W_T \right)}_{\text{independent to }W_T} \\ &\stackrel{\mathcal{D}}{=} \mathcal{N} \left(\frac{t}{T}W_T,\frac{t(T-t)}{T} \right) \stackrel{\mathcal{D}}{=} \frac{t}{T}W_T + \sqrt{\frac{t(T-t)}{T}} \cdot \mathcal{N}(0,1) \end{align}$$ then $$\begin{align} \mathbb{E} \left(\left.\int_0^Tf'(W_t)dt \right|W_T\right) &=\int_0^T\mathbb{E} \left(\left.f'(W_t) \right|W_T\right) dt\\ &=\int_0^T\mathbb{E} \left(\left.f'\left(\frac{t}{T}W_T + \sqrt{\frac{t(T-t)}{T}} \cdot Z\right) \right|W_T\right) dt\\ &=\int_0^T\int_{z \in \mathbb{R}} f'\left(\frac{t}{T}W_T + \sqrt{\frac{t(T-t)}{T}} \cdot z\right)\varphi(z) dz dt\\ &=\int_{z \in \mathbb{R}}\varphi(z) \left(\int_0^Tf'\left(\underbrace{\frac{t}{T}W_T + \sqrt{\frac{t(T-t)}{T}} \cdot z}_{=:g(t,T,W_T,z)=: g(t)} \right) dt\right)dz\\ &=\int_{z \in \mathbb{R}}\varphi(z) \left(\frac{f(g(T))}{g'(T)} + \int_0^T\left(f(g(t))\cdot \frac{g''(t)}{(g'(t))^2}\right) dt\right)dz \end{align}$$ where $\varphi(x)$ is the density function of $\mathbb{N}(0,1)$ and $$\begin{align} &g(t): = g(t,T,W_T,z) = \frac{t}{T}W_T + \sqrt{\frac{t(T-t)}{T}} \cdot z \\ &g'(t) = \frac{W_T}{T} + \frac{T-2t}{2\sqrt{tT(T-t)}}\cdot z \\ &g''(t)= -\frac{1}{4}\cdot\left(\frac{T}{t(T-t)}\right)^{3/2}z \tag{2} \end{align}$$ Because $g'(t) \xrightarrow{t\to T} -\infty$, then $f(g(t))/g'(t) \xrightarrow{t\to T} 0$, we deduce that: $$ \mathbb{E} \left(\left.\int_0^Tf'(W_t)dt \right|W_T\right)= \int_{z \in \mathbb{R}}\varphi(z) \left(\int_0^T\left(f(g(t))\cdot \frac{g''(t)}{(g'(t))^2}\right) dt\right)dz\tag{3}$$ Finally, from $(1),(3)$ we deduce that $$\color{red}{\mathbb{E} \left(X|W_T \right) = F(W_T) - \frac{1}{2}\int_{z \in \mathbb{R}} \left( \int_0^T \left( f(g(t))\cdot \frac{g''(t)}{(g'(t))^2}\right) dt\right)\varphi(z)dz }$$ where the function $g$ and its derivatives are defined by $(2)$.