Regression of stochastic integral on Wiener process

Question

This question is a follow-up from the following: conditional expectation of stochastic integral so I won't repeat myself regarding assumptions and notation.

Using Brownian bridge approach, we know that ${\mathbb E}[W_t|W_T]=\frac{t}{T}W_T$. This is compatible with a regression decomposition of $W_t$ on $W_T$, such as:

$$ W_t = \beta^W_t W_T + \epsilon $$ for t $\leq T$, where $\epsilon \sim \mathcal{N}(0,1)$ is an independent noise and $\beta^W_t$ can be interpreted as a standard OLS estimator, indeed

$$ \beta^W_t = \frac{{\mathbb Cov}(W_t,W_T)}{{\mathbb Var}(W_T)} = \frac{{\mathbb E}[W_t W_T]}{{\mathbb E}[W^2_T]} = \frac{t}{T} $$

In question conditional expectation of stochastic integral, we showed that the conditional expectation of the stochastic integral of a deterministic function $\sigma_t$ $$ M_t = \int_0^t \sigma_s dW_s $$ w.r.t. to the Wiener process at $T \geq t$ can be written as

$$ {\mathbb E}[M_t|W_T] = \frac{\int^t_0 \sigma_s ds}{T} W_T $$

By analogy, we extend the above regression decomposition as

$$ M_t = \beta^M_t W_T + \epsilon $$

with

$$ \beta^M_t = \frac{\int^t_0 \sigma_s ds}{T} $$

Now, $\beta^M_t$ can be properly interpreted as an OLS estimator as long as

$$ \beta^M_t = \frac{{\mathbb Cov}(M_t,W_T)}{{\mathbb Var}(W_T)} = \frac{\int^t_0 \sigma_s ds}{T} $$

that is to say, as long as the covariance between the stochastic integral $M_t$ and the Wiener $W_T$ is

$$ {\mathbb Cov}(M_t,W_T) = \int^t_0 \sigma_s ds $$

which is the conjecture we'd like to prove.

Answer 1

By definition,

$$ {\mathbb Cov}(M_t,W_T) = {\mathbb E}[M_t W_T] - {\mathbb E}[M_t] {\mathbb E}[W_T] = {\mathbb E}[M_t W_T] $$

since ${\mathbb E}[M_t] = {\mathbb E}[W_T] = 0 $. We now consider the representation of $M_t$ in terms of $W_t$ as suggested in this answer

$$ M_t = \sigma_t W_t - \int^t_0 \dot{\sigma}_s W_s ds $$

where we are assuming that $\sigma_t$ is regular enough such that $\dot{\sigma}_t \stackrel{def}{=}\frac{d \sigma}{dt}$ is well defined. We can live with that.

Therefore, we can write ($t \leq T)$:

\begin{align} {\mathbb E}[M_tW_T] & = {\mathbb E}\left[\left(\sigma_t W_t - \int^t_0 \dot{\sigma}_s W_s ds \right) W_T \right] \\ & = \sigma_t {\mathbb E}[W_t W_T] - \int^t_0 \dot{\sigma}_s {\mathbb E}[W_s W_T] ds \\ & = \sigma_t t - \int^t_0 \dot{\sigma}_s s ds \\ & = \sigma_t t - \left[\sigma_t t - \int^t_0 \sigma_s \cdot 1 ds \right] \\ &= \int^t_0 \sigma_s ds \end{align}

where integration by parts has been used in the next-to-last line. This proves the conjecture.