# Regression of stochastic integral on Wiener process

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.

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.