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

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


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

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.


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.