How to obtain this exponential martingale? A question concerning a paper from Stroock and Varadhan

77 Views Asked by At

In page 364 of the article Diffusion processes with continuous coefficients I (Stroock Varadhan - 1969), one finds in lemma 3.5:

enter image description here

enter image description here

The question is: how do we prove that $$ Y^s_\theta(t) = \exp \big\{ \langle \theta, \eta(t)\rangle - \frac{1}{2} \int_s^t \langle \theta, \sigma^*(u) a(u) \sigma(u)\rangle \, du \big\} ?$$ **Attempt ** In the proof of lemma 3.5 one reads enter image description here

So we come back to theorem 3.2 enter image description here

Now we try to see how to obtain the martingale property of $$ Y^s_\theta(t) = \exp \big\{ \langle \theta, \eta(t)\rangle - \frac{1}{2} \int_s^t \langle \theta, \sigma^*(u) a(u) \sigma(u)\rangle \, du \big\} $$

as a consequence of Theorem 3.2

First we note that

$$\langle \theta, \eta(t) \rangle = \sum_{j} \theta_j \eta_j(t) $$

but since $\eta(t) = \int_s^t\sigma(u)d \xi(u)$

$$ \eta_j(t) = \sum_k \int_s^t \sigma_{j,k}(u) d\xi_k(u)$$ so

$$ \langle \theta, \eta(t) \rangle = \sum_j \sum_k \theta_j\int_s^t \sigma_{j,k}(u) d\xi_k(u) \\ = \sum_k \int_s^t \sum_j \sigma*_{k,j}(u)\theta_j d\xi_k(u)\\ = \sum_k \int_s^t (\sigma*(u)\theta)_k d\xi_k(u)\\ $$

So applying theorem 3.2 we get $$ Y^s_\theta(t) = \exp \big\{ \langle \theta, \eta(t)\rangle - \frac{1}{2} \int_s^t \langle \sigma^*(u)\theta, a(u) \sigma^*(u)\theta\rangle \, du \big\}\\ = \exp \big\{ \langle \theta, \eta(t)\rangle - \frac{1}{2} \int_s^t \langle \theta, \sigma(u) a(u) \sigma^*(u)\theta\rangle \, du \big\} $$

So this result is a bit different than what is stated in theorem 3.2.

Is this a typo?