3
$\begingroup$

I am trying to compute the following expectation: $$ M_T = \mathbb E\left[W_T\int_0^T\,t\,d W_t \right] $$

where $0<t<T$ and $W = (W_t)_{t\geq 0}$ is a standard Brownian Motion started at $0$. I already know the solution is given by $M_T = \int_0^T t\,dt$ thanks to one of the answers given in this other question.

This looks trivial, but integration by parts is not working for me (I seem to be getting to the same place over and over) and alternative efforts like finding an expression that is equal in law have not being very fruitful.

I have seen in the forum that this case might be similar to this other question, but instead this last one involves working with sines and cosines, thus I have the feeling that in my particular case the solution is much simpler.

Is there an strategy to solve products of stochastic integrals and a function of brownian motion?

$\endgroup$
  • $\begingroup$ @user7530 looking at the tags of the question and the title of the question should render that question answered. I will however add more information. $\endgroup$ – Adam Feb 5 '15 at 9:08
5
$\begingroup$

Obviously, we can write

$$M_T = \mathbb{E} \left[ \left( \int_0^T 1 \, dW_t \right) \cdot \left( \int_0^T t \, dW_t \right) \right],$$

i.e. we are interested in the expectation of a product of stochastic integrals. Using the identity

$$a \cdot b = \frac{1}{4} ((a+b)^2-(a-b)^2) \tag{1}$$

for

$$a := \int_0^T 1 \, dW_t \qquad \quad b := \int_0^T t \, dW_t$$

we get

$$M_T = \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1+t) \, dW_t \right)^2 \right] - \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1-t) \, dW_t \right)^2 \right].$$

Applying Itô's isometry yields

$$\begin{align*} M_T &= \frac{1}{4} \mathbb{E} \left( \int_0^T (1+t)^2 \, dt \right) - \frac{1}{4} \mathbb{E} \left( \int_0^T (1-t)^2 \, dt \right) \\ &= \int_0^T \frac{1}{4} ((1+t)^2-(1-t)^2) \, dt \\ &\stackrel{(1)}{=} \int_0^T t \, dt. \end{align*}$$

In fact, we have shown the following (more general) statement:

Let $f,g \in L^2([0,T] \otimes \mathbb{P})$ be progressively measurable. Then $$\mathbb{E} \left[ \left( \int_0^T f(t) \, dW_t \right) \cdot \left( \int_0^T g(t) \, dW_t \right) \right] = \mathbb{E} \int_0^T g(t) \cdot f(t) \, dt.$$

Note that for $f=g$ this is the (standard version of) Itô's isometry.

$\endgroup$
2
$\begingroup$

Ito's Isometry:

$$ \mathbb{E}\left[W_T\cdot\int\limits_{0}^{T}t\,\mathrm dW_t\right]=\mathbb{E}\left[\int\limits_{0}^{T}1\,\mathrm dW_t\cdot\int\limits_{0}^{T}t\,\mathrm dW_t\right] =\mathbb{E}\left[\int\limits_{0}^{T}(1\cdot t)\,\mathrm d\langle W\rangle_t\right] =\int\limits_{0}^{T}t\,\mathrm dt\,. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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