$M_t=\int_0^t e^{-3W_s}dW_s$ properties

Question

Let $W_t$ standard Wiener process and $M_t=\int_0^t e^{-3W_s}dW_s$
a) Is $M_t$ well defined? Is it a martingale? Is it integrable with a square? Justify.
b) Determine $\left< M \right>_t$.
c) Determine the covariance function for the process $Y_t=\int_0^t e^{W_s}dM_s$. Why is $Y_t$ well defined?

I have a lot of questions to this task:

1. Is $M_t$ well defined when $\mathbb EM_t<\infty$? Is this enough or do I need to check any other conditions?
2. So how to calculate $\mathbb EM_t=\mathbb E \left( \int_0^t e^{-3W_s}dW_s \right)$?
3. $M_t$ is martingale if $\mathbb EM_t<\infty$ and $\forall_{s\le t; s,t \in T} \mathbb E[M_t | \mathcal F_s]=M_s$.
   My try: $$\mathbb E\left [M_t |\mathcal F_s \right]=\mathbb E \left[ M_t-M_s+M_s |\mathcal F_s \right]=\mathbb E \left[ \int_0^t e^{-3W_u}dW_u-\int_0^s e^{-3W_u}dW_u+M_s \bigg|\mathcal F_s \right]=$$ $$=\mathbb E \left[ 0+\int_s^t e^{-3W_u}dW_u+M_s \bigg|\mathcal F_s \right]$$ It is correct? Is $\mathbb E \left[ \int_s^t e^{-3W_u}dW_u \right]$ equal or not equal $0$?
4. $M_t$ is integrable with square when $\mathbb E |M_t|^2=\int_0^t \mathbb E (e^{-6W_s})ds<\infty$. How I can check this?
5. $<M_t>=\int_0^t \left( e^{-3W_s} \right)^2dW_s$ How to calculate it?
6. I have absolutely no idea how to solve c)

Answer 1

1. Is $M_t$ well defined when $\mathbb EM_t<\infty$? Is this enough or do I need to check any other conditions?

Any random variable is well-defined when its expectation is bounded Does finite expectation imply bounded random variable?.

In fact, $M_{t}$ is an $L_{2}$-object by Itô isometry

$$E[M_{t}^{2}]=\int_{0}^{t}e^{\frac{6}{2}s}ds<\infty.$$

2. So how to calculate $\mathbb EM_t=\mathbb E \left( \int_0^t e^{-3W_s}dW_s \right)$?

Every Itô integral has zero expectation: Itō Integral has expectation zero or more directly here.

3. $M_t$ is martingale if $\mathbb EM_t<\infty$ and $\forall_{s\le t; s,t \in T} \mathbb E[M_t | \mathcal F_s]=M_s$.
   My try: $$\mathbb E\left [M_t |\mathcal F_s \right]=\mathbb E \left[ M_t-M_s+M_s |\mathcal F_s \right]=\mathbb E \left[ \int_0^t e^{-3W_u}dW_u-\int_0^s e^{-3W_u}dW_u+M_s \bigg|\mathcal F_s \right]=$$ $$=\mathbb E \left[ 0+\int_s^t e^{-3W_u}dW_u+M_s \bigg|\mathcal F_s \right]$$ It is correct? Is $\mathbb E \left[ \int_s^t e^{-3W_u}dW_u \right]$ equal or not equal $0$?

By the independence of increments we have

$$\mathbb E \left[ \int_s^t e^{-3W_u}dW_u \bigg|\mathcal F_s\right]$$

$$=e^{-3W_{s}}\mathbb E \left[ \int_s^t e^{-3(W_u-W_{s})}dW_u \bigg|\mathcal F_s\right]$$

$$=e^{-3W_{s}}\mathbb E \left[ \int_s^t e^{-3(W_u-W_{s})}dW_u \right]$$

$$=0.$$

4. $M_t$ is integrable with square when $\mathbb E |M_t|^2=\int_0^t \mathbb E (e^{-6W_s})ds<\infty$. How I can check this?

This follows by Itô isometry. (besides various sde textbooks, it is explained here too.)

5. $<M_t>=\int_0^t \left( e^{-3W_s} \right)^2dW_s$ How to calculate it?

That's not right. Here you actually integrate against the quadratic variation of $W_{s}$ which is $d<W_{s}>=ds$,eg. studied in Quadratic variation of Ito integral : Relationship between Ito Integral and Brownian Motion and Quadratic variation of ito integral.

6. I have absolutely no idea how to solve c)

We have

$$Y_{t}=\int_0^t e^{-2W_u}dW_u $$

and so as above we use zero expectation and Itô isometry for $t\geq s$

$$E[Y_{t}Y_{s}]=0+E[Y_{s}^{2}]=\int_0^{min(s,t)} e^{-2u}du.$$

The well-defined follows as above because $Y_{t}\in L_{2}$.