1
$\begingroup$

I need to prove that if $u: [0,T]\rightarrow \mathbb{R}$ is a deterministic square integrable function then stochastic exponential process defined :

$M_{t} = exp(-\int_0^t \! u(s) \, \mathrm{d}W_{s} - \frac{1}{2}\int_0^t \! u(x)^{2} \, \mathrm{d}s) \qquad\forall t \in [0,T]$

is integrable:

$ \mathbb{E}(|M_{t}|) < \infty\qquad\forall t \in [0,T]$

I'm kind of lost on this, even though it looks very elementary. I do not know how should I estimate the expressions with stochastic integral in exponent. I would appreciate any help.

edit:

I just came up with an idea, but I need someone to validate it. Because $|M_{t}| = M_{t}$ and the fact that exponent is convex, we can use Jensen inequality to obtain:

$\mathbb{E}(M_{t})^{2} \leq \mathbb{E}(M_{t}^{2})$,

then, because $M_{t}$ is log-normal with mean $\mu = - \frac{1}{2}\int_0^t \! u(x)^{2} \, \mathrm{d}s$ and variance $\sigma^{2} = \int_0^t \! u(x)^{2} \, \mathrm{d}s$ it means that its second moment is

$\mathbb{E}(M_{t}^{2}) = exp(2\mu + 2\sigma^{2}) = exp(\int_0^t \! u(x)^{2} \, \mathrm{d}s) \leq exp(\int_0^T \! u(x)^{2} \, \mathrm{d}s) < \infty $.

And : $\mathbb{E}(M_{t})^{2} < \infty \Rightarrow \mathbb{E}(|M_{t}|) < \infty$

Is that correct?

$\endgroup$
  • 1
    $\begingroup$ Yeah, sounds good. An alternative is using appromximation by step functions. $\endgroup$ – saz Mar 23 '15 at 7:11

Your Answer

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

Browse other questions tagged or ask your own question.