2
$\begingroup$

I have the following equation : $$ E[e^{(\theta +1) X_T} | \mathcal{F}_t] = e^{X_t} E[e^{\theta X_T} | \mathcal{F}_t] $$ where $X_t$ is a Levy process, $\mathcal{F_t}$ the filtration at time $t$, $T > t$, and $\theta$ some number. I am trying to find a solution for $\theta$ in closed form (it probably has to involve the charateristic function $E(e^{iuX_t})$ of the Levy process)

What I have tried:

Since $e^{(\theta+1)X_t}$ is $\mathcal{F}_t$-measurable, it can be written $$ e^{(\theta +1 )X_t} E[e^{(\theta +1)(X_T-X_t)} | \mathcal{F}_t] = e^{X_t} E[e^{\theta X_T} | \mathcal{F}_t]$$ By stationarity of increments of the Levy process we have $$ e^{\theta} E[e^{(\theta +1)X_{T-t}} | \mathcal{F}_t] = E[e^{\theta X_T} | \mathcal{F}_t] $$ Now by making the substitution $\theta +1 = iu$ we rewrite the equation in terms of characteristic functions: $$ e^{\theta} e^{(T-t)\psi(u)} = e^{t\psi(u)}E(e^{-X_T}|\mathcal{F}_t) $$ Where $\psi$ is the characteristic exponent (given in closed form by the Levy-Khintchine formula). This is almost what I need, except the extra expectation.

$\endgroup$
1
$\begingroup$

Let's start from the beginning. Using the independence and stationarity of the increments, we see that

$$\begin{align*} \mathbb{E}\bigg(e^{(\theta+1) X_T} \mid \mathcal{F}_t \bigg) &= e^{(\theta+1)X_t} \mathbb{E}\bigg(e^{(\theta+1) (X_T-X_t)} \mid \mathcal{F}_t \bigg) \\ &= e^{(\theta+1)X_t} e^{(T-t) \psi(-\imath (\theta+1))} \tag{1} \end{align*}$$

where $\psi$ denotes the characteristic exponent of the Lévy process. Similarly,

$$e^{X_t} \mathbb{E}\bigg(e^{\theta X_T} \mid \mathcal{F}_t \bigg) = e^{(\theta+1) X_t} \mathbb{E}\bigg(e^{\theta (X_T-X_t)} \mid \mathcal{F}_t \bigg) = e^{(\theta+1)X_t} e^{(T-t) \psi(-\imath \theta)}. \tag{2}$$

This means that $\theta$ has to solve the equation

$$\psi(-\imath (\theta+1)) = \psi(-\imath \theta).$$

For example, if $(X_t)_t$ is a Brownian motion, then the solution is given by $\theta = -1/2$.

$\endgroup$
8
  • $\begingroup$ Yes but aren't these steps that I already performed? We end up with $e^{\theta}e^{(T-t)\psi (u) = E[e^{\theta X_T} | \mathcal{F}_t}]$, which is where I am stuck. $\endgroup$
    – user126540
    Mar 18 '14 at 16:14
  • $\begingroup$ @user126540 Sorry, misunderstood your question. I'll think about it. $\endgroup$
    – saz
    Mar 18 '14 at 16:27
  • $\begingroup$ Beautiful. Although one question about the relation $E[exp(X_{T-t}) | \mathcal{F}_t] = E[exp(X_{T-t})]$. Let us say T-t > t. Intuitively, shouldn't it instead be something like $E[exp(X_{T-t}) | \mathcal{F}_t] = exp(X_t)+E[exp(X_{T-t})]$? $\endgroup$
    – user126540
    Mar 18 '14 at 18:19
  • $\begingroup$ @user126540 Why should it? If $T-t>t$, then $\exp(X_{T-t})$ is $\mathcal{F}_t$-measurable, i.e. $$\mathbb{E}(\exp(X_{T-t}) \mid \mathcal{F}_t) = \exp(X_{T-t}).$$ $\endgroup$
    – saz
    Mar 18 '14 at 18:27
  • $\begingroup$ Guess I'll have to take a closer look at filtrations. Thanks! $\endgroup$
    – user126540
    Mar 18 '14 at 18:31

Your Answer

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