Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_t$ be a Levy Process and $e^{X_t}$ the corresponding exponential Levy process. Using the Esscher transform for a change of measure for which the Radon-Nykodym derivative is $$\frac{d\mathbb{Q}}{d\mathbb{P}} = \frac{e^{\theta X_T}}{E[e^{\theta X_T}]},$$

I am looking to find the Esscher parameter $\theta$ such that the measure $\mathbb{Q}$ is risk neutral, i.e. such that the following equation is satisfied: $$ E^{\mathbb{Q}}[e^{X_T} \vert \mathcal{F}_t] = e^{X_t} $$ where $T>t$ and $\mathcal{F}_t$ is the filtration at time t. My goal is to find an explicit formula for $\theta$ in terms of characteristic functions of the Levy process.

What I have tried: Using Bayes' rule $$ E^{\mathbb{Q}}[X \vert \mathcal{F}] = \frac{E^{\mathbb{P}}[ X f \vert \mathcal{F}]}{E^{\mathbb{P}} [f \vert \mathcal{F}]} $$ where $f$ is a Radon-Nykodym derivative $dQ/dP$, we get $$ E^{\mathbb{P}} \left[ \frac{e^{\theta X_T}}{E^{\mathbb{P}}[e^{\theta X_T}]} e^{X_T} \bigg| \mathcal{F}_t \right]\frac{1}{ E^{\mathbb{P}} \left[ \frac{e^{\theta X_T}}{E^{\mathbb{P}}[e^{\theta X_T}]} \big| \mathcal{F}_t \right]} = e^{X_t} \Leftrightarrow\\ E^{\mathbb{P}} [e^{(\theta +1) X_T} | \mathcal{F}_t] = e^{X_t} E^{\mathbb{P}}[e^{\theta X_T} | \mathcal{F}_t]$$ Since $e^{(\theta+1)X_t}$ is $\mathcal{F}_t$-measurable, this can be written $$ e^{(\theta +1 )X_t} E^{\mathbb{P}}[e^{(\theta +1)(X_T-X_t)} | \mathcal{F}_t] = e^{X_t} E^{\mathbb{P}}[e^{\theta X_T} | \mathcal{F}_t]$$ By stationarity of increments of the Levy process this can be written $$ e^{\theta} E^{\mathbb{P}}[e^{(\theta +1)X_{T-t}} | \mathcal{F}_t] = E^{\mathbb{P}}[e^{\theta X_T} | \mathcal{F}_t] $$ 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. This is almost what I need, except the extra expectation. What to do with it? I have a somewhat limited knowledge of filtrations for continuous time models so I am not sure whether the above calculations are correct either.

share|improve this question
are you sure that your application of the Bayes' rule is correct ? – Probilitator Mar 19 '14 at 11:08
also how do you arrive at $e^{X_t}$ in the second equation ? – Probilitator Mar 19 '14 at 11:10
pretty sure about the Bayes' rule (the calculations involving it are basically mere algebraic operations). The $e^{X_t}$ comes from the risk neutral valuation principle that $X_t = E^Q [X_T | \mathcal{F}_t]$ where $T$ is the end date. I have removed the discount factor for simplicity. – Slug Pue Mar 19 '14 at 15:54
okey now I see it - I will think some more on the topic :) – Probilitator Mar 19 '14 at 16:13
according to your reference it should be $$ E^{\mathbb{Q}}[X \vert \mathcal{F}] = \frac{E^{\mathbb{P}}[ X f \vert \mathcal{F}]}{E^{\mathbb{P}} [f \vert \mathcal{F}]} $$ This is different from what you wrote above - but you apply it correctly later on :) – Probilitator Mar 20 '14 at 14:05
up vote 1 down vote accepted

I got a solution to this problem by posting an excerpt of it at math.stackexchange: http://math.stackexchange.com/questions/716242/equation-involving-expectations-of-levy-processes

share|improve this answer

In the paper OPTION PRICING BY ESSCHER TRANSFORMS the authors explore this topic extensively and provie equations that enable the calculation of the risk neutral $\theta$.

Also note that you can easily deal with the expectation in $$ e^{\theta} e^{(T-t)\psi(u)} = e^{t\psi(u)}E(e^{-X_T}|\mathcal{F}_t) $$

if the process $X_t$ itself has nice properties. One could solve it in the GBM cases. A solution should also be attainable if the process' transition density is known explicitly.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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