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.

  • $\begingroup$ are you sure that your application of the Bayes' rule is correct ? $\endgroup$ Mar 19, 2014 at 11:08
  • $\begingroup$ also how do you arrive at $e^{X_t}$ in the second equation ? $\endgroup$ Mar 19, 2014 at 11:10
  • $\begingroup$ 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. $\endgroup$
    – Slug Pue
    Mar 19, 2014 at 15:54
  • $\begingroup$ okey now I see it - I will think some more on the topic :) $\endgroup$ Mar 19, 2014 at 16:13
  • 1
    $\begingroup$ 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 :) $\endgroup$ Mar 20, 2014 at 14:05

2 Answers 2


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


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.


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