# Risk neutral Esscher transform of exponential Levy processes

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.

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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

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.