4
$\begingroup$

I am trying to derive a characteristic function (in Levy-Khintchine form) of a compound Poisson process $X_T$ under a risk neutral measure $\mathbb{Q}$, using the Esscher transfrom to change the original measure. The jumps of the compound Poisson are normally distributed with $\sigma=1$, $\mu=0$, and we set the jump intensity $\lambda=1$.

I am having trouble getting it to work "right", i.e. that $E_{\mathbb{Q}}[e^{X_T}]=1$ under the martingale condition.

Using Cont/Tankov -Financial Modelling with Jump Processes (section 9.5) we find that the Esscher trasnformed characteristic exponent is $$ \psi_Q(u) = iu\int_{-1}^1x(e^{\theta x}-1)F(dx)+\int_{\mathbb{R}}(e^{iux}-1)e^{\theta x}F(dx) $$ where $F$ is the normal distribution function. Now, under the martingale condition we should have $\psi_Q(-i)=0$. However, using* $$ \text{mgf}(\theta +1)=\text{mgf}(\theta) $$ to find the risk neutral Esscher parameter (mgf is the moment-gnereating function of $X_T$), I get $\theta = -1/2$, and with this we have $$ \psi_Q(-i) = \int_{-1}^1x(e^{-x/2}-1)F(dx) +0 $$ which is non-zero.

Is it possible that I have misunderstood something important here? I have checked my math quite thoroughly at nearly every step.

*This equation is taken from the paper Gerber and Shiu - Option pricing by Esscher transform

$\endgroup$
1
+300
$\begingroup$

You seem to be missing the truncation term in the jump part. It is indeed absent in the characteristic function of the compound Poisson by finiteness of the Levy measure, but one cannot necessarily remove it from the measure-changed characteristic function.

Instead of

$$\int_{\mathbb{R}}(e^{iux}-1)e^{\theta x}F(dx)$$ it should be

$$\int_{\mathbb{R}}(e^{iux}-1-x_{|x| \leq1})e^{\theta x}F(dx)$$

$\endgroup$

Your Answer

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

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