Characteristic Function of Variance Gamma Distribution I am having trouble on proofing the characteristic function of Variance Gamma distribution.
The VG model is obtained from the normal distribution by mixing on the variance parameter. Let $R_t$ be the return and suppose the distribution of  $\log(R_t)$ is normal with mean $\mu$ and a random variance $\sigma^2V$. Both $\mu$ and  $\sigma^2$ are known constants. With the distribution of $V$ is taken to be gamma with $c$ and $\gamma$ as the parameters. So the density function of $V$ is
$$g(v)=\frac{c^\gamma v^{\gamma-1}e^{-cv}}{\Gamma(\gamma)}$$
If $X=\log(R_t)-\mu$, then according to Madan and Seneta (1990) the density of $X$ is:
$$f(x)=\int_0^\infty[{e^{-x^2/2\sigma^2v}}/(\sigma\sqrt{2v\pi})]g(v)\,dv$$
It is told in the journal that there is no closed form for $f(x)$. However, the characteristic function of $X$ has the closed form by conditioning on $V$, given by $$\phi{_X}(u)=[1+(\sigma^2v/m)(u^2/2]^{-m^2/v}$$
where $m=\gamma/c$ and $v=\gamma/c^2$ are both mean and the variance of $g(v)$ respectively.
I have tried to do it by several integration techniques, but stuck on the double integral calculation and the complex integration. One of my friend suggest me to solved it using Maple software, but we still don't know which integration method that could solve this problem.
 A: Note that $$X \mid V \sim \operatorname{Normal}(0,\sigma^2 V).$$  Thus, $$\varphi_X(t) = \operatorname{E}[e^{itX}] = \operatorname{E}[\operatorname{E}[e^{itX} \mid V]] = \operatorname{E}[\varphi_{X \mid V}(t)] = \operatorname{E}[e^{-\sigma^2 t^2 V/2}] = M_V(\sigma^2 t^2/2),$$ where $M_V$ is the moment generating function of $V$.  For a gamma distribution with shape $\gamma$ and rate $c$, we have $$M_V(t) = (1 - t/c)^{-\gamma},$$ hence $$\varphi_X(t) = \left(1 + \frac{\sigma^2 t^2}{2c} \right)^{-\gamma}.$$  When expressed in terms of the mean $m = \gamma/c$ and variance $v = \gamma/c^2$ of $V$, we immediately obtain the given result, since $\gamma = m^2/v$ and $c = m/v$.  If you are hoping to obtain this result through direct integration, I would advise against it, as it is unnecessarily tedious.  But if you insist, we would need to evaluate $$\varphi_X(t) = \int_{x=-\infty}^\infty \int_{v=0}^\infty e^{itx} \frac{e^{-x^2/(2\sigma^2 v)}}{\sqrt{2\pi v} \sigma} \frac{c^\gamma v^{\gamma - 1} e^{-cv}}{\Gamma(\gamma)} \, dv \, dx.$$  This is accomplished by interchanging the order of integration and evaluating first $$\int_{x=-\infty}^\infty \frac{1}{\sqrt{2\pi}} \exp\left( itx - \frac{x^2}{2\sigma^2 v} \right) \, dx = \sigma \sqrt{v} e^{-\sigma^2 t^2 v/2}, \quad \sigma, v > 0.$$  Then the resulting integral with respect to $v$ is $$\frac{c^\gamma}{\Gamma(\gamma)} \int_{v = 0}^\infty v^{\gamma-1} e^{-(c - \sigma^2 t^2/2)v} \, dv = c^\gamma\left(c + \frac{\sigma^2 t^2}{2}\right)^{-\gamma},$$ and the result follows.  But again, why?  This is simply reproducing the work of deriving a Normal characteristic function and a Gamma MGF.
