$\Bbb{E}(e^{\lambda X})=e^{\lambda^2/2}$ impliesthat $\Bbb{E}(e^{zX})$ is holomorphic and well-defined.

Question

I am reading Brownian Motion, Martingales, and Stochastic Calculus. The author starts with Gaussian Random Variables.

He say that

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The complex Laplace transform of X is then given by $\Bbb{E}(e^{zX})=e^{z^2/2}$ for all $z\in \Bbb{C.}$ To get this formula (and also to verify that the complex Laplace transform is well defined), consider first the case when $z=\lambda\in \Bbb{R}$: $$\Bbb{E}(e^{\lambda X})=\ldots=e^{\lambda^2/2}$$

This calculation ensures that $\Bbb{E}(e^{zX})$ is well-defined for $z\in \Bbb{C}$, and defines a holomorphic function on $\Bbb{C}$.

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Question:

Does it means that if we consider real and imaginary parts separately then it follow that $\Bbb{E}(e^{zX})$ is well-defined and as the result for both real and imaginary parts is a holomorphic function it follow that $\Bbb{E}(e^{zX})$ is holomorphic ?

Answer 1

A short answer. Both real and imaginary parts of the m.g.f. are well-defined. But they are not holomorphic separately.

A longer answer. Let $X$ be a $\mathbb{R}$-valued random variable such that its m.g.f. $M_X(s) = \mathbb{E}[e^{sX}]$ converges for $s \in I$ for some open interval $I$ containing $0$. Then utilizing the fact that $|e^z| = e^{\operatorname{Re}(z)}$, on the strip $I \times \mathbb{R}$ we have

$$\forall z = s + i\xi \in I \times \mathbb{R} \ : \qquad \mathbb{E}\left[\left|e^{zX}\right|\right] = \mathbb{E}\left[e^{sX}\right] = M_X(s) < \infty$$

and thus $M_X$ is well-defined on $I \times \mathbb{R}$. Next, for any closed piecewise $C^1$ curve $\gamma : [0, 1] \to I \times \mathbb{R}$,

\begin{align*} \int_{0}^{1} \mathbb{E}\left[ \left| e^{\gamma(t) X} \gamma'(t) \right| \right] \, dt &\leq \int_{0}^{1} \mathbb{E}\left[ e^{\operatorname{Re}(\gamma(t)) X} |\gamma'(t)| \right] \, dt \\ &\leq \operatorname{length}(\gamma) \cdot \sup_{z \in \gamma}M_X(\operatorname{Re}(z)) < \infty. \end{align*}

This allows us to apply the Fubini's theorem to obtain

$$ \int_{\gamma} M_X(z) \, dz = \mathbb{E}\left[ \int_{\gamma} e^{zX} \, dz \right] = 0. $$

So by the Morera's theorem, $M_X$ is holomorphic on $I \times \mathbb{R}$.