Composition of Characteristic Function / Fourier Transformation

Question

I have to prove that if $\Phi$ is a characteristic function, then $\psi(t)=e^{\lambda (\Phi (t) - 1)}$ is also a characteristic function.

I guess I'm supposed to use Bochner-Khinchin's theorem, showing that $\psi$ is a positive-semidefinite function, but I'm not sure how.

Also, $\psi(t)=h(\Phi (t))$ where $h(x)=e^{\lambda(x-1)}$ and $h$ is continuous. Is there an assertion about the composition of characteristic functions with continuous or convex functions?

Answer 1

Recall two statements; the first one is obvious from the very definiton of positive definiteness:

Lemma 1: Let $\Phi_1$ and $\Phi_2$ be positive definite functions and $\lambda \geq 0$ a constant. Then $\lambda \Phi_1$ and $\Phi_1+\Phi_2$ are positive definite functions.

Lemma 2: For any characteristic function $\Phi$ and any $n \in \mathbb{N}$, $\Phi^n$ is a characteristic function; in particular $\Phi^n$ is positive definite.

Proof of Lemma 2: Since $\Phi$ is a characteristic function, there exists a random variable $X$ such that $\Phi(t) = \mathbb{E}e^{i tX}$. Without loss of generality, we may assume that there exists an independent random variable $Y$ such that $Y \sim X$ (otherwise we enlarge the probability space using a product construction). Then

$$\mathbb{E}e^{i t (X+Y)} = \mathbb{E}e^{i t X} \mathbb{E}e^{i t Y} = \Phi(t)^2$$

which shows that $\Phi^2$ is a characteristic function. By induction, we find that $\Phi^n$ is a characteristic function for all $n \in \mathbb{N}$; hence, by Bochner's theorem, positive definite. This finishes the proof.

Now use these two statements to prove the assertion:

1. Prove that for each $n \in \mathbb{N}$, $$\phi_n(t) := \sum_{k=0}^n \frac{1}{k!} (\lambda \Phi(t))^k$$ is positive definite.
2. Using that the pointwise limit of positive definite functions is positive definite (provided the limit exists), show that $\phi(t) = e^{\lambda \Phi(t)}$ is positive definite.
3. Applying Bochner's theorem conclude that $\psi(t)=e^{\lambda (\Phi(t)-1)}$ is a characteristic function.

Answer 2

Let $\{X_i, i\ge 1\}$ be i.i.d. rvs with common characteristic function $\Phi(t)$ and $Y$ be a independent Poisson distributed r.v with $E(Y)=\lambda$, then the characteristic function of $Z=\sum\limits_{i=1}^Y X_i$ is the following, $$ E[e^{itZ}]=E[E[e^{it\sum_{i=1}^YX_i}\mid Y]]=\sum_{k=1}^\infty E[e^{it\sum_{i=1}^kX_i}]P(Y=k)=\sum_{k=1}^\infty[\Phi(t)]^k\frac{\lambda^k}{k!}e^{-\lambda} = e^{\lambda(\Phi(t)-1)}.$$