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3

Considering that $\phi(0)=1$ and that $\phi_k(0)=1$ for every $k$, one sees that the condition $$\sum_kp_k=1$$ is necessary. To prove that is is also sufficient and to avoid most of the technicalities, consider some random variables $N$ and $(X_k)$ defined on the same probability space, $N$ independent of $(X_k)$, each $X_k$ with characteristic function ...


5

Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ...


0

I think it is best to use the definition of weak convergence, denoted "$\Rightarrow$": $\mu_n \Rightarrow \mu$ if and only if $\int f d\mu_n \rightarrow \int f d\mu$ for all bounded and continuous $f$. If $x_n \rightarrow x$ and $f \in C_b(\mathbb{R})$, then \begin{equation} \int f(y) \delta_{x_n}(y) = f(x_n) \rightarrow f(x) = \int f(y) \delta_x(y), ...


0

Let $F,F_n$ denote the CDF's of probability measures $\delta_{x}$ and $\delta_{x_n}$. These are the characteristic functions of the sets $[x,\infty)$ and $[x_n,\infty)$. Then: $$\delta_{x_{n}}\stackrel{w}{\rightarrow}\delta_{x}\iff\forall y\neq x\; F_{n}\left(y\right)\rightarrow F\left(y\right)$$ This because $F$ is continuous at $y$ if and only if $y\neq ...


0

I'll consider a generalization. First, observe that $\phi(t):=\exp(-2 z\sin^2 t)=e^{-z}\exp(z \cos 2t)$ is a $\pi$-periodic function and so has a Fourier series. (The question above is the case of $z=1$.) To be explicit, recall the Jacobi-Anger expansion of $e^{z \cos t}$ (see equation 10.35.3 in the DLMF) to write $\phi(t)=\sum\limits_{k=-\infty}^\infty ...


-1

Actually the answer was MUCH more complicated, for anyone interested it is possible to find a proof in Sato, Lèvy processes and infinitely divisible distributions, in chapter 2


1

In general, the continuity of a positive definite function at zero implies continuity everywhere. See for example Section 2 of Roger Horn's paper ``QUADRATIC FORMS IN HARMONIC ANALYSIS AND THE BOCHNER-EBERLEIN THEOREM'', Proc. AMS, 1975. I believe your result is a special case.


1

Denoting $\varphi\left(t\right)=\mathbb{E}e^{itX_{1}}$ and assuming that $N\sim Poisson\left(\lambda\right)$ : $\mathbb{E}\left[e^{it\sum_{j=1}^{N}X_{j}}\right]=\sum_{n=0}^{\infty}\mathbb{E}\left[e^{it\sum_{j=1}^{N}X_{j}}\mid ...


2

Suppose $X_i$'s characteristic function is $f(t) = E(e^{itX_i})$, then \begin{align} E(e^{it\sum_{i=1}^N X_i}) &= E(E(e^{it\sum_{i=1}^N X_i}|N)) \\ & = E(\prod_{i=1}^NE(e^{itX_i})) \\ & = E(f^N(t)) \\ & = e^{-\lambda}\sum_{k=0}^{+\infty}\dfrac{\lambda^k}{k!}f^{k}(t)\\ & = e^{-\lambda}e^{\lambda f(t)} \\ & = e^{(f(t)-1)\lambda} ...



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