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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} ...


2

Yes. The characteristic function $\phi(s) = {\mathbb E}[\exp(i s\cdot X)]$ is the complex conjugate of the Fourier transform of the density, and Fourier transform is one-to-one on $L^1$. If it's real analytic, $\phi(s)$ is determined by the coefficients of its series expansion at $s=0$, which are the moments.


0

Not only this one but also we have: For subsets A and B of a non empty set X: $$χ_ϕ=0 and χ_X=1$$ $$A⊆B ⇔ χ_A≤χ_B$$ $$χ_{A^c }=1-χ_A$$ $$χ_{A∩B}=χ_A χ_B=min(χ_A,χ_B )$$ $$χ_{A∪B}=χ_A+χ_B-χ_{A∩B}=max(χ_A,χ_B )$$ $$χ_{A-B}=χ_A-χ_{A∩B}.$$ $$χ_{A△B}=χ_A+χ_B-2χ_{A∩B}$$ $$χ_{A×B}=χ_A χ_B$$


2

Your reasoning is perfectly valid. The given answer is wrong and your answer is correct. One easily checks that if $x\in A\cap B$ then $\chi_C(x)=1+1-1=1$, so $A\cap B\subseteq C$ contrary to what the given answer suggests. Indeed, the characteristic function of $A\triangle B$ is $$\begin{array}{ll} \chi_{A\triangle B}(x) & ...



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