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We denote by $\phi$ the characteristic function of the random variable $X1_A/\mathbb{P}(A)$. Then we have $$\phi(t)=\mathbb E\left(\mathbb 1_A\exp\left(it\frac X{\mathbb{P}(A)}  \right)\right)+1-\mathbb{P}(A).$$ I'm not sure that this can be simplified more or written as a function of the characteristic function of $X$ and the indicator function of $A$. ...


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Let $Z=g(Y,X_1,X_2 \cdots)=\sum_{i=1}^Y X_i$. The characteristic function is $$\Phi_Z(t)= E[\exp (i t Z)]=E[E[\exp (i t Z)|Y]]$$ where the later is a consecuence of the tower property: $E[g(X,Y)]=E[E[g(X,Y)|Y]]$ Then $$\Phi_Z(t)=E[ E[\prod_{i=1}^Y \exp{( i t X_i)}|Y]]=E[ \Phi_X(t)^Y]$$ where $\Phi_X(t)$ is the CF of a Poisson (we've used here the ...


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Your proof is nearly perfect. Simply note that $$ \Vert \phi - \chi \Vert \geq \max \{ |\phi(v_1) - \chi(v_1)|, |\phi(v_2) - \chi(v_2)|\}\\ = \max\{ |c_0 - 0|, |c_0 - 1|\} \geq 1/2. $$ Here, we used that $v_1 \in \Bbb{Q}$ and hence $\chi(v_1) = 0$ and $v_2 \in \Bbb{R}\setminus\Bbb{Q}$, hence $\chi(v_2) = 1$ and we also used $\phi(v_1) = \phi(v_2) = c_0$ ...


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By Lévy's truncation inequality, $$\mathbb{P}(|X_n| \geq R) \leq 7R \int_0^{\frac{1}{R}} |\text{Re}(1-\phi_n(t))| \, dt.$$ Using the given estimate for the characteristic functions, we find $$\mathbb{P}(|X_n| \geq R) \leq c \cdot \frac{1}{R^{1+\delta}}$$ for some constant $c=c(A,\delta)$. Since $$\mathbb{E}(|X_n|^p) = \int_{0}^{\infty} \mathbb{P}(|X_n| ...


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First, we have to show that the second moment is finite. Notice that for any $x$, $2n^2(1-\cos(x/n))\to x^2$, hence $$\mathbb E[X^2]=\mathbb E\left[\liminf_{n\to \infty}2n^2(1-\cos(X/n))\right].$$As a consequence of Fatou's lemma, we obtain that $$\mathbb E[X^2]\leqslant \liminf_{n\to \infty}2n^2\mathbb E\left[(1-\cos(X/n))\right].$$ Since $$2\mathbb ...


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The minimal and characteristic polynomials of (a square matrix) $A$ are both invariant under transposition of$~A$. For the characteristic polynomial this is because the determinant is invariant, and for the minimal polynomial because clearly $P[A^T]=P[A]^T$ for every (complex) polynomial $P$. So instead of $T^*$ you can work with its transpose $\overline ...


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I am not quite sure about how to use the uniform continuity to prove this conclusion. Then don't. Maybe we should apply some techniques in Fourier analysis. Hmmm... And maybe we should look for explicit simple counterexamples? Try $X$ such that $P(X=1)=1$, and $s=0$, $t\to0$.


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This is an old question, but I still like to write an answer. First note that for your c.f. it holds that $\phi_x(\omega)=\phi_x(\omega+2\pi)$, indicating that your random variable takes only value in $\mathbb{Z}$ and hence is of lattice type for which a simple inversion formula is $f_X(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i \omega ...


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As David Mitra suggests in the comments, Apply Littlewood's first principle to find a finite collection of open intervals whose union $U$ satisfies $\mu(U\Delta E)<\varepsilon$. Take $F$ to be $(U\Delta E)^C$ and define the step function to be $1$ on $U$ and $0$ off $U$.


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As a consequence of Taylor's formula, we have for each $x$ the inequality $$\left|e^{ix}-1+ix-\frac{x^2}2\right|\leqslant\min\left\{\frac{|x|^3}6,\frac{2x^2}2\right\}.$$ Using this with $x=u\cdot X(\omega)$, we have \begin{align} |\varphi(u)|&\leqslant \mathbb E\left[1-iX+\frac{X^2}2\right]+ \mathbb ...


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Some hints: expressing $\phi$ by an integral involving $f'$, we obtain that $$t\cdot |\phi(t)|\leqslant \left|\int_{-\infty}^{\infty}f'(x)e^{itx}\mathrm dx\right|;$$ the Fourier transform of an integrable function vanishes at infinity. To see this, approximate the function by characteristic functions of finite unions of intervals.



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