# Tag Info

1

The first equality that you don't understand is indeed a MacLaurin series expansion of the characteristic function, using the link between the moments of a random variable and the derivatives of the characteristic function. See https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29#Moments . So Shalop is right, the $\sigma$ is a ...

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This is a weighted (with positive weights and sum $1$) sum of functions of the form $t\mapsto e^{iat}$, where $a$ is a real number. If $\varphi_X(t)=\sum_{j=1}^np_je^{ia_jt}$ where $p_j\geqslant 0$ and $\sum_{j=1}^np_j=1$, then $\mathbb P(X=a_j)=p_j$ for each $j\in \{1,\dots,n\}$ (the converse is also true).

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It should be \begin{align} \int_{\mathbb R^2} f(x,y)&=\int_{\mathbb R}\left(\int_{\mathbb R}\chi_{(A-x) \cap B}(y)dy\right)dx \\ &=\int_{\mathbb R}\left(\int_{B}\chi_{(A-x) }(y)dy\right)dx \\ &=\int_{\mathbb R}\left(\int_{B}\chi_{A }(y+x)dy\right)dx \\ &=\int_{B}\left(\int_{\mathbb R}\chi_{A }(y+x)dx\right)dy \\ &=\int_{B}m(A)dy \\ ...

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You are correct, here is a hint to solve the integral $$\int e^{ax} x^{n} = \frac{e^{ax}}{a^{n+1}} \left[ (ax)^n - n(ax)^{n-1}+ n(n-1)(ax)^{n-2} - \dots + (-1)^nn! \right]$$ and your final result should be $$\frac{(-1)^n}{(e^{it}-2)^n}$$

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Let $X,Y$ be iid Cauchy random variables. Then $$\mathbb P((X,Y)\in A)=\frac{1}{\pi^2}\int_{(x,y)\in A}\frac{dx\ dy}{(1+x^2)(1+y^2)}.$$ Let $U=XY$. It follows that $$\mathbb P(U\in B)=\frac{1}{\pi^2}\int_{xy\in B}\frac{dx\ dy}{(1+x^2)(1+y^2)}.$$ Let $u=xy$. For fixed $x>0$, we have $y=u/x$ and therefore $dy=du/x$. Since the distribution of $X$ is ...

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You must employ a Lebesgue integration as $\chi_C$, the characteristic function of $C$, is not classically or even Riemann integrable because its set of discontinuities,which is $C$ , is uncountable.The measure $\mu (I \backslash C)=\mu([0,1]\backslash C)= 1$ so $\mu (C)=0$ so $\chi_C (x)=0$ almost everywhere (which means except on a set of measure 0 ...

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I assume you already know what the cantor set is. In that case, the characteristic function is the function $f$ with $f(x)=1$ for $x \in C$ and $f(x)=0$ otherwise.

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You can calculate $E[\exp(zZ)]$ for any arbitrary complex number $z$. Note that all odd moments of $Z$ are $0$, while the even moments are given by $E[Z^{2k}] = \frac{(2k)!}{2^k k!}$. The power series of the exponential function and the dominated convergence theorem yield the following equality: $$E[\exp(zZ)] = \sum \limits_{l = 0}^\infty \frac{1}{l!} z^l ... 0 Hint: use A_n=\bigcup_{i=1}^nA_i. as A_n \supseteq A_i \forall i \in\{1,2,3,....,n\} 2 Let x\in X. i) if x \in \cup_n A_n, then 1_{\cup_n A_n}(x)=1. Now in this case there is k_0\in \mathbb{N} such that x\in A_k for every k>k_0 (since the sequence is increasing), so the left hand side is a constant sequence with constant value 1 for k>k_0. Now it remains to consider ii) x\notin \cup_n A_n. I suggest you try that case ... 1 Hint: Your function g(x,t) is 1 precisely when t\leq x, otherwise 0. Therefore, it is 1 precisely when x\geq t, otherwise zero. Therefore, it coincides with the function g(x,t) = \chi_{[t,\infty)}(x). Can you go on now? 1 In general, the characteristic function will not be integrable. Because if it was, it would follow (via Fourier inversion) that the PDF is actually a continuous function. So any discontinuous PDF (i.e. not a.e. to a continuous function) will do, for example 1_{[0,1]}. Conversely, a cheap criterion is the following: If the PDF f has k derivatives ... 2 Theorem(or Exercise): X be a r.v. with distri. func. F(.) and c.f. \phi(.) and suppose that E|X|^n<\infty for some integer n\geq1. THEN \forall k=1,...,n 1) \phi has uniformly continuous derivatives \phi^{(k)} and$$\phi^{(k)}(t)=i^k E[X^k.e^{itX}]  2) $E(X^k)=\frac{\phi^{(k)}(0)}{i^k}$ Partial converse: $X$ be a r.v. with distri. ...

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