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0

This approach can work, we have to use an expression of cumulative distribution function. There is an approach bases on the expression of the characteristic function: we have $\varphi(t)=\sum_{n\in\mathbb Z}\mathbb P\{X=n\}e^{itn}.$ Since the series $\sum_{n\in\mathbb Z}\mathbb P\{X=n\}$ is convergent, we can switch the series and the integral. This has ...

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It's probably better to use the cumulative distribution function, since it behaves better with respect to the maximum of an independent sequence. We have $$\mu\{n(1-M_n)\leqslant t\}=\mu\{1-t/n\leqslant M_n\}=1-(1-t/n)^n$$ and we recognize a well-known limit.

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Actually, $f$ is a simple function. To be concrete, its standard representation is $$f=-\chi_{[-1,0)}+\chi_{[1,2)}+2\,\chi_{[2,3)}+3\,\chi_{[3,3]}.$$ And hence its Lebesgue integral is \begin{align*} \int_{\mathbb{R}} f\,\mathrm{d}\mu&=-1\cdot \mu([-1,0))+1\cdot \mu([1,2))+2\cdot \mu([2,3))+3\cdot \mu([3,3])\\ &=-1\times 1+1\times 1+2\times ...

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Direct evaluation of the characteristic function for $X \sim N(0,1)$ Observing that the expected value of a measurable function of $X$, $g(X)$, given that $X$ has a probability density function $f_X(x)$, is given by the inner product of $f_X$ and $g$: $$\Bbb{E}(g(X)) = \int_{-\infty}^\infty g(x) f_X(x)\, \operatorname{d}x .$$ evaluating for $X\sim ... 1 Hint: Use theorem 3.3.9 followed by 3.3.8 in Durrett's text. (I assume you are working from this text, as your last two questions are directly from it) 1 We can avoid a further use of Taylor expansion using the inequality $$-x-\frac{x^2}2\leqslant \ln(1-x)\leqslant -x,\quad 0\leqslant x\lt 1.$$ The notation "o" can be replaced by an integral remainder, thought it's not a problem. 0 Hint: the characteristic function of$X_1+\dots+X_n$is indeed$\left(\frac{\sin t}t\right)^n$. The inversion formula for an integrable characteristic function ($n\geqslant 2$) will give the result (cut the integral$\int_{-\infty}^0+\int_0^\infty$and use the substitution$s=-t\$ in the first integral).

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