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some minor misprints
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edit approved Jun 19, 2016 at 7:15
Denis Denisov
edit approved Jun 19, 2016 at 7:15
Denis Denisov
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Let us consider the one-dimensional case. By the LevyLévy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $t\to 0$$\theta\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see that the expectation is only determined by the “big jumps term” (provided there is neither a drift term nor a BrownionBrownian motion term).

Let us consider the one-dimensional case. By the Levy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $t\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see that the expectation is only determined by the “big jumps term” (provided there is neither a drift term nor a Brownion motion term).

Let us consider the one-dimensional case. By the Lévy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $\theta\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see that the expectation is only determined by the “big jumps term” (provided there is neither a drift term nor a Brownian motion term).

Slight rephrasing, improved punctuation
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Pedro Lauridsen Ribeiro
Pedro Lauridsen Ribeiro
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Let us consider the one-dimensional case. By the Levy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $t\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see that the expectation is only determined by the “big jumps term”.  (ifprovided there is noneither a drift term andnor a Brownion motion term).

Let us consider the one-dimensional case. By the Levy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $t\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see the expectation is only determined by the “big jumps term”.(if there is no drift term and Brownion motion term)

Let us consider the one-dimensional case. By the Levy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $t\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see that the expectation is only determined by the “big jumps term”  (provided there is neither a drift term nor a Brownion motion term).

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Peisen Li
Peisen Li
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Let us consider the one-dimensional case. By the Levy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $t\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see the expectation is only determined by the “big jumps term”.(if there is no drift term and Brownion motion term)