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I have exam tomorrow will you help me to solve this problem

  1. Given $Y_n$ are random variables with characteristic function $T_n$, show they are weakly converging to zero iff there is a $\delta>0$ so that $T_n(t)\to1$ for $|t|<\delta$.
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    This is a consequence of the continuity theorem for characteristic functions, so I assume you may not use that in your proof. What can you use?2011-04-26

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To prepare yourself for the proof, you should sketch the function $\psi(z)= 1-\sin(z)/z$. This is a non-negative, bounded, continuous function and $\psi(z)=0$ only at $z=0$. Most importantly, $\psi$ is bounded away from zero outside any neighborhood of the origin. Intuitively, if the average $\mathbb{E}(\psi(Y))$ is small then $Y$ must be concentrated near zero.

More rigorously, we first note that the definition of characteristic function and Fubini's theorem give us
$ {1\over 2\delta}\int_{-\delta}^\delta (1-T_n(t))\, dt = \mathbb{E}(\psi(\delta Y_n)).$

Let $\delta>0$ be as in the problem statement, and fix $\varepsilon>0$. Then $c=\inf(\psi(\delta y): |y|>\varepsilon)$ is strictly positive and $ {1\over 2\delta}\int_{-\delta}^\delta (1-T_n(t))\, dt \geq c\, \mathbb{P}(|\delta Y_n|>\varepsilon).$ The left hand side goes to zero as $n\to\infty$, and thus so does the right hand side.

This shows that $\delta Y_n\to 0$ in probability as $n\to\infty$, and hence also $Y_n$.