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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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It's not clear what's to be shown; you'll need to write it slightly more carefully. –  joriki Apr 26 '11 at 14:20
    
show that if Yn are random variable with characteristic funtion Tn which weakly convering to zero iff there is a delta>zero so that Tn(t)-->1 for mod t<delta. –  Bapuram sapure Apr 26 '11 at 14:23
    
Is there supposed to be another question after that? Because you labelled the question with "1" and I didn't want to change that. However, if there is no other question, it looks a bit silly. –  Raskolnikov Apr 26 '11 at 14:32
    
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? –  Nate Eldredge Apr 26 '11 at 15:10

1 Answer 1

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 $Y_n$.

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