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Thread: Some problems

  1. #1
    Member
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    Jan 2006
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    Gdansk, Poland
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    Some problems

    1

    Prove that if $\displaystyle X_n \rightarrow X$ in probability than $\displaystyle X_n \Rightarrow X$ (in distribution).

    My approach:

    $\displaystyle F_{X_n}(t) = P\{X_n \le t \} = $
    $\displaystyle = P\{X_n \le t,\quad X \le t+\epsilon \} + P\{X_n \le t, \quad X > t+\epsilon \} $
    $\displaystyle \le P\{X \le t+\epsilon \} + P\{X_n \le t, \quad X > t+\epsilon \}$

    Because $\displaystyle X_n \rightarrow X$ in probability $\displaystyle \lim_{n \rightarrow \infty} P\{X_n \le t, \quad X > t+\epsilon \} = 0 $

    Thus:
    $\displaystyle \limsup_{n \rightarrow \infty}F_{X_n}(t) \le F_{X}(t+\epsilon)$

    Hence: $\displaystyle \limsup_{n \rightarrow \infty}F_{X_n}(t) \le F_{X}(t)$

    I was unable to prove the other inequality unless X is absolutly continuous.
    Does anybody know how to do that?


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    2.

    Let $\displaystyle Y_n$ be random variables with characteristic functions $\displaystyle \phi_n$, $\displaystyle n = 1, 2,...$
    Prove that $\displaystyle Y_n \Rightarrow 0$ (in distribution) if there exists $\displaystyle \delta > 0$ such that $\displaystyle \lim_{n\rightarrow\infty} \phi_n(t) = 1$ for $\displaystyle |t| < \delta$.

    I was unable to solve it. The only reasonable idea I had was to assume that all moments of the variables exist and prove thay all converge to 0 (than convergense of distribution follows, I think)

    Can you help me with this one too?
    Last edited by albi; Jun 9th 2008 at 09:47 AM. Reason: Error in proof
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  2. #2
    Member
    Joined
    Jan 2006
    From
    Gdansk, Poland
    Posts
    117

    1st problem

    Finally I solved the first one. I thought $\displaystyle X_n \Rightarrow X$ (in distribution) means that $\displaystyle F_{X_n}(t) \rightarrow F_X(t)$ FOR ALL t.

    While it is the real definition is $\displaystyle F_{X_n}(t) \rightarrow F_X(t)$ FOR ALL t WHERE F_X is CONTINUOUS.

    So the solution is:

    $\displaystyle F_{X}(t - \epsilon) = P\{X \le t - \epsilon \} =$
    $\displaystyle = P\{X \le t-\epsilon,\quad X_n \le t \} + P\{X \le t-\epsilon, \quad X_n > t \}$
    $\displaystyle \le P\{X_n \le t \} + P\{X \le t-\epsilon, \quad X_n > t \}$

    Because $\displaystyle X_n \rightarrow X$ in probability $\displaystyle \lim_{n \rightarrow \infty} P\{X \le t-\epsilon, \quad X_n > t \} = 0$

    Thus:
    $\displaystyle \liminf_{n \rightarrow \infty}F_{X_n}(t) \ge F_{X}(t-\epsilon)$

    Hence (FX is continuous in t): $\displaystyle \liminf_{n \rightarrow \infty}F_{X_n}(t) \ge F_{X}(t)$

    Finally

    $\displaystyle \limsup_{n \rightarrow \infty}F_{X_n}(t) \le F_{X}(t) \le \liminf_{n \rightarrow \infty}F_{X_n}(t) $

    Which solves the problem!

    The second one is still unsolved!
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