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Thread: [SOLVED] "Easy" Convergence Problems

  1. #1
    Mar 2008
    Acolman, Mexico

    [SOLVED] "Easy" Convergence Problems

    Hello, this is supposed to be an easy problem, but I am lost.

    Let $\displaystyle X_{n}$ be a sequence of random variables with

    $\displaystyle p_{n}(x) = \left\{ %always put [\] {
    1,& x=2+\frac{1}{n}\\
    0,& elsewhere

    Why $\displaystyle \lim_{n\rightarrow \infty} p_{n}=0$ for all values of $\displaystyle x$? is it because x just approaches 2? how would you state that more formally?

    Also, I don't get this.
    The cdf of $\displaystyle X_n$ is
    F_{n}(x) = \left\{
    0,& x < 2+\frac{1}{n}\\
    1,& x \geq 2+\frac{1}{n}

    Why $\displaystyle \lim_{n \rightarrow \infty} F_{n}(x) = \left\{
    0,& x \leq 2\\
    1,& x > 2
    \right.$ ?? why the inequalities change?

    Thanks in advance.
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  2. #2
    Jul 2008
    For the first part:

    pick an x. The sequence $\displaystyle p_n(x)$ is non-zero at most 1 time. So there exists some m such that $\displaystyle p_n(x)=0$ for $\displaystyle n>m$. So the tail of the sequence converges to 0, since it is identically 0, and thus so does the whole sequence.

    For the second part, look at the limit in 3 cases: $\displaystyle x<2,\;x=2,\;x>2$

    For $\displaystyle x<2:\,F_n(x)\rightarrow 0$ since $\displaystyle F_n(x)=0$ for all n.

    For $\displaystyle x>2:\,F_n(x)\rightarrow 1$ since again for a given x there is some m at which $\displaystyle F_n(x)=1$ for all n>m.

    For $\displaystyle x=2:\,F_n(x)\rightarrow 0$ since $\displaystyle F_n(x)=0$ for all n.
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