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Math Help - [SOLVED] "Easy" Convergence Problems

  1. #1
    Member
    Joined
    Mar 2008
    From
    Acolman, Mexico
    Posts
    118

    [SOLVED] "Easy" Convergence Problems

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

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

    p_{n}(x) = \left\{  %always put [\] {<br />
\begin{array}{cl}<br />
1,& x=2+\frac{1}{n}\\<br />
0,& elsewhere<br />
\end{array}<br />
\right.

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

    Also, I don't get this.
    The cdf of X_n is
    <br />
F_{n}(x) = \left\{<br />
\begin{array}{cl}<br />
0,& x < 2+\frac{1}{n}\\<br />
1,& x \geq 2+\frac{1}{n}<br />
\end{array}<br />
\right.

    Why \lim_{n \rightarrow \infty} F_{n}(x) = \left\{<br />
\begin{array}{cl}<br />
0,& x \leq 2\\<br />
1,& x > 2<br />
\end{array}<br />
\right. ?? why the inequalities change?

    Thanks in advance.
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  2. #2
    Member
    Joined
    Jul 2008
    Posts
    138
    For the first part:

    pick an x. The sequence p_n(x) is non-zero at most 1 time. So there exists some m such that p_n(x)=0 for 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: x<2,\;x=2,\;x>2

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

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

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