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Thread: Convergence in Probability Questions

  1. #1
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    Convergence in Probability Questions

    I am stuck with this questions. I am supposed to use the following definition to prove them:
    $\displaystyle X_n \stackrel{\mbox{P}}{\longrightarrow} X $ if $\displaystyle \forall \epsilon > 0$

    $\displaystyle \lim_{n \to \infty} P[|X_n-X| \geq \epsilon]=0$
    or equivalently
    $\displaystyle \lim_{n \to \infty} P[|X_n-X| < \epsilon]=1$

    1. Prove the following. Suppose $\displaystyle X_n \stackrel{\mbox{P}}{\longrightarrow} a $ and $\displaystyle g$ is a real function continuous at $\displaystyle a$. Then $\displaystyle g(X_n) \stackrel{\mbox{P}}{\longrightarrow} g(X) $

    2. Let $\displaystyle \{a_n\}$ be a sequence of real numbers. Hence, we can also say that $\displaystyle \{a_n\}$ is a sequence of constant (degenerate) random variables. Let $\displaystyle a$ be a real number. Show that $\displaystyle a_n \longrightarrow a$ is equivalent to $\displaystyle a_n \stackrel{\mbox{P}}{\longrightarrow} a $
    (is this proving that convergence in probability is equivalent as pointwise converge in sequences??)

    Thanks in advance.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by akolman View Post
    I am stuck with this questions. I am supposed to use the following definition to prove them:
    $\displaystyle X_n \stackrel{\mbox{P}}{\longrightarrow} X $ if $\displaystyle \forall \epsilon > 0$

    $\displaystyle \lim_{n \to \infty} P[|X_n-X| \geq \epsilon]=0$
    or equivalently
    $\displaystyle \lim_{n \to \infty} P[|X_n-X| < \epsilon]=1$

    1. Prove the following. Suppose $\displaystyle X_n \stackrel{\mbox{P}}{\longrightarrow} a $ and $\displaystyle g$ is a real function continuous at $\displaystyle a$. Then $\displaystyle g(X_n) \stackrel{\mbox{P}}{\longrightarrow} g(X) $
    Remember the definition (or property) of the continuity :
    f is continuous $\displaystyle \Leftrightarrow \forall \delta>0, \exists \epsilon>0,~ \forall x,~ |x-c|<\epsilon \Rightarrow |f(x)-f(c)|<\delta$
    (I inverted delta and epsilon from the usual definition so that it wouldn't confuse with the epsilon you're using in the text)

    This means that the set : $\displaystyle \{\omega ~:~ |X_n(\omega)-a|< \epsilon \} \subseteq \{\omega ~:~ |g(X_n(\omega)-g(a)|< \delta \}$


    One of the consequences of the Kolmogorov axiomatic is that if $\displaystyle A \subseteq B$, then $\displaystyle \mathbb{P}(A) \leqslant \mathbb{P}(B)$

    Then use the second axiom of Kolmogorov ($\displaystyle \mathbb{P}(A)\leqslant 1$) to prove that $\displaystyle \lim_{n \infty} \mathbb{P}(|g(X_n)-g(a)|<\delta)=1$

    It should be easy from here


    (is this proving that convergence in probability is equivalent as pointwise converge in sequences??)
    Yes.

    I haven't been thinking very long on it, but my first idea would be, as above, to use the epsilon definition of $\displaystyle a_n \to a$
    Last edited by Moo; Feb 25th 2009 at 11:27 AM.
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  3. #3
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    Thank you very much. I got confused with the epsilons.
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  4. #4
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    I am sorry to bother you again, but I don't get this crucial step.

    Quote Originally Posted by Moo View Post
    Hello,

    Remember the definition (or property) of the continuity :
    f is continuous $\displaystyle \Leftrightarrow \forall \delta>0, \exists \epsilon>0,~ \forall x,~ |x-c|<\epsilon \Rightarrow |f(x)-f(c)|<\delta$
    (I inverted delta and epsilon from the usual definition so that it wouldn't confuse with the epsilon you're using in the text)

    This means that the set : $\displaystyle \{\omega ~:~ |X_n(\omega)-a|< \epsilon \} \subseteq \{\omega ~:~ |g(X_n(\omega))-g(a)|< \delta \}$
    Why is the set $\displaystyle \{\omega ~:~ |X_n(\omega)-a|< \epsilon \} \subseteq \{\omega ~:~ |g(X_n(\omega))-g(a)|< \delta \}$??

    Thanks in advance.

    EDIT: typo corrected.
    Last edited by akolman; Feb 25th 2009 at 08:37 PM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    Typo....with $\displaystyle g(x_n(\omega))$
    Quote Originally Posted by moo View Post
    hello,

    remember the definition (or property) of the continuity :
    F is continuous $\displaystyle \leftrightarrow \forall \delta>0, \exists \epsilon>0,~ \forall x,~ |x-c|<\epsilon \rightarrow |f(x)-f(c)|<\delta$
    (i inverted delta and epsilon from the usual definition so that it wouldn't confuse with the epsilon you're using in the text)

    this means that the set : $\displaystyle \{\omega ~:~ |x_n(\omega)-a|< \epsilon \} \subseteq \{\omega ~:~ |g(x_n(\omega))-g(a)|< \delta \}$


    one of the consequences of the kolmogorov axiomatic is that if $\displaystyle a \subseteq b$, then $\displaystyle \mathbb{p}(a) \leqslant \mathbb{p}(b)$

    then use the second axiom of kolmogorov ($\displaystyle \mathbb{p}(a)\leqslant 1$) to prove that $\displaystyle \lim_{n \infty} \mathbb{p}(|g(x_n)-g(a)|<\delta)=1$

    it should be easy from here (rofl)



    yes.

    I haven't been thinking very long on it, but my first idea would be, as above, to use the epsilon definition of $\displaystyle a_n \to a$ (nod)
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  6. #6
    Moo
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    Quote Originally Posted by matheagle View Post
    Typo....with $\displaystyle g(x_n(\omega))$
    Why is it a typo ?
    Or more exactly... where is the typo ? ^^

    Quote Originally Posted by akolman View Post
    I am sorry to bother you again, but I don't get this crucial step.

    Why is the set $\displaystyle \{\omega ~:~ |X_n(\omega)-a|< \epsilon \} \subseteq \{\omega ~:~ |g(X_n(\omega))-g(a)|< \delta \}$??

    Thanks in advance.

    EDIT: typo corrected.
    Okay, so imagine there is an $\displaystyle \omega$ such that $\displaystyle |X_n(\omega)-a|< \epsilon$
    The implication means that with that same $\displaystyle \omega$, $\displaystyle |g(X_n(\omega))-g(a)|<\delta$

    Thus the set of $\displaystyle \omega$ such that $\displaystyle |X_n(\omega)-a|< \epsilon$ is included in the set of $\displaystyle \omega$ such that $\displaystyle |g(X_n(\omega))-g(a)|<\delta$.
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  7. #7
    MHF Contributor matheagle's Avatar
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    It was a minor typo, you left out the second ).
    I thought I pointed that out.
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  8. #8
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    Thanks again Moo.

    I'm still halfway with this problem

    2. Let $\displaystyle \{a_n\}$ be a sequence of real numbers. Hence, we can also say that $\displaystyle \{a_n\}$ is a sequence of constant (degenerate) random variables. Let $\displaystyle a$ be a real number. Show that $\displaystyle a_n \Rightarrow a$ is equivalent to $\displaystyle a_n \stackrel{\mbox{P}}{\Rightarrow} a $

    I was able to prove that

    Show that $\displaystyle a_n \longrightarrow a$ $\displaystyle \Rightarrow$
    $\displaystyle a_n \stackrel{\mbox{P}}{\longrightarrow} a $

    But I don't know what to do to go the other way around.

    If I assume $\displaystyle a_n \stackrel{\mbox{P}}{\longrightarrow} a $.

    Then, $\displaystyle \forall \epsilon_{1}>0, \forall \epsilon_{2}, \exists N>0,~ \forall n > N, ~P(|a_n-a|<\epsilon_{1}) < \epsilon_{2}$

    Is there a way to make $\displaystyle P(|a_n-a|<\epsilon_{1})=0$?

    So I can get $\displaystyle |a_n-a|<\epsilon_{1}$, for $\displaystyle n>N^{*},~ \forall \epsilon_{1}>0$. Can I get a $\displaystyle N^{*}$ that makes this happen?

    Or how can I go with this proof. Thanks again for your patience.
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