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Thread: convergences in probability

  1. #1
    Junior Member
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    convergences in probability

    Hi

    I am having difficulty in understanding a convergence question. I was wondering if someone could please explain how I get started?

    Thanks
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  2. #2
    MHF Contributor
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    Re: convergences in probability

    two random objects converge in probability if in the limit they have the same density or distribution function.

    In this case it should be pretty clear that

    $\begin{align*}
    &\lim \limits_{n \to \infty}~P_{X_n} = \\ \\

    &\lim \limits_{n \to \infty}~\left \{ \dfrac{n-2}{n},~\dfrac 1 n,~\dfrac 1 n \right \} = \\ \\

    &\{1,~0,~0\}

    \end{align*}$

    The distribution of $\theta$ is just $P[X=\theta]=1$

    which is seen to be identical to the limiting distribution of $X_n$

    and thus $X_n$ does converge to $\theta$ in probability

    $X_n$ will converge in mean square to $\theta$ if

    $\lim \limits_{n\to\infty}~E[(X_n - \theta)^2] = 0$

    $\begin{align*}

    &\lim \limits_{n\to\infty}~E[(X_n - \theta)^2] = \\ \\

    &\lim \limits_{n\to \infty}~\left(\dfrac{n-2}{n}(\theta - \theta)^2 + \dfrac 1 n (\theta - (\theta+1))^2 + \dfrac 1 n (\theta - (\theta+n))^2\right) = \\ \\

    &\lim \limits_{n\to \infty}~ \left(\dfrac 1 n (1^2 + n^2)\right) = \\ \\

    &\lim \limits_{n\to \infty}~ n + \dfrac 1 n = \infty

    \end{align*}$

    $X_n$ does not converge to $\theta$ in mean square
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  3. #3
    Junior Member
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    Re: convergences in probability

    The way you've written it makes a lot of sense. I just couldn't make sense of it from the lecture notes.

    Thank you
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