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Thread: Show the limit of a discrete markov chain's distribution converge to a continuous one

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    Show the limit of a discrete markov chain's distribution converge to a continuous one





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    Re: Show the limit of a discrete markov chain's distribution converge to a continuous

    How do you get the expression in part b)?

    If you are following the hints, you should try to evaluate the CDF of Y^{(N)} first and take N \to \infty. Although I have not calculate it, the expression you put down here does not like the CDF of a geometric distribution.

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    Re: Show the limit of a discrete markov chain's distribution converge to a continuous



    One of the main things bothering me is that the denominator inside the brackets is N whilst the exponent is x, the two being different meaning I can't use what I thought i needed to use in my previous post
    Last edited by Maharero; 10-05-2015 at 01:50 PM.

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    Re: Show the limit of a discrete markov chain's distribution converge to a continuous

    You have not get the CDF of Y^{(N)} correctly.

    The CDF of \frac {T_1^{(N)}} {N} is not dividing the CDF of T_1^{(N)} by N.


    A simple way to transform it is to consider

    F_{Y^{(N)}}(x) = \Pr\{Y^{(N)} \leq x \}

    = \Pr\left\{\frac {T_1^{(N)}} {N} \leq x \right\}

    = \Pr\{T_1^{(N)} \leq Nx \}

    = F_{T_1^{(N)}}(Nx)

    so now it is in terms of the geometric CDF you just found.

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    Maharero (10-05-2015)

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    Re: Show the limit of a discrete markov chain's distribution converge to a continuous


    Thank you for teaching me that, it makes sense. I worked through the problem like before and now was able to reach the CDF of the exponential distribution as I wanted

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