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

BGM

TS Contributor
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.

Maharero

New Member
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

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BGM

TS Contributor
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.

Maharero

New Member
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