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

BGM

TS Contributor
#2
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 [math] Y^{(N)} [/math] first and take [math] N \to \infty [/math]. Although I have not calculate it, the expression you put down here does not like the CDF of a geometric distribution.
 
#3
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
#4
Re: Show the limit of a discrete markov chain's distribution converge to a continuous

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

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


A simple way to transform it is to consider

[math] F_{Y^{(N)}}(x) = \Pr\{Y^{(N)} \leq x \} [/math]

[math] = \Pr\left\{\frac {T_1^{(N)}} {N} \leq x \right\} [/math]

[math] = \Pr\{T_1^{(N)} \leq Nx \} [/math]

[math] = F_{T_1^{(N)}}(Nx) [/math]

so now it is in terms of the geometric CDF you just found.
 
#5
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