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Limiting Distribution 
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#1
Oct2011, 04:55 PM

P: 8

1. The problem statement, all variables and given/known data
Suppose P(X_{n} = i) = [itex]\frac{n+i}{3n+6}[/itex], for i=1,2,3. Find the limiting distribution of X_{n} 2. Relevant equations 3. The attempt at a solution I first found the MGF by Expectation(e^{tx}) which resulted in e^{tx}([itex]\frac{n+1}{3n+6}[/itex] + [itex]\frac{n+2}{3n+6}[/itex] + [itex]\frac{n+3}{3n+6}[/itex]) I then took the limit as n[itex]\rightarrow[/itex] ∞ which gives me 2e^{tx} Did I do this problem correctly? Is that the limiting distribution of Xn? Thanks. 


#2
Oct2011, 06:03 PM

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P: 5,083

2) You don't need the MGF; in fact, using the MGF is doing it the hard way. RGV 


#3
Oct2011, 08:32 PM

P: 8

Do I have to calculate the CDF?
After calculating the CDF and doing the limit as n goes to infinity, I get 1. Did I calculate the MGF incorrectly? Should I have done E(e^ti) instead? 


#4
Oct2111, 02:15 AM

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P: 5,083

Limiting Distribution
RGV 


#5
Oct2111, 04:22 AM

P: 30

No, you got it wrong. If you want to use the MGF method to find the limiting distribution, then, first, find the MGF of Xn for each n, which is
[itex]MGF_{X_n}(t)=E[e^{tX_n}]=\sum_{k=1}^3 e^{tk}P(X_n=k)[/itex]. Then, find the limit of the MGF as n tends to infinity. Finally, find some random variable X with MGF equal to [itex]\lim_{n\rightarrow\infty}MGF_{X_n}(t)[/itex]. My suggestion is, instead using the MGF method, find the cumulative probability function, Fn(i)=P(Xn<=i) for each i=1,2,3. Then, limit this function as n tends to infinity. The limit is the cumulative probability function of the limiting distribution. 


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