# Limiting Distribution

by stevenham
Tags: distribution, limiting
 P: 8 1. The problem statement, all variables and given/known data Suppose P(Xn = i) = $\frac{n+i}{3n+6}$, for i=1,2,3. Find the limiting distribution of Xn 2. Relevant equations 3. The attempt at a solution I first found the MGF by Expectation(etx) which resulted in etx($\frac{n+1}{3n+6}$ + $\frac{n+2}{3n+6}$ + $\frac{n+3}{3n+6}$) I then took the limit as n$\rightarrow$ ∞ which gives me 2etx Did I do this problem correctly? Is that the limiting distribution of Xn? Thanks.
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P: 5,083
 Quote by stevenham 1. The problem statement, all variables and given/known data Suppose P(Xn = i) = $\frac{n+i}{3n+6}$, for i=1,2,3. Find the limiting distribution of Xn 2. Relevant equations 3. The attempt at a solution I first found the MGF by Expectation(etx) which resulted in etx($\frac{n+1}{3n+6}$ + $\frac{n+2}{3n+6}$ + $\frac{n+3}{3n+6}$) I then took the limit as n$\rightarrow$ ∞ which gives me 2etx Did I do this problem correctly? Is that the limiting distribution of Xn? Thanks.
1) What you wrote is not the MGF.
2) You don't need the MGF; in fact, using the MGF is doing it the hard way.

RGV
 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?
 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 $MGF_{X_n}(t)=E[e^{tX_n}]=\sum_{k=1}^3 e^{tk}P(X_n=k)$. Then, find the limit of the MGF as n tends to infinity. Finally, find some random variable X with MGF equal to $\lim_{n\rightarrow\infty}MGF_{X_n}(t)$. 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.