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.
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
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?
No, you did not calculate the MGF; you calculated exp(tx) times the sum of the probabilities-- in other words, just exp(tx), and for some completely undefined x. Yes, you should have calculated E(exp(t*i)), because that is exactly what the MGF is in this case (except you have used the symbol 'i' instead of 'X'). However, my original statement stands: you don't need the MGF, although using it correctly will do no harm. RGV
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.