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Limiting Distribution

by stevenham
Tags: distribution, limiting
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stevenham
#1
Oct20-11, 04:55 PM
P: 8
1. The problem statement, all variables and given/known data
Suppose P(Xn = i) = [itex]\frac{n+i}{3n+6}[/itex], 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([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 2etx

Did I do this problem correctly? Is that the limiting distribution of Xn?

Thanks.
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Ray Vickson
#2
Oct20-11, 06:03 PM
Sci Advisor
HW Helper
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P: 5,208
Quote Quote by stevenham View Post
1. The problem statement, all variables and given/known data
Suppose P(Xn = i) = [itex]\frac{n+i}{3n+6}[/itex], 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([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 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
stevenham
#3
Oct20-11, 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?

Ray Vickson
#4
Oct21-11, 02:15 AM
Sci Advisor
HW Helper
Thanks
P: 5,208
Limiting Distribution

Quote Quote by stevenham View Post
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
canis89
#5
Oct21-11, 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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