Limiting Distribution of Xn | Probability Homework Solution

[stevenham]

Homework Statement

Suppose P(X_n = i) = $\frac{n+i}{3n+6}$, for i=1,2,3.
Find the limiting distribution of X_n

Homework Equations

The Attempt at a Solution

I first found the MGF by Expectation(e^tx)
which resulted in e^tx($\frac{n+1}{3n+6}$ + $\frac{n+2}{3n+6}$ + $\frac{n+3}{3n+6}$)
I then took the limit as n$\rightarrow$ ∞ which gives me 2e^tx

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

Thanks.

 

[Ray Vickson]

Homework Statement

Suppose P(X_n = i) = $\frac{n+i}{3n+6}$, for i=1,2,3.
Find the limiting distribution of X_n

Homework Equations

The Attempt at a Solution

I first found the MGF by Expectation(e^tx)
which resulted in e^tx($\frac{n+1}{3n+6}$ + $\frac{n+2}{3n+6}$ + $\frac{n+3}{3n+6}$)
I then took the limit as n$\rightarrow$ ∞ which gives me 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

 

[stevenham]

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]

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]

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.