# Statistics - Discrete Markov Chains

1. Apr 30, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

[Broken]

2. Relevant equations

3. The attempt at a solution

I'm not really sure if this belongs here or in the precalculus mathematics section. I had to take calculus before taking this class so I'm putting it here.

I'm confused about part (b). I don't really understand how I'm supposed to find $P(X_{n+2}=0|X_{n}=2)$ because I don't know what state n+1 is. Thanks for any help.

Last edited by a moderator: May 6, 2017
2. Apr 30, 2013

### micromass

Staff Emeritus
3. Apr 30, 2013

### GreenPrint

$P(X_{n+2}=0) = P(X_{n+2}=0|X_{n}=2)P(X_{n}=2) + P(X_{n+2}=0|X_{n}=1)P(X_{n}=1) + P(X_{n+2}=0|X_{n}=0)P(X_{n}=0)$
$P(X_{n+2}=0|X_{n}=2)P(X_{n}=2) = P(X_{n+2}=0) - P(X_{n+2}=0|X_{n}=1)P(X_{n}=1) - P(X_{n+2}=0|X_{n}=0)P(X_{n}=0)$

I'm not sure how this helps.

4. Apr 30, 2013

### micromass

Staff Emeritus
You're using the wrong events.

Do this

$$\begin{eqnarray*} P(X_{n+2} = 0~\vert~X_n=2) & = & P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=0)P(X_{n+1}=0~\vert~X_n=2)\\ & & + P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=1)P(X_{n+1}=1~\vert~X_n=2)\\ & & + P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=2)P(X_{n+1}=2~\vert~X_n=2) \end{eqnarray*}$$

5. Apr 30, 2013

### Ray Vickson

If you had been told that $X_0 = 2$ would you have been able to work out the probability that $X_2 = 0?$ Have you really never seen how to get multi-step transition probabilities?

Note: I am waiting for answers to these questions before offering more help.

Last edited by a moderator: May 6, 2017
6. May 2, 2013

### GreenPrint

Well for b) I got .21 and believed that I solved the problem correctly. I don't know exactly what c is even asking me. Find P(X_1 = 0). What exactly are the alphas? Like what do they represent? Alpha 1 = probability x equals zero is .25.

I have indeed never seen how to get multi-step transition probabilities =( but i believe i figured it out correctly and got the answer.