Statistics - Discrete Markov Chains

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Homework Statement





Homework Equations





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.
 
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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.
 
You're using the wrong events.

Do this

\begin{eqnarray*}<br /> P(X_{n+2} = 0~\vert~X_n=2) <br /> &amp; = &amp; P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=0)P(X_{n+1}=0~\vert~X_n=2)\\<br /> &amp; &amp; + P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=1)P(X_{n+1}=1~\vert~X_n=2)\\<br /> &amp; &amp; + P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=2)P(X_{n+1}=2~\vert~X_n=2)<br /> \end{eqnarray*}
 
GreenPrint said:

Homework Statement





Homework Equations





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.

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.
 
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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.

Thanks for your help guys.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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