1

I needed to find the limiting distribution of the matrix

$$\pmatrix{ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0}$$

Instead of $\pi$ I'll use $A, B$ and $C$ because that might be easier to read.

To solve this, I got

$$A = B + \frac{1}{2}C$$ $$B = \frac{1}{2}C$$ $$C = A$$

Now to solve this, I ended up getting $A = C$ and I got $A = 2B$. I also need to satisfy $A + B + C = 1$. So, I said let $A = \frac{1}{4}, B = \frac{1}{2}, C = \frac{1}{4}$, which satisfies all my equations. But in the answers they say that $A = \frac{1}{5}, B = \frac{2}{5}, C = \frac{1}{5}$, which also satisfy the equations.

How did I get mine wrong? Obviously it matters what the limiting distribution is, so how am I supposed to tell which is the correct one?

Kaish
  • 6,126

2 Answers2

3

Both answers are wrong...

The system of three equations can be readily rewritten as $A=2B=C$ hence the only solution such that $A+B+C=1$ is $(A,B,C)=(\frac25,\frac15,\frac25)$.

If $(A,B,C)=(\frac14,\frac12,\frac14)$ (your solution), then $A\ne2B\ne C$.

If $(A,B,C)=(\frac15,\frac25,\frac15)$ (their solution), then $A+B+C\ne1$.

Did
  • 279,727
  • Huh? How did you get this? If $A = 2b$ and $A = \frac{2}{5}, B = \frac{1}{5}$ then isn't $A = \frac{B}{2}$? – Kaish Jan 22 '13 at 20:16
  • Well... no. $ $ – Did Jan 22 '13 at 20:57
  • Ok, I managed to get $A = C$ and $A = 2B$, but how do I then decide the values for $A, B$ and $C$? – Kaish Jan 22 '13 at 21:19
  • Also, how doesn't it? If I put in $\frac{2}{5}$ in my calculator, then divide by $2$, this gives me $\frac{1}{5} = B$.

    EDIT: Ohhh, so $\frac{A}{2} = B$ then?

     – Kaish Jan 22 '13 at 21:20
  • Please concentrate: if $A=C$ and $A=2B$, then $A+B+C=(2B)+B+(2B)=5B$ hence, since $A+B+C=1$, one gets $B=$ $____$. – Did Jan 22 '13 at 21:23
  • $5B = 1 \implies B = \frac{1}{5}$. Then do $2A = 1 - \frac{1}{5} \implies A = \frac{2}{5} = C$. – Kaish Jan 22 '13 at 21:33
1

Their answers are the essentially same as yours, just not normalized--they don't satisfy $A+B+C = 1$. As such, I would consider your answer "more correct" than theirs.

(Though, as Did points out, they're both wrong...)

Jonathan Christensen
  • 3,870
  • 15
  • 21
  • It says this is for a "limiting distribution" not stationary, mistake in my post, does this change things? – Kaish Jan 22 '13 at 19:30
  • No, it shouldn't make a difference in this case. Limiting and stationary distributions are not always the same, but here they will be, and it still needs to satisfy $A+B+C=1$. – Jonathan Christensen Jan 22 '13 at 19:32
  • This bit about them "not always being the same thing" has confused me because they want me to work out both in the next question, but I can't find out the difference. Can you tell me or give me a good link or something that will help me find out please? – Kaish Jan 22 '13 at 19:35
  • @Kaish This old question may be helpful. – Jonathan Christensen Jan 22 '13 at 19:42
  • That just tells me about equilibrium distribution, so that could still be eithe limiting or stationary right? It doesn't clarify any difference between them. – Kaish Jan 22 '13 at 20:03
  • @Kaish whoops, that wasn't the question I meant to link. Try this one. – Jonathan Christensen Jan 22 '13 at 20:11