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I have the following set of equations

$$\pi_1 = \pi_3 + [1 - \alpha(1 - p)]\pi_4$$ $$\pi_2 = \alpha(1 - p)\pi_4$$ $$\pi_3 = \alpha(1 - p)]\pi_1$$ $$\pi_4 = [1 - \alpha(1 - p)]\pi_1 + \pi_2$$ $$\pi_1 + \pi_2 + \pi_3 + \pi_4 = 1$$

Using this, I'm supposed to get the answer

$$\pi_1 = \pi_4 = \frac{1}{2 + 2\alpha(1 - p)} \hspace{1.5cm} \pi_2 = \pi_3 = \frac{\alpha(1 - p)}{2 + 2\alpha(1 - p)} $$

But I can't seem to do this. I always just end up with all the $\pi$'s equaling each other.

How do I do this question

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Does this have anything to do with Markov Chains? –  Inquest Jan 19 '13 at 22:51
    
What have you tried? –  Code-Guru Jan 19 '13 at 22:55
    
One reason that you could have all the $\pi$ equalling each other, is that's what happens when $\alpha(1-p)=1$. –  Calvin Lin Jan 19 '13 at 23:05
    
@Inquest Yes. It's to find the stationary distribution –  Kaish Jan 20 '13 at 11:01
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2 Answers 2

up vote 2 down vote accepted

In all honesty, "Just do it."

For example, if you substitute the third equation into the first, you get that

$$ [1-\alpha(1-p)](\pi_1-\pi_4)=0$$

For now, I'm going to assume that the coefficient is non-zero, which gives $\pi_1 = \pi_4$. The second and third equation then give that $\pi_2=\pi_3$. Substituting this into the last equation, we get that $\pi_1 = \frac {1}{ 2 +2\alpha(1-p)}$ (which is equal to $\pi_4$). And the value of $\pi_3 = \pi_2$ drops out immediately.

Now, if the coefficient is 0, then $[1-\alpha(1-p)]=0$, which gives $\pi_1=\pi_3$, $\pi_2=\pi_4$, and you have infinitely many solutions subject to $\pi_1+\pi_2 = \frac {1}{2}$.

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We have to make this easier to read, so please forgive my rewriting of terms.

Let: $s = \pi_1, t = \pi_2, u = \pi_3, v = \pi_4, w = \alpha(1 - p)$.

Rewriting your system yields:

$\tag 1 s = u + (1 - w)v$

$\tag 2 t = wv$

$\tag 3 u = ws$

$\tag 4 v = (1 - w)s + wv$

$\tag 5 s + t + u + v = 1$

From $(4)$, we collect like terms on each side of the equation by subtracting $wv$ and get: $$\tag 6 v(1 - w) = (1 - w)s, \text{hence} ~v = s$$

Now, just substitute all the values you have into (5), yielding:

$u + (1-w)v + wv + ws + (1-w)s + wv = 1$, and collecting like terms and simplifying, we get:

$u + v + s + wv = 1$, and from $(3)$ we have $u = ws$, so we get:

$ws + s + v + wv = 1$

$s(1+w) + v(1+w) = 1$, or

$(s+v) = \frac{1}{1+w}$, but we know from (6) that $v = s$ and now have:

$2s = \frac{1}{1+w}$, or

$s = \frac{1}{2(1+w)}$

We now can easily substitute into $(2)$, yielding:

$t = w v = w s = w \frac{1}{2(1+w)}$, and

Substituting into $(3)$, yields:

$u = w s = w \frac{1}{2(1+w)}$

Regards

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You should deal with the $w=1$ case separately, since $w\neq 1$ wasn't mentioned, and this leads to additional answers. In particular, if OP is using $w=1$, then they will get all the $\pi_i$ being the same, which was their initial problem. –  Calvin Lin Jan 19 '13 at 23:06
    
@CalvinLin: Thanks for the comment. From (6), I would have thought that would be clear, but all I am showing is the π1,π2,π3,π4 issue the OP is asking about. Regards –  Amzoti Jan 19 '13 at 23:17
    
+1, Amzoti!! $\ddot\smile^{\ddot\smile}$ –  amWhy May 8 '13 at 1:09
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