Example: Social Class Problem
Given TPM:
\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ 0 & \frac{1}{2} & \frac{1}{2} \end{bmatrix}$$ Solving $\pi = \pi \mathbf{P}$ for the [[3 Reference/def-stationary-distribution_202603280834\|stationary distribution]]: $$\begin{bmatrix} \pi_{1} & \pi_{2} & \pi_{3} \end{bmatrix} = \begin{bmatrix} \pi_{1} & \pi_{2} & \pi_{3} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ \end{bmatrix}$$ ## Matching Components\begin{aligned} \pi_1 &= \frac{1}{2}\pi_1 + \frac{1}{4}\pi_2 \ \pi_2 &= \frac{1}{2}\pi_1 + \frac{1}{2}\pi_2 + \frac{1}{2}\pi_3 \ \pi_3 &= \frac{1}{4}\pi_2 + \frac{1}{2}\pi_3 \end{aligned}
## Solution From equation 1: $\frac{1}{2}\pi_1 = \frac{1}{4}\pi_2 \Rightarrow \pi_2 = 2\pi_1$ From equation 3: $\frac{1}{2}\pi_3 = \frac{1}{4}\pi_2 \Rightarrow \pi_3 = \frac{1}{2}\pi_2 = \pi_1$ Substituting into $\pi_1 + \pi_2 + \pi_3 = 1$: $$\pi_1 + 2\pi_1 + \pi_1 = 1 \Rightarrow 4\pi_1 = 1 \Rightarrow \pi_1 = \frac{1}{4}$$ Therefore: $\pi_2 = \frac{1}{2}$, $\pi_3 = \frac{1}{4}$ ## Related - [[3 Reference/def-stationary-distribution_202603280834\|Stationary Distribution]] - [[3 Reference/theorem-limiting-distribution_202603280833\|Limiting Distribution Theorem]]