Discussion
Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3) :
\[ P = \begin{bmatrix} 0.3 & \alpha & \beta \\ 0.4 & \gamma & 0.1 \\ \delta & 0.2 & \varepsilon \end{bmatrix} \]
If P is a doubly stochastic matrix, the value of \(\beta\) :
Answer: D
\(\beta\) is 0.4.
In a doubly stochastic matrix every row and every column sums to 1. Work through the entries that are forced:
\[ \begin{aligned} \text{column 1: } 0.3 + 0.4 + \delta = 1 &\Rightarrow \delta = 0.3 \\ \text{row 3: } 0.3 + 0.2 + \varepsilon = 1 &\Rightarrow \varepsilon = 0.5 \\ \text{column 3: } \beta + 0.1 + 0.5 = 1 &\Rightarrow \beta = 0.4 \end{aligned} \]
As a check, row 1 then gives \(\alpha = 0.3\) and column 2 gives \(\gamma = 0.5\), with row 2 summing to 1. So \(\beta\) can be computed and equals exactly 0.4.
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