基于转移矩阵证明$p_{1,1}^{(n)}$概率表达式：随机过程问题

Alright, let's wrap up the full induction proof you started for this symmetric 4-state Markov chain. You already nailed the base cases, so let's build on that.

Step 1: Confirm the Base Cases #

First, let's formalize the base cases you already checked:

• n=0: The 0-step transition probability ( p_{1,1}^{(0)} = 1 ) (we're already at state 1). Plugging into the formula:
  [
  0.25 + 0.75(1-4\alpha)^0 = 0.25 + 0.75 \times 1 = 1
  ]
  Perfect, that matches exactly.
• n=1: From the given transition matrix, ( p_{1,1}^{(1)} = 1 - 3\alpha ). Using the formula:
  [
  0.25 + 0.75(1-4\alpha) = 0.25 + 0.75 - 3\alpha = 1 - 3\alpha
  ]
  That's exactly what we get from ( P ), so the base cases hold solid.

Step 2: Inductive Hypothesis & Step #

Now let's use mathematical induction to cover all ( n \geq 0 ):

• Inductive Hypothesis: Assume for some integer ( k \geq 1 ), the formula is true:
  [
  p_{1,1}^{(k)} = 0.25 + 0.75(1-4\alpha)^k
  ]

• Inductive Step: We need to prove this holds for ( k+1 ). Use the Chapman-Kolmogorov equation, which links n-step transitions to smaller steps:
  [
  p_{1,1}^{(k+1)} = \sum_{j=1}^4 p_{1,j}^{(k)} p_{j,1}^{(1)}
  ]
  Break down this sum using the chain's symmetry:

  • For ( j=1 ): The term is ( p_{1,1}^{(k)} \times (1-3\alpha) ) (directly from the transition matrix).
  • For ( j=2,3,4 ): Due to the chain's symmetry, ( p_{1,2}^{(k)} = p_{1,3}^{(k)} = p_{1,4}^{(k)} ). Let's call this common value ( q ). Each of these terms is ( q \times \alpha ), so total for these three states is ( 3q\alpha ).

  Since all k-step transition probabilities from state 1 must sum to 1:
  [
  p_{1,1}^{(k)} + 3q = 1 \implies q = \frac{1 - p_{1,1}^{(k)}}{3}
  ]
  Substitute this back into the Chapman-Kolmogorov equation:
  [
  p_{1,1}^{(k+1)} = p_{1,1}^{(k)}(1-3\alpha) + 3 \times \frac{1 - p_{1,1}^{(k)}}{3} \times \alpha
  ]
  Simplify the expression step by step:
  [
  p_{1,1}^{(k+1)} = p_{1,1}^{(k)}(1-3\alpha) + (1 - p_{1,1}^{(k)})\alpha
  ]
  Expand and combine like terms:
  [
  p_{1,1}^{(k+1)} = p_{1,1}^{(k)}(1-4\alpha) + \alpha
  ]
  Now plug in our inductive hypothesis:
  [
  p_{1,1}^{(k+1)} = \left0.25 + 0.75(1-4\alpha)^k\right + \alpha
  ]
  Expand and simplify the constant terms:
  [
  0.25(1-4\alpha) + \alpha = 0.25 - \alpha + \alpha = 0.25
  ]
  This leaves us with:
  [
  p_{1,1}^{(k+1)} = 0.25 + 0.75(1-4\alpha)^{k+1}
  ]
  That's exactly the formula we needed to prove for ( k+1 )!

Step 3: Conclusion #

By mathematical induction, the formula ( p_{1,1}^{(n)} = 0.25 + 0.75(1-4\alpha)^n ) holds for all non-negative integers ( n \geq 0 ).

内容的提问来源于stack exchange，提问作者Ryan Honea