Intelligent Systems II · TEE · papers

Q4(b) Transition matrix properties and two-step probabilities [10]

Re-examination · 25 Jun 2024

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

Question. Q4(b) [10 marks · 2024-06-25]

(1) Discuss how state transitions are represented by a transition matrix. (2) Cite and illustrate any 3 properties. (3) Given P=[[1/2,1/2],[3/4,1/4]] and initial row distribution p⁽⁰⁾=[1/4,3/4], define and find (a) P₂₁⁽²⁾, (b) P₁₂⁽²⁾, (c) P⁽²⁾, (d) p₁⁽²⁾.

Answer

Given

P and p⁽⁰⁾ as printed; row-vector convention.

To find

Two-step conditional transition probabilities, the full two-step matrix and unconditional state 1 probability at time 2.

Formula and why it applies

Pᵢⱼ is next-state j probability given current-state i. Entries lie in[0,1], rows sum to 1, and the n-step matrix is Pⁿ for a time-homogeneous chain. Row sums need not equal column sums; a transition matrix need not be symmetric.

(P^2)_{ij}=\sum_kP_{ik}P_{kj},\qquad p^{(2)}=p^{(0)}P^2LaTeX: (P^2)_{ij}=\sum_kP_{ik}P_{kj},\qquad p^{(2)}=p^{(0)}P^2

Solution

Step 1. Sum every possible two-step path
EntryVia state 1Via state 2Sum
(P²)11(1/2)(1/2)=1/4(1/2)(3/4)=3/85/8
(P²)12(1/2)(1/2)=1/4(1/2)(1/4)=1/83/8
(P²)21(3/4)(1/2)=3/8(1/4)(3/4)=3/169/16
(P²)22(3/4)(1/2)=3/8(1/4)(1/4)=1/167/16
Step 2. Write the transition matrix and requested entries
P^2=\begin{bmatrix}5/8&3/8\\9/16&7/16\end{bmatrix}LaTeX: P^2=\begin{bmatrix}5/8&3/8\\9/16&7/16\end{bmatrix}

Therefore (a) P₂₁⁽²⁾=9/16 and (b) P₁₂⁽²⁾=3/8. Each entry conditions on a specific starting state.

Step 3. Mix over the supplied initial distribution
p_1^{(2)}=\frac14\frac58+\frac34\frac9{16}=\frac{10+27}{64}=\frac{37}{64}=0.578125LaTeX: p_1^{(2)}=\frac14\frac58+\frac34\frac9{16}=\frac{10+27}{64}=\frac{37}{64}=0.578125
p_2^{(2)}=1-37/64=27/64=0.421875LaTeX: p_2^{(2)}=1-37/64=27/64=0.421875
Step 4. Keep matrix and distribution notation separate

P² is 2×2 and gives two-step probabilities for either starting state. p⁽²⁾ is one row of two probabilities after averaging those starting states using p⁽⁰⁾. If part (c) was intended as the distribution rather than capital-P matrix, the supplied distribution is [37/64,27/64]. Both objects are shown.

Final answer. P₂₁⁽²⁾=9/16; P₁₂⁽²⁾=3/8; P²=[[5/8,3/8],[9/16,7/16]]; p⁽²⁾=[37/64,27/64].