# Q6(b) Two-step walking/cycling Markov transition [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q6b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q6b/

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

**Question. Q6(b)** [10 marks · 2025-04-28]

A student walks to school, or he rides his bicycle. If he walks to school one day, then the next day he will walk or cycle with equal probability. But if he bicycles one day, then the probability that he will walk next day is 1/4.

- Express this information in a transition matrix.
- if it is assumed that the initial day is Monday, write a matrix that gives probabilities of a transition from Monday to Wednesday.

### Answer

### Given

State orderW=walk, C=cycle. One transition is one day.

### To find

P for one day andP² for Monday→Wednesday.

### Formula and why



$$
(P^2)_{ij}=\sum_kP_{ik}P_{kj}
$$



There are two transitions, Monday→Tuesday and Tuesday→Wednesday. Multiply along each possible intermediate Tuesday state and add the alternatives.

### Step 1. Complete the one-day rows

From walking:W→W=1/2 andW→C=1/2. From cycling:C→W=1/4, soC→C=1−1/4=3/4.



$$
P=\begin{bmatrix}1/2&1/2\\1/4&3/4\end{bmatrix}
$$



### Step 2. Compute all four two-day entries

| Monday→Wednesday | Sum over Tuesday W or C | Result |
| --- | --- | --- |
| W→W | (1/2)(1/2)+(1/2)(1/4)=1/4+1/8 | 3/8 |
| W→C | (1/2)(1/2)+(1/2)(3/4)=1/4+3/8 | 5/8 |
| C→W | (1/4)(1/2)+(3/4)(1/4)=1/8+3/16 | 5/16 |
| C→C | (1/4)(1/2)+(3/4)(3/4)=1/8+9/16 | 11/16 |

### Step 3. Write the requested matrix



$$
P^2=\begin{bmatrix}3/8&5/8\\5/16&11/16\end{bmatrix}=\begin{bmatrix}0.375&0.625\\0.3125&0.6875\end{bmatrix}
$$



Rows still specify the Monday state; columns specify Wednesday. Each row sums to 1. If Monday's actual state is not given, this conditional matrix is the complete answer; an unconditional Wednesday distribution additionally requires Monday's starting distribution.