# Q1(d) Markov chain diagram and path probability [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-final/q1d

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-final--q1d/

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

**Question. Q1(d)** [5 marks · 2024-04-17]

Consider the Markov chain with S={1,2,3} and transition matrix below. (a) Draw the state transition diagram. (b) The scan prints “P(X₁=1)=(X₂=2)=1/4”; calculate P(X₁=3, X₂=2, X₃=1).



$$
P=\begin{bmatrix}1/2&1/4&1/4\\1/3&0&2/3\\1/2&1/2&0\end{bmatrix}
$$



### Answer

#### Given

The printed matrix; ambiguous condition reproduced above.

#### To find

All nonzero transitions and the joint probability of the path 3→2→1.

#### Formula and why it applies

Rows are current states and columns are next states. A specified path probability is initial-state probability multiplied by each transition along that path.



$$
P(X_1=3,X_2=2,X_3=1)=P(X_1=3)P_{32}P_{21}
$$



#### Solution

##### Step 1. Draw the matrix as arrows

Swipe across the diagram to read its labels ↔

![Redrawn transition diagram. Each arrow goes from the current state to the next state; labels are probabilities.](https://deep-learning-study-room.vercel.app/is2/fig/legacy24-three-state.d3ab43c06c.svg)

Diagram: Redrawn transition diagram. Each arrow goes from the current state to the next state; labels are probabilities.. Labels: legacy24-three-state | 0.5 | 0.25 | 0.25 | 0.333333 | 0.666667 | 0.5 | 0.5 | 1 | 2 | 3

Redrawn transition diagram. Each arrow goes from the current state to the next state; labels are probabilities.

| From | Outgoing destinations |
| --- | --- |
| 1 | 1 with 1/2; 2 with 1/4; 3 with 1/4 |
| 2 | 1 with 1/3; 3 with 2/3 |
| 3 | 1 with 1/2; 2 with 1/2 |

**Source / assumption note.** The source drops the second P and uses X₂ in the second condition. Read literally as P(X₂=2)=1/4, the answer differs from the common intended initial-distribution reading P(X₁=2)=1/4. Both are calculated explicitly; do not silently change the subscript.

##### Step 2. Solve the literal reading

Let p₁=P(X₁=1)=1/4 and p₃=P(X₁=3). To be in state 2 at time 2, arrive from state 1 or state 3; transition 2→2 is zero.



$$
\frac14=p_1P_{12}+p_2P_{22}+p_3P_{32}=\frac14\frac14+0+\frac12p_3
$$





$$
\frac12p_3=\frac14-\frac1{16}=\frac3{16}\quad\Rightarrow\quad p_3=\frac38
$$





$$
P(3,2,1)=\frac38\cdot\frac12\cdot\frac13=\frac1{16}=0.0625
$$



##### Step 3. Show the alternative if the second subscript was a typo

If the intended statement was P(X₁=1)=P(X₁=2)=1/4, then P(X₁=3)=1−1/4−1/4=1/2.



$$
P(3,2,1)=\frac12\cdot\frac12\cdot\frac13=\frac1{12}\approx0.0833333
$$



**Final answer.** Literal time-2 condition gives 1/16. Corrected initial-state condition gives 1/12; the source wording must be clarified.