Discussion
Let \(\{X_{n}\}_{0}^{\infty}\) be a Markov chain with \(P[X_{0} = 0] = P[X_{0} = 1] = \dfrac{1}{4}\) and \(P[X_{0} = 2] = \dfrac{1}{2}\). The t.p.m. is given by :
| 0 | 1 | 2 | |
|---|---|---|---|
| 0 | 1/2 | 1/3 | 1/6 |
| 1 | 0 | 1/3 | 2/3 |
| 2 | 1/2 | 0 | 1/2 |
Then \(P[X_{1} = 1]\) is given by :
Answer: D
\(P[X_{1} = 1] = \) 1/6.
Multiply the initial distribution by the column of the transition matrix for state 1:
\[ \begin{aligned} P[X_{1} = 1] &= \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{2}\cdot 0 \\ &= \tfrac{1}{12} + \tfrac{1}{12} = \tfrac{1}{6} \end{aligned} \]
1/3 is just the transition probability from 0 (or 1) to 1, ignoring the starting distribution, and 1/4 is the initial probability of state 1.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development