ASRB Agricultural Research Service (ARS) - Agricultural Statistics scientist · India · Agricultural Scientists Recruitment Board (ICAR)
Original practice questions written to the published syllabus · not past papersRandom processes in time: Markov chains and their classification and limits, Poisson and birth-death processes, branching processes and Brownian motion, applied to population growth, epidemics, queues and weather sequences such as wet- and dry-day chains.
In the exams: GATE ST 'Stochastic Processes' (heavy); UPSC ISS Statistics I (present); ISI admission (present); CSIR-NET Unit 4 (present); light in AICE-JRF Ag Stat
How our questions for ASRB Agricultural Research Service spread across the units of Stochastic Processes. Difficulty: 5 easy, 24 medium, 4 hard.
| Markov-chain models of rainfall and crop sequences | 9 | |
| Queueing models M/M/1 and M/M/c | 5 | |
| Renewal processes (introduction) | 4 | |
| Markov chains: transition matrices and Chapman-Kolmogorov equations | 3 | |
| Pure birth, pure death and birth-death processes | 3 | |
| Branching processes and extinction probability | 3 | |
| Limiting and stationary distributions | 2 | |
| Poisson process and its properties | 2 | |
| Brownian motion and its basic properties | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Rain events occur as a Poisson process with rate 2 per week, and amounts per event have mean 5 mm and E(Y²) = 34 mm². What is the variance of total rainfall over 10 weeks?
D. 680
Compound Poisson: Var(S(t)) = λtE(Y²) = 20 × 34 = 680 mm². Using Var(Y) = 9 instead of E(Y²) gives 180; the mean is 100 mm.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
A biocontrol agent has per-individual birth rate 0.3 and death rate 0.2 per day (linear birth-death). What is the probability of eventual extinction if 5 individuals are released?
D. 0.132
For λ > μ, extinction from i individuals has probability (μ/λ)ⁱ = (2/3)⁵ = 0.132. 0.667 is the risk from a single individual.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
A single introduction has extinction probability 0.5. What is the extinction probability with 10 independent founders?
A. 0.0010
Independent lines must all die out: qᵏ = 0.5¹⁰ = 0.0010.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
In an M/M/1 queue with μ = 6 per hour, the arrival rate rises from 4 to 5.5 per hour. How does the mean time in the system change?
B. From 0.5 hour to 2 hours
W = 1/(μ − λ): 1/2 hour at λ = 4, 1/0.5 = 2 hours at λ = 5.5. Waiting explodes as ρ approaches 1.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
In a monsoon wet/dry chain, P(wet | dry yesterday) = 0.2 and P(wet | wet yesterday) = 0.6. What is the long-run probability of a wet day?
A. 0.333
π(W) = p₀₁/(1 − p₁₁ + p₀₁) = 0.2/0.6 = 0.333. 0.4 averages the two conditional probabilities.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
In a wet/dry chain with P(wet | wet) = 0.6, what is the mean length of a wet spell?
B. 2.5 days
Wet-spell length is geometric with mean 1/(1 − p₁₁) = 1/0.4 = 2.5 days. 1/0.6 uses the persistence probability itself.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
With P(wet | dry) = 0.2, what is the probability that a dry spell, once started, lasts at least 10 days?
D. 0.134
After the first dry day, 9 more dry days are needed: (1 − p₀₁)⁹ = 0.8⁹ = 0.134. 0.8¹⁰ = 0.107 counts one day too many.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
How can a first-order rainfall chain be tested against a second-order chain?
D. A G² test or AIC on transition counts
The second-order chain conditions on two previous days; G² compares the nested models with df equal to the extra free probabilities, or AIC/BIC choose between them.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
What does the renewal-reward theorem give?
B. Long-run reward per unit time = E(reward per cycle)/E(cycle length)
With iid cycles, total reward up to t divided by t converges almost surely to E(R)/E(X).
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
Field surveys estimate infectious periods by measuring animals that are infected on the survey day. What bias results?
D. Long periods are over-sampled
Units sampled at a fixed time are more likely to be in long intervals (the inspection paradox), so naive averages overestimate the mean duration.
Further reading: Karlin and Taylor, A First Course in Stochastic Processes; Ross, Introduction to Probability Models; Medhi, Stochastic Processes
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