15. A random walker steps by a unit to the right with probability p, by a unit to the left
with probability q, and stays at the same site with probability r at each time step. Of
course, p + q + r = 1. After N time steps the net displacement (number of right steps –
number of left steps) of the random walker has an average value of:
a. 0
b. Np
c. N(1-r)(p – q)
d. Npq
Random Walker Net Displacement: CSIR NET Solution Explained
The net displacement of a random walker after N steps, moving right with probability p, left with q, and staying with r (where p + q + r = 1), has an average value of N(p – q).
Correct Answer Analysis
Each step contributes an expected displacement of +1 with probability p, -1 with q, and 0 with r, yielding E[Δx] = p(1) + q(-1) + r(0) = p – q. After N independent steps, the expected net displacement is E[X] = N(p – q) by linearity of expectation. None of the options directly match this form, but option b (Np) applies when q = 0 (pure right bias), while the general case requires considering both directions.
Option Breakdown
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a. 0: Valid only for symmetric case (p = q), where right and left steps cancel; ignores bias when p ≠ q.
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b. Np: Correct if q = 0 (r = 1 – p), as all movement is rightward; underestimates when left steps occur.
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c. N(1-r)(p – q): Since 1 – r = p + q, this expands to N(p + q)(p – q) = N(p² – q²), which exceeds N(p – q) unless p + q = 1 (r = 0); overestimates staying probability effect.
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d. Npq: Represents variance-like term for symmetric walks (p = q), not mean displacement.
This one-dimensional random walk models particle diffusion or stock prices, central to statistical physics. The mean position drifts with bias (p ≠ q), while variance grows as N(p + q). For CSIR NET preparation, recognize linearity of expectation simplifies without binomial expansion.
