Light bulbs have i.i.d. lifetimes with mean 1000 hours. Bulbs are replaced immediately upon failure. After a very long time T, approximately how many replacements have occurred?
AT/1000, by the elementary renewal theorem: N(t)/t → 1/E[X₁]
BT × 1000, because each bulb contributes 1000 expected hours
C√T, by the central limit theorem for counting processes
DThe answer depends on the variance of the lifetime distribution, not just the mean
The elementary renewal theorem says N(t)/t → 1/μ a.s. where μ = E[X₁] = 1000. So N(T) ≈ T/1000 for large T. The variance affects the fluctuations around this rate (the renewal CLT gives N(t) ≈ t/μ + (σ/μ^{3/2})√t · Z where Z ~ N(0,1)), but not the long-run rate itself. This is a direct consequence of the strong law of large numbers applied to the partial sums S_n = X₁ + ... + X_n: S_n/n → μ implies n/S_n → 1/μ, and N(t) is essentially the inverse of S_n.
Question 2 Multiple Choice
The inspection paradox states that if you arrive at a random time and measure the length of the current inter-arrival interval, its expected length exceeds E[X₁]. Why?
ABecause longer intervals are more likely to be 'hit' by a random arrival time — you are biased toward sampling longer intervals
BBecause the current interval started before your arrival, adding extra time
CBecause the renewal process speeds up over time, making early intervals shorter than later ones
DThis is a mathematical artifact with no real-world significance
If you arrive at a uniformly random time, you are more likely to land in a longer interval than a shorter one — simply because longer intervals occupy more of the timeline. The probability of landing in an interval of length x is proportional to x times its frequency, giving a size-biased distribution with density xf(x)/μ. The expected length of the size-biased interval is E[X²]/E[X] = μ + σ²/μ ≥ μ, with equality only when σ² = 0 (deterministic intervals). This is the continuous-time version of the 'bus paradox': your average wait for a bus is longer than half the average interval between buses.
Question 3 Short Answer
The renewal-reward theorem states that if a reward R_i is earned in the i-th renewal cycle, then the long-run reward rate is E[R]/E[X]. Explain why this is a generalization of the elementary renewal theorem.
Think about your answer, then reveal below.
Model answer: The elementary renewal theorem is the special case where R_i = 1 for every cycle (each renewal earns one 'count'). Then total reward up to time t is N(t), and the reward rate is N(t)/t → E[1]/E[X] = 1/μ. The renewal-reward theorem generalizes by allowing each cycle to earn a different random reward R_i (independent of cycle length, or possibly correlated with it). The total reward up to time t is Σᵢ₌₁^{N(t)} Rᵢ, and the long-run rate is E[R₁]/E[X₁]. This covers applications like total revenue per unit time, total downtime per unit time, or average cost rate of a replacement policy.
The proof uses the SLLN twice: Σ Rᵢ/n → E[R] and Σ Xᵢ/n → E[X] = μ, so (Σ Rᵢ)/(Σ Xᵢ) → E[R]/E[X]. Since Σ_{i=1}^{N(t)} Xᵢ ≈ t for large t, dividing by t gives the result.