Stochastic Processes · 免费预览
年龄与剩余寿命
使用中物品的年龄和剩余寿命收敛到大小偏置分布，其密度为分布尾部除以均值；指数寿命的无记忆性使年龄与剩余寿命相互独立；而在时间平均中，长寿命物品所占比例偏高，这就是检验悖论。
1. Two clocks inside every renewal
Two clocks inside every renewal
2. The general limiting law
The general limiting law
3. The paradox of the long-lived bulb
The paradox of the long-lived bulb
4. Why commuters wait more than half a gap
Why commuters wait more than half a gap
5. Definitions at a random time
At time t, the last renewal happened at T sub N of t and the next at T sub N plus 1. What are the age and the residual life?
6. The tail-over-mean law
What is the limiting distribution of the residual life for a renewal process with finite mean gap?
7. Memorylessness as independence
Why do exponential gaps make the age and residual life independent in the limit?
8. The shape under uniform gaps
Gaps are uniform on the interval from 0 to b. What does the limiting density of the age look like?
9. The heart of the paradox
Why does the item in use at a random time average longer than a typical item?
10. The commuter's complaint
Buses arrive with gaps uniform on the interval from 0 to 10 minutes. What is the expected wait of a commuter arriving at a random time?
11. Landing exactly on Go
In Monopoly a token moves by the sum of two dice, so the number of squares moved has mean 7. Compute the long-run fraction of trips around the board on which the token lands exactly on Go, and write the answer as a fraction or decimal.
12. The bus you are on
Bus gaps are uniform on the interval from 0 to 10 minutes, so the mean gap is 5 and the second moment is 100 over 3. Compute the expected total lifetime of the gap a random commuter lands in, E of tau squared over E of tau, and write the answer as a fraction or decimal.
13. A memoryless elderly bulb
A bulb's lifetime is exponential with rate one fifth per hour, and the system has run long enough to equilibrate. Compute the limiting probability that the bulb in use has already survived more than 5 hours, and write the answer as a decimal.