Memorylessness
For exponential and geometric waits, the past does not change the future.
What is it?
An exponential satisfies ; the geometric distribution is the discrete version.
Why is it useful?
It collapses questions about "already waited" or "remaining time" into the original distribution.
When should you think of it?
Use it when waiting times are exponential or geometric and you are told time has already passed.
Think of this method when…
- Exponential lifetimes or Poisson arrivals
- "Given it has lasted hours…"
- Repeated independent trials until first success
- Remaining time after an observation
Core idea
Conditioned on survival so far, the remaining wait has the same distribution as a fresh one.
Step-by-step playbook
- 1Confirm the waiting time is exponential or geometric.
- 2Discard the elapsed time.
- 3Answer using the fresh distribution.
Worked example
Worked example
A bulb's lifetime is exponential with mean 1000 hours. It has already worked for 500 hours. What is its expected remaining life?
- 1
By memorylessness the remaining lifetime is again exponential with mean 1000.
- 2
The 500 elapsed hours are irrelevant.
- Answer: hours
Common mistakes
- Applying it to uniform or normal waiting times.
- Thinking a machine that has run long is "due" to fail.
Would you use this method?
A process has a uniform lifetime on and has survived 5 units. Is the remaining life still uniform on ?
Practice this method
41 QuantSolve questions are tagged with this method.
Practice this method