Updated 12 September 2026

Exponential distribution

The exponential distribution models the waiting time until the next event in a homogeneous Poisson process. A constant event rate and independent counts in disjoint intervals are model assumptions; real request traffic or bus arrivals need not satisfy them. Its rate parameter $\lambda$ sets how quickly events happen on average.

What to notice

• Higher $\lambda$ shifts mass toward zero — events happen sooner. The distribution compresses toward the left.
• Mean vs median. Because the distribution is right-skewed, the mean (= $1/\lambda$) always exceeds the median (= $\ln 2/\lambda$). Most waiting times are shorter than average.
• Unbounded right tail. Even with high $\lambda$, long waiting times have nonzero probability. Rare but very long waits always exist.

The memoryless property

Among proper continuous distributions on nonnegative waiting times with positive survival at every finite time, the exponential is characterized by this property for every s,t≥0:

If you’ve already waited s minutes for the bus, the remaining wait has the same distribution as if you had just arrived. The bus has no memory of how long you’ve been there.

This makes the exponential both analytically tractable and sometimes an unrealistic model: many real-world waits do have memory (a scheduled bus is not equally likely to arrive at every instant).

Condition on the lifetimes that survived

With λ=1 per second, the survival function is S(t)=exp(−t) and the CDF is F(t)=1−S(t). The event “another t seconds after already surviving s” has probability S(s+t)/S(s). For s=t=1, this is exp(−2)/exp(−1)=exp(−1), about 36.79%. The simulation keeps only lifetimes exceeding s in the denominator, so a small surviving group can give a visibly noisy estimate even when the model identity is exact.

The same identity follows from Poisson counts: waiting beyond t means observing zero events in a length-t interval. Thus P(T>t)=P(N(t)=0)=exp(−λt). Rate has units events per second; scale 1/λ has units seconds.

A counterexample with increasing hazard

The Weibull comparison uses survival S(t)=exp(−(λt)^k), with shape k and scale 1/λ. At k=2, λ=1 and s=t=1, a fresh lifetime exceeds one second with probability exp(−1), but a lifetime that already survived one second lasts another with probability exp(−3), about 4.98%. Surviving the past changes the future because the hazard increases with age.

At k=1 the curves coincide with the exponential. At k<1 the hazard decreases and surviving longer increases the chance of another fixed interval. Matching scale does not match mean across shapes. These are synthetic lifetime models, not evidence about any particular device or arrival service.

Reference and next step

NIST: Weibull distribution gives the survival, hazard, and scale convention used in the comparison. Continue to Weibull to explore changing hazards or gamma to wait for several Poisson events.