Poisson Process Expected Events & Rate Calculator
In a homogeneous Poisson point process with memoryless independent arrivals at rate λ per unit time, the expected number of events (and variance) over window t is μ = λ × t.
Expected arrivals in window t (also equals Poisson variance σ²)
Mean event rate per single time unit
Duration of the exposure interval
Step-by-Step Calculation
In a homogeneous Poisson point process with memoryless independent arrivals at rate λ per unit time, the expected number of events (and variance) over window t is μ = λ × t.
Real-World Reference Values
Standard deviation of count is σ = √μ = √(λt)
Probability of zero failures/arrivals during interval t
Follows an Exponential(λ) distribution
20-Row Comparison Table
Expected Event Count (Mean & Variance) (μ) · Arrival Rate per Unit Time (λ)
| # | Arrival Rate per Unit Time (λ) | Expected Event Count (Mean & Variance) (μ) | Δ % |
|---|---|---|---|
| 01 | 2.5 Current | 30 | 0.00% |
| 02 | 72.5009 | 870.0113 | +2800.04% |
| 03 | 145.0009 | 1,740.01 | +5700.04% |
| 04 | 217.5008 | 2,610.01 | +8600.03% |
| 05 | 290.0008 | 3,480.01 | +11500.03% |
| 06 | 362.5007 | 4,350.01 | +14400.03% |
| 07 | 435.0006 | 5,220.01 | +17300.03% |
| 08 | 507.5006 | 6,090.01 | +20200.02% |
| 09 | 580.0005 | 6,960.01 | +23100.02% |
| 10 | 652.5005 | 7,830.01 | +26000.02% |
| 11 | 725.0004 | 8,700.00 | +28900.02% |
| 12 | 797.5003 | 9,570.00 | +31800.01% |
| 13 | 870.0003 | 10,440.00 | +34700.01% |
| 14 | 942.5002 | 11,310.00 | +37600.01% |
| 15 | 1,015.00 | 12,180.00 | +40500.01% |
| 16 | 1,087.50 | 13,050.00 | +43400.00% |
| 17 | 1,160.00 | 13,920.00 | +46300.00% |
| 18 | 1,232.50 | 14,790.00 | +49200.00% |
| 19 | 1,305.00 | 15,660.00 | +52100.00% |
| 20 | 1,377.50 | 16,530.00 | +54999.99% |
Frequently Asked Questions
What is the standard deviation of a Poisson count?+
Because the variance of a Poisson distribution equals its mean μ, the standard deviation (shot noise) is simply √μ = √(λt).
What is the probability of observing zero events in window t?+
By the Poisson PMF P(k) = μ^k e^(-μ) / k!, the probability of k = 0 events is e^(-μ).
How is the Poisson distribution connected to the Exponential distribution?+
If discrete event counts follow a Poisson process with rate λ, the continuous waiting time between consecutive events follows an Exponential distribution with mean 1/λ.