Exponential Distribution Calculator
Enter a rate parameter and a value to compute the probability density (PDF), cumulative probability (CDF), survival probability, and all key statistics for the exponential distribution. Switch between rate (lambda) and mean (1/lambda) input, add a second value for interval probability, or use the inverse mode to find the x that gives a target probability.
What is the exponential distribution?
The exponential distribution models the time between successive independent events that occur at a constant average rate. It is the continuous-time counterpart to the discrete geometric distribution and plays a central role in queuing theory, reliability engineering, and survival analysis. If phone calls arrive at a call center at an average rate of 3 per hour, the waiting time between consecutive calls follows an exponential distribution with lambda = 3.
PDF, CDF, and the survival function
The probability density function (PDF) is f(x) = lambda * e^(-lambda * x). It gives the relative likelihood that the waiting time equals exactly x, and it decreases monotonically from lambda at x = 0. The cumulative distribution function (CDF) is F(x) = 1 - e^(-lambda * x), giving the probability that the event occurs by time x. The survival function S(x) = e^(-lambda * x) is the complement, giving the probability of still waiting at time x. To find the probability of the event falling in an interval, subtract the CDF values: P(x1 < X < x2) = F(x2) - F(x1).
Mean, median, and the memoryless property
The mean waiting time is 1 / lambda. The median is ln(2) / lambda, which is always shorter than the mean because the right-skewed distribution has a long tail. Importantly, the standard deviation equals the mean (both equal 1 / lambda), making the coefficient of variation exactly 1. The exponential distribution is the only continuous distribution with the memoryless property: P(X > s + t | X > s) = P(X > t). This means the probability of waiting an additional time t does not depend on how long you have already waited, a property that distinguishes it from distributions with aging or wear effects.
Inverse CDF and practical applications
The inverse CDF (quantile function) answers the question: "How long must I wait to be p% certain the event has occurred?" The answer is x = -ln(1 - p) / lambda. For example, with lambda = 0.5, to be 90% certain: x = -ln(0.1) / 0.5 = 4.61 units. Common applications include modeling the lifespan of components under constant hazard (reliability engineering), inter-arrival times in queuing systems (call centers, network packets), and time-to-event in survival analysis when the hazard rate is assumed constant.
Exponential distribution: key formulas
| Quantity | Formula | Description |
|---|---|---|
| f(x) = lambda * e^(-lambda * x) | Probability density at x | |
| CDF | F(x) = 1 - e^(-lambda * x) | P(X <= x) |
| Survival | S(x) = e^(-lambda * x) | P(X > x) |
| Interval | F(x2) - F(x1) | P(x1 < X < x2) |
| Inverse CDF | x = -ln(1-p) / lambda | Quantile function |
| Mean | mu = 1 / lambda | Expected value |
| Median | ln(2) / lambda | 50th percentile |
| Variance | sigma^2 = 1 / lambda^2 | Spread |
| Std Dev | sigma = 1 / lambda | Equal to the mean |
All formulas assume a rate parameter lambda > 0 and x >= 0.
Frequently asked questions
What does the rate parameter lambda mean?
Lambda (the Greek letter for the rate parameter) is the average number of events per unit of time or distance. If buses arrive three times per hour on average, lambda = 3. The mean waiting time between buses is then 1 / lambda = 1/3 hour, or about 20 minutes.
How is the exponential distribution related to the Poisson distribution?
If the number of events in a fixed time window follows a Poisson distribution with rate lambda, then the waiting time between consecutive events follows an exponential distribution with the same rate lambda. The two distributions are two views of the same Poisson process: one counting events, the other measuring gaps between them.
What is the memoryless property and why does it matter?
The memoryless property states that P(X > s + t | X > s) = P(X > t): given that you have already waited s units of time, the probability of waiting an additional t units is the same as the probability of waiting t units from the very start. This is realistic for phenomena without aging, like radioactive decay, but unrealistic for contexts where wear or fatigue increase failure rates over time.
Why is the standard deviation equal to the mean for the exponential distribution?
Both the mean and the standard deviation equal 1 / lambda, so the coefficient of variation (standard deviation / mean) is exactly 1. This is a defining characteristic of the exponential family that does not hold for the normal, Gamma, or Weibull distributions. It implies that the distribution is always highly right-skewed, with the skewness coefficient equal to 2.
When should I use the Gamma distribution instead?
Use the Gamma distribution when you are modeling the total waiting time until the k-th event in a Poisson process. The sum of k independent exponential random variables with the same rate follows a Gamma distribution with shape k and rate lambda. The exponential distribution is the special case where k = 1.
How do I interpret the PDF value at a specific x?
For a continuous distribution like the exponential, the probability of X equaling any exact value is zero. The PDF value f(x) is a density: the probability that X falls in a tiny interval around x is approximately f(x) times the width of that interval. High PDF values indicate regions where outcomes are more concentrated. The total area under the PDF curve always equals 1.