Compute P(X < x₁), mean, standard deviation — or solve the inverse problem
This page provides two calculators for the exponential distribution \( X \sim \mathrm{Exp}(\lambda) \):
Rate parameter. \( \lambda > 0 \) is the average number of events per unit time. Larger \( \lambda \) means events are more frequent (shorter mean waiting time). The exponential distribution is the continuous analogue of the geometric distribution — both are memoryless.
The exponential distribution models the time between events in a Poisson point process. Its probability density function is:
\[ f(x;\lambda) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0 \\[1em] 0 & x < 0 \end{cases} \]
| Quantity | Formula |
|---|---|
| Cumulative probability | \( P(X < x_1) = \displaystyle\int_0^{x_1} \lambda e^{-\lambda x}\,dx = 1 - e^{-\lambda x_1} \) |
| Mean \( \mu \) | \( \mu = \dfrac{1}{\lambda} \) |
| Variance \( \sigma^2 \) | \( \sigma^2 = \dfrac{1}{\lambda^2} \) |
| Standard deviation \( \sigma \) | \( \sigma = \dfrac{1}{\lambda} \) |
| Inverse / quantile | \( x_1 = -\dfrac{\ln(1-p)}{\lambda} \) |
Memoryless property. \( P(X > s + t \mid X > s) = P(X > t) \) for all \( s, t \ge 0 \). If you've already waited \(s\) units, the probability of waiting at least \(t\) more doesn't depend on \(s\). This is the continuous analogue of the geometric distribution.
Edge cases. For the forward calculator, if \(x_1 < 0\), \(P(X < x_1) = 0\). For the inverse, \(p = 1\) would give \(x_1 = \infty\) — so the inverse is defined for \(p \in [0, 1)\).
Given \( \lambda \) and \( x_1 \), compute the mean, standard deviation, and \( P(X < x_1) \).
Given \( \lambda \) and a target probability \( p \in [0, 1) \), find the value \( x_1 \) such that \( P(X < x_1) = p \). This computes the \( p \)-th quantile.
Mean and standard deviation:
\[ \mu = \sigma = \frac{1}{1} = 1 \]Cumulative probability:
\[ P(X < 2) = 1 - e^{-1 \cdot 2} = 1 - e^{-2} \approx 1 - 0.1353 = 0.8647 \]Answer: \( \mu = 1 \), \( \sigma = 1 \), \( P(X < 2) \approx 0.8647 \).
Solve \( 1 - e^{-x_1} = 0.25 \):
\[ e^{-x_1} = 0.75 \;\Rightarrow\; x_1 = -\ln(0.75) \approx 0.2877 \]Check: \( 1 - e^{-0.2877} \approx 0.25 \; \checkmark \)
A radioactive source emits particles at a rate \( \lambda = 0.5 \) per second. What is the probability of waiting less than 4 seconds for the next emission?
For \( \lambda = 0.5 \), find the median waiting time (the value \(x_1\) such that \(P(X < x_1) = 0.5\)).
Note: the exponential median is \( \ln(2)/\lambda \approx 0.693/\lambda \), which is smaller than the mean \(1/\lambda\).
For \( X \sim \mathrm{Exp}(\lambda) \), show that \( P(X > 5 \mid X > 2) = P(X > 3) \).
By definition:
\[ P(X > 5 \mid X > 2) = \frac{P(X > 5)}{P(X > 2)} = \frac{e^{-5\lambda}}{e^{-2\lambda}} = e^{-3\lambda} = P(X > 3) \; \checkmark \]This is the memoryless property: past waiting time gives no information about future waiting time.
For \(X \sim \mathrm{Exp}(\lambda)\) with rate \(\lambda > 0\): \(f(x) = \lambda e^{-\lambda x}\) for \(x \ge 0\), and 0 for \(x < 0\).
\(\mu = 1/\lambda\), \(\sigma^2 = 1/\lambda^2\), and \(\sigma = 1/\lambda\). The mean and standard deviation are equal — a distinctive property.
\(P(X > s + t \mid X > s) = P(X > t)\) for all \(s, t \ge 0\). Having waited \(s\) units already doesn't change the distribution of additional waiting time.
Given \(p \in [0, 1)\), solve \(1 - e^{-\lambda x_1} = p\) for \(x_1\): \(x_1 = -\ln(1-p)/\lambda\). The boundary \(p = 1\) gives \(x_1 = \infty\), so it's excluded.
Last updated: October 20, 2026