Exponential distribution
The exponential distribution is a continuous probability distribution defined on the non-negative real numbers, with probability density function f(x; \lambda) = \lambda e^{-\lambda x} for x \geq 0 and rate parameter \lambda > 0, modeling the waiting time until the first event in a Poisson process where events occur continuously and independently at a constant average rate \lambda.[1] It is distinguished by its memoryless property, which states that the conditional probability of the waiting time exceeding x + y given that it has already exceeded x equals the unconditional probability of exceeding y, for x, y > 0.[2] This property implies a constant hazard rate of \lambda, making the exponential distribution the only continuous distribution with a failure rate independent of time.[2] Key statistical properties include a mean of $1/\lambda and variance of $1/\lambda^2, with the cumulative distribution function given by F(x; \lambda) = 1 - e^{-\lambda x} for x \geq 0.[3][2] The moment-generating function is M(t) = \lambda / (\lambda - t) for t < \lambda.[3] These characteristics position the exponential distribution as a foundational model in stochastic processes, where it serves as the interarrival time distribution for the Poisson process.[1] The exponential distribution finds extensive applications across disciplines, including queueing theory for modeling customer arrival intervals, reliability engineering for constant-failure-rate components such as electronic systems, and survival analysis for lifetimes or time-to-event data in biological and medical contexts.[1][2][4] It also approximates processes like radioactive decay and photon emissions, where events follow a Poisson pattern.[2]Definitions
Probability Density Function
The probability density function (PDF) of the exponential distribution with rate parameter \lambda > 0 is given by f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0, and f(x; \lambda) = 0 for x < 0.[1][5][6] This PDF exhibits an exponential decay, beginning at a maximum value of \lambda when x = 0 and asymptotically approaching 0 as x increases to infinity, resulting in a right-skewed curve that is strictly positive over the non-negative real line.[5][6] The support is confined to x \geq 0, reflecting the distribution's application to non-negative quantities such as durations or waiting times.[1][5] The parameter \lambda represents the instantaneous rate of occurrence of an event, where higher values of \lambda correspond to a steeper initial decay and more frequent events on average.[1][6] In the context of a Poisson process, the exponential distribution arises naturally as the distribution of inter-arrival times between successive events, with \lambda denoting the average rate of the process.[1][5]Cumulative Distribution Function
The cumulative distribution function (CDF) of an exponential random variable X with rate parameter \lambda > 0 is given by F(x; \lambda) = P(X \leq x). It is derived by integrating the probability density function (PDF) f(t) = \lambda e^{-\lambda t} for t \geq 0 from 0 to x: F(x; \lambda) = \int_0^x \lambda e^{-\lambda t} \, dt = \left[ -e^{-\lambda t} \right]_0^x = 1 - e^{-\lambda x}, \quad x \geq 0, with F(x; \lambda) = 0 for x < 0.[7][1] This CDF exhibits key properties: F(0; \lambda) = 0, \lim_{x \to \infty} F(x; \lambda) = 1, and it is strictly increasing and continuous on [0, \infty). The PDF can be recovered as the derivative of the CDF, f(x; \lambda) = \frac{d}{dx} F(x; \lambda).[7][8] In probability calculations, the CDF directly computes P(X \leq x) = F(x; \lambda). The complementary survival function, S(x; \lambda) = P(X > x) = 1 - F(x; \lambda) = e^{-\lambda x} for x \geq 0, quantifies the probability of exceeding x.[7] Graphically, the CDF traces a smooth S-shaped curve, originating at (0, 0) and asymptotically approaching 1 as x grows, reflecting the accumulation of probability over the positive real line.[9]Parameterizations
The exponential distribution is commonly parameterized using a rate parameter \lambda > 0, which represents the number of events per unit time, such as arrivals or failures. In this standard rate parameterization, the mean of the distribution is $1/\lambda.[10][11] An equivalent scale parameterization uses \beta > 0, defined as the mean lifetime or expected value, where \beta = 1/\lambda. The probability density function in this form is given by f(x; \beta) = \frac{1}{\beta} e^{-x/\beta}, \quad x \geq 0. [10][11] The conversion between parameterizations is straightforward: \lambda = 1/\beta, ensuring equivalence in the moments and cumulative distribution function across both forms.[10][12] The rate parameterization with \lambda is prevalent in modeling Poisson processes, where it directly corresponds to the intensity of event occurrences.[13] In contrast, the scale parameterization with \beta is more common in reliability engineering, emphasizing durations like time to failure under constant hazard rates.[14][15] The rate form is typically preferred for high-frequency events, such as queueing arrivals, while the scale form suits analyses of prolonged durations, like component lifetimes.[12][14]Moments and Basic Properties
Mean, Variance, and Higher Moments
The expected value of an exponential random variable X with rate parameter \lambda > 0 is given by E[X] = \frac{1}{\lambda}. This follows from the definition of the probability density function f(x) = \lambda e^{-\lambda x} for x \geq 0, where the mean is computed as the integral \int_0^\infty x \lambda e^{-\lambda x} \, dx. Using integration by parts, let u = x and dv = \lambda e^{-\lambda x} \, dx, so du = dx and v = -e^{-\lambda x}, yielding \left[ -x e^{-\lambda x} \right]_0^\infty + \int_0^\infty e^{-\lambda x} \, dx = 0 + \frac{1}{\lambda} = \frac{1}{\lambda}.[16] In the scale parameterization, where the density is f(x) = \frac{1}{\beta} e^{-x/\beta} for scale parameter \beta > 0, the mean is E[X] = \beta, with \beta = 1/\lambda. The variance is \text{Var}(X) = E[X^2] - (E[X])^2. First, E[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x} \, dx, which by repeated integration by parts (or recognizing it as the second moment of a Gamma(2, \lambda) distribution) equals \frac{2}{\lambda^2}. Thus, \text{Var}(X) = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}. In scale form, \text{Var}(X) = \beta^2.[16][17] The higher-order moments are E[X^k] = \frac{k!}{\lambda^k} for positive integer k. This general formula arises from the integral \int_0^\infty x^k \lambda e^{-\lambda x} \, dx = \lambda \cdot \frac{\Gamma(k+1)}{\lambda^{k+1}} = \frac{k!}{\lambda^k}, since \Gamma(k+1) = k! for integer k, leveraging the Gamma function representation of the exponential distribution as a special case of the Gamma(1, $1/\lambda) family. Alternatively, the moment-generating function M(t) = \frac{\lambda}{\lambda - t} for t < \lambda yields the k-th moment as the k-th derivative evaluated at t=0, confirming the factorial form. In scale parameterization, E[X^k] = k! \beta^k.[16] The coefficient of variation, defined as \text{CV}(X) = \frac{\sqrt{\text{Var}(X)}}{E[X]}, equals 1 for the exponential distribution, since \sqrt{1/\lambda^2} / (1/\lambda) = 1. This unit value indicates that the standard deviation equals the mean, reflecting the high relative variability inherent in the distribution's heavy right tail.[18] The skewness, measuring asymmetry, is \gamma_1 = \frac{E[(X - \mu)^3]}{\sigma^3} = 2, where \mu = E[X] and \sigma^2 = \text{Var}(X). This positive value of 2 underscores the exponential distribution's right-skewed nature, computed using the third central moment derived from E[X^3] = \frac{6}{\lambda^3}: E[(X - \mu)^3] = E[X^3] - 3\mu E[X^2] + 2\mu^3 = \frac{6}{\lambda^3} - 3 \cdot \frac{1}{\lambda} \cdot \frac{2}{\lambda^2} + 2 \left(\frac{1}{\lambda}\right)^3 = \frac{2}{\lambda^3}, so \gamma_1 = \frac{2/\lambda^3}{(1/\lambda^2)^{3/2}} = 2.[17][18]Median and Quantiles
The median of an exponential random variable with rate parameter \lambda > 0 is the value m such that the cumulative distribution function F(m) = 0.5, given by m = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}.[19][20] The general quantile function, or inverse cumulative distribution function, for the exponential distribution is Q(p) = -\frac{\ln(1-p)}{\lambda} for p \in (0,1), which provides the value x such that F(x) = p.[19][21] This follows from solving the cumulative distribution function equation $1 - e^{-\lambda x} = p for x, yielding e^{-\lambda x} = 1 - p, \lambda x = -\ln(1-p), and thus x = -\frac{\ln(1-p)}{\lambda}.[21] The first quartile is Q(0.25) = -\frac{\ln(0.75)}{\lambda} \approx \frac{0.288}{\lambda} and the third quartile is Q(0.75) = -\frac{\ln(0.25)}{\lambda} = \frac{\ln 4}{\lambda} \approx \frac{1.386}{\lambda}.[19] The interquartile range, the difference between the third and first quartiles, is therefore \frac{\ln 3}{\lambda} \approx \frac{1.099}{\lambda}.[19] Due to the positive skewness of the exponential distribution, the median is less than the mean, with \frac{\ln 2}{\lambda} < \frac{1}{\lambda}.[19]Key Characteristics
Memorylessness Property
The memoryless property of the exponential distribution states that the conditional probability of the random variable X exceeding a sum s + t, given that it already exceeds s, equals the unconditional probability of exceeding t, for all s, t > 0:P(X > s + t \mid X > s) = P(X > t). [6] This property can be proven using the survival function, which for an exponential random variable with rate parameter \lambda > 0 is P(X > x) = e^{-\lambda x} for x > 0. Substituting into the conditional probability yields
P(X > s + t \mid X > s) = \frac{P(X > s + t)}{P(X > s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t). [22] Among continuous distributions supported on the positive reals, the exponential distribution is the only one exhibiting this memoryless property.[23] The proof involves showing that the memoryless condition implies the survival function satisfies S(s + t) = S(s)S(t), whose general solution for continuous cases is the exponential form S(x) = e^{-\lambda x}.[24] The memoryless property implies that the distribution exhibits no aging: the expected remaining lifetime is independent of the time already elapsed, making it suitable for modeling phenomena where past duration does not influence future behavior.[25] This independence of elapsed time from remaining lifetime underscores the distribution's lack of "memory" of prior events.[26] The memoryless property forms the foundation for continuous-time Markov chains, where holding times in states follow exponential distributions to ensure the Markov property—that future states depend only on the current state, not the history.[27]