Exponential Distribution

Modeling the time until an event occurs in a Poisson process.

The "Time Between Events" Distribution

The Exponential distribution is a continuous probability distribution that models the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate.

It is the continuous analogue of the Geometric distribution. While the Geometric distribution models the number of trials until the first success, the Exponential distribution models the amount of time until the next event. In finance, it's used to model the time between trades, the time until a bond defaults, or the duration until a stock price hits a certain level.

Interactive Exponential Distribution
Adjust the rate parameter (λ) to see how the shape of the distribution changes. A higher rate means events happen more frequently, so the probability of a short waiting time is high.
Mean (μ\mu): 0.67
Variance (σ2\sigma^2): 0.44

Core Concepts

Probability Density Function (PDF)
The PDF is defined by a single rate parameter, λ.
f(x;λ)=λe−λxf(x; \lambda) = \lambda e^{-\lambda x}
  • x≥0x \ge 0 is the time variable.
  • λ>0\lambda > 0 (lambda) is the rate parameter, the average number of events per unit of time.
Mean (μ\mu): 0.67
Variance (σ2\sigma^2): 0.44
Cumulative Distribution Function (CDF)
The CDF gives the probability that the event occurs on or before time x. It is derived by integrating the PDF.
F(x)=1−e−λxF(x) = 1 - e^{-\lambda x}

The CDF, F(x)F(x), is the integral of the PDF, f(t)f(t), from 0 up to xx. This gives us the total probability accumulated up to time xx.

Mean (μ\mu): 0.67
Variance (σ2\sigma^2): 0.44
Key Derivations
Understanding how the CDF, Mean, and Variance are derived from the PDF.

Deriving the CDF

We find the CDF by integrating the PDF from 0 to x.

Step 1: Set up the Integral

The CDF, F(x)F(x), is the integral of the PDF, f(t)f(t), from 0 up to xx.

F(x)=∫0xλe−λtdtF(x) = \int_{0}^{x} \lambda e^{-\lambda t} dt

Step 2: Calculate the Integral

We find the antiderivative of λe−λt\lambda e^{-\lambda t}.

∫λe−λtdt=−e−λt\int \lambda e^{-\lambda t} dt = -e^{-\lambda t}

Step 3: Apply the Limits of Integration

Now we evaluate the antiderivative at the limits xx and 00.

F(x)=[−e−λt]0x=(−e−λx)−(−e−λ⋅0)F(x) = \left[-e^{-\lambda t}\right]_{0}^{x} = (-e^{-\lambda x}) - (-e^{-\lambda \cdot 0})
=−e−λx−(−1)= -e^{-\lambda x} - (-1)

Deriving the Expected Value (Mean)

We use integration by parts: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

Step 1: Set up the Integral

The expected value E[X]E[X] is the integral of x⋅f(x)x \cdot f(x) over its domain.

E[X]=∫0∞x⋅λe−λxdxE[X] = \int_{0}^{\infty} x \cdot \lambda e^{-\lambda x} dx

Step 2: Apply Integration by Parts

Let u=xu=x and dv=λe−λxdxdv = \lambda e^{-\lambda x} dx. Then du=dxdu=dx and v=−e−λxv = -e^{-\lambda x}.

E[X]=[−xe−λx]0∞−∫0∞(−e−λx)dxE[X] = \left[ -x e^{-\lambda x} \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-\lambda x}) dx

Step 3: Evaluate the Terms

The first term [−xe−λx]0∞\left[ -x e^{-\lambda x} \right]_{0}^{\infty} evaluates to 0. (The limit as x→∞x \to \infty is 0 by L'Hôpital's rule, and the value at 0 is 0).

The second term becomes a simple integral:

E[X]=∫0∞e−λxdx=[−1λe−λx]0∞E[X] = \int_{0}^{\infty} e^{-\lambda x} dx = \left[ -\frac{1}{\lambda} e^{-\lambda x} \right]_{0}^{\infty}
=(0)−(−1λe0)=1λ= (0) - (-\frac{1}{\lambda} e^0) = \frac{1}{\lambda}

This makes intuitive sense: if events happen at a rate of λ=2\lambda=2 per hour, you would expect to wait, on average, 1/21/2 an hour between them.

Deriving the Variance

We use the formula Var(X)=E[X2]−(E[X])2Var(X) = E[X^2] - (E[X])^2. We already have E[X]E[X], so we need to find E[X2]E[X^2].

Step 1: Find E[X²]

This requires applying integration by parts twice.

E[X2]=∫0∞x2λe−λxdxE[X^2] = \int_{0}^{\infty} x^2 \lambda e^{-\lambda x} dx

Let u=x2u = x^2 and dv=λe−λxdxdv = \lambda e^{-\lambda x} dx. Then du=2xdxdu = 2x dx and v=−e−λxv = -e^{-\lambda x}.

E[X2]=[−x2e−λx]0∞+∫0∞2xe−λxdxE[X^2] = \left[ -x^2 e^{-\lambda x} \right]_{0}^{\infty} + \int_{0}^{\infty} 2x e^{-\lambda x} dx

The first term is 0. The remaining integral is 2λ∫0∞xλe−λxdx\frac{2}{\lambda} \int_{0}^{\infty} x \lambda e^{-\lambda x} dx. We recognize the integral part as E[X]E[X], which we know is 1/λ1/\lambda.

E[X2]=2λ⋅E[X]=2λ⋅1λ=2λ2E[X^2] = \frac{2}{\lambda} \cdot E[X] = \frac{2}{\lambda} \cdot \frac{1}{\lambda} = \frac{2}{\lambda^2}

Step 2: Calculate Variance

Var(X)=E[X2]−(E[X])2=2λ2−(1λ)2=2λ2−1λ2Var(X) = E[X^2] - (E[X])^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2}
Final Variance Formula
Var(X)=1λ2Var(X) = \frac{1}{\lambda^2}

Applications