Skip to the content

Topics Covered

PDF CDF Moments MGF / CF / CGF Skewness & Kurtosis Memoryless Property
On this page
  1. 1. Definition & PDF
  2. 2. Distribution Function (CDF)
  3. 3. Moments via MGF
  4. 4. Other Generating Functions
  5. 5. Skewness & Kurtosis
  6. 6. Memoryless (Lack-of-Memory) Property
  7. 7. Weibull Distribution
  8. 8. Laplace (Double-Exponential) Distribution
  9. 9. Worked Examples
  10. Key Take-aways

1. Definition & PDF

DEFINITION

A continuous random variable \(X\) follows an Exponential distribution with parameter \(\lambda > 0\) if its PDF is

\[ f(x) \;=\; \begin{cases} \lambda e^{-\lambda x}, & x \ge 0\\ 0, & x < 0. \end{cases} \]

Notation: \(X \sim \text{Exp}(\lambda)\). The parameter \(\lambda\) is the rate; the mean is \(1/\lambda\).

It models the time between events in a Poisson process — e.g., time between two consecutive arrivals.

λ 0 PDF: λe−λx
Decreasing from λ; long right tail
1 0 CDF: 1 − e−λx
Approaches 1 asymptotically

2. Distribution Function (CDF)

\[ F(x) \;=\; \int_0^x \lambda e^{-\lambda t}\,dt \;=\; 1 - e^{-\lambda x}, \quad x \ge 0. \]

Hence \(P(X > x) = e^{-\lambda x}\).

3. Moments via MGF

MGF \[ M_X(t) \;=\; \int_0^\infty e^{tx} \lambda e^{-\lambda x}\,dx \;=\; \dfrac{\lambda}{\lambda - t}, \quad t < \lambda. \]

Differentiate:

\(M'(t) = \dfrac{\lambda}{(\lambda-t)^2};\; M'(0) = 1/\lambda = E(X)\).

\(M''(0) = 2/\lambda^2 = E(X^2)\).

\[ E(X) = \dfrac{1}{\lambda}, \qquad \text{Var}(X) = \dfrac{1}{\lambda^2}. \]

Mean = SD for exponential — a defining property.

4. Other Generating Functions

\[ \phi_X(t) = \dfrac{\lambda}{\lambda - it}, \qquad K_X(t) = \ln \lambda - \ln(\lambda - t). \]

Higher central moments: \(\mu_r = \int_0^\infty (x - 1/\lambda)^r \lambda e^{-\lambda x}\,dx\).

\(\mu_2 = 1/\lambda^2;\; \mu_3 = 2/\lambda^3;\; \mu_4 = 9/\lambda^4\).

5. Skewness & Kurtosis

\[ \beta_1 = \dfrac{\mu_3^2}{\mu_2^3} = 4, \qquad \beta_2 = \dfrac{\mu_4}{\mu_2^2} = 9. \] \[ \gamma_1 = 2, \qquad \gamma_2 = 6. \]

Strongly right-skewed and leptokurtic. (Independent of \(\lambda\).)

6. Memoryless (Lack-of-Memory) Property

STATEMENT

The Exponential distribution is the only continuous distribution with the lack-of-memory property:

\[ P(X > s + t \mid X > s) \;=\; P(X > t), \quad s, t \ge 0. \]

Proof. Recall the exponential density \(f(x) = \lambda e^{-\lambda x}\) for \(x \ge 0\).

  1. Find the tail probability. Integrate the density from \(x\) to \(\infty\): \[ P(X > x) = \int_{x}^{\infty} \lambda e^{-\lambda u}\,du = \Big[-e^{-\lambda u}\Big]_{x}^{\infty} = 0 - (-e^{-\lambda x}) = e^{-\lambda x}. \]
  2. Apply the definition of conditional probability. Since \(s+t \ge s\), the event \(\{X > s+t\}\) already lies inside \(\{X > s\}\), so their intersection is \(\{X > s+t\}\) and \[ P(X > s+t \mid X > s) = \dfrac{P(X > s+t)}{P(X > s)}. \]
  3. Substitute the tail from step 1. \[ P(X > s+t \mid X > s) = \dfrac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda(s+t)-(-\lambda s)} = e^{-\lambda t}. \]
  4. Recognise the result. By step 1, \(e^{-\lambda t} = P(X > t)\). Hence \(P(X > s+t \mid X > s) = P(X > t)\) — the lack-of-memory property. \(\blacksquare\)

Interpretation: Past waiting time is irrelevant — the future wait has the same distribution.

7. Weibull Distribution

The Weibull distribution generalises the exponential by allowing the failure rate to change with time — the workhorse of reliability and life-testing. With shape \(k > 0\) and scale \(\lambda > 0\):

\[ f(x) = \dfrac{k}{\lambda}\left(\dfrac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^{k}},\quad x \ge 0; \qquad F(x) = 1 - e^{-(x/\lambda)^{k}}. \]

\(k = 1\) gives the exponential (constant hazard); \(k > 1\) an increasing hazard (wear-out); \(k < 1\) a decreasing hazard (infant mortality). Mean \(= \lambda\,\Gamma(1 + 1/k)\).

EXAMPLE

For \(k = 2,\ \lambda = 10\): \(P(X > 10) = e^{-(10/10)^{2}} = e^{-1} = 0.368\).

Weibull density (λ = 1): shape k sets the hazard 0 0.5 1 1.5 2 2.5 3 k = 0.5 (infant mortality) k = 1 (exponential) k = 1.5 k = 3.4 (≈ normal)
Fig 7.1 — The shape \(k\) reshapes the curve entirely: \(k<1\) decreases from infinity (early-life failures), \(k=1\) is the exponential (constant hazard), \(k>1\) rises to a peak (wear-out), and \(k\approx3.4\) is nearly symmetric.

8. Laplace (Double-Exponential) Distribution

The Laplace distribution is a two-sided exponential — an exponential decay on each side of a centre \(\mu\):

\[ f(x) = \dfrac{1}{2b}\,e^{-|x-\mu|/b}, \quad -\infty < x < \infty,\ b > 0. \]

It is symmetric about \(\mu\) (so mean = median = mode = \(\mu\)), with variance \(2b^{2}\) and heavier tails than the normal. The difference of two i.i.d. exponentials follows a Laplace distribution.

Laplace vs. normal (equal variance) -6 -4 -2 0 2 4 6 sharp peak Laplace (b=1) Normal (σ²=2)
Fig 8.1 — Matched to the same variance (\(\sigma^2=2b^2\)), the Laplace shows a sharp cusp at the centre and heavier tails than the normal — more probability both very close to the mean and far out in the tails.

9. Worked Examples

EXAMPLE 1

Time (in hours) between phone calls received at a help desk follows Exp(2). Find probability that the next call comes (i) within 30 min, (ii) after more than 1 hour.

\(\lambda = 2\) per hour. (i) \(P(X \le 0.5) = 1 - e^{-1} = 0.632\).
(ii) \(P(X > 1) = e^{-2} = 0.135\).

EXAMPLE 2 (Memoryless)

Lifetime of a bulb is Exp with mean 1000 hours. The bulb has already been used for 500 hours. Find probability it lasts more than 700 additional hours.

By memoryless: \(P(X > 1200 | X > 500) = P(X > 700) = e^{-700/1000} = e^{-0.7} \approx 0.4966\).

Key Take-aways

← Previous: Unit 1 Next: Unit 3 — Gamma & Beta →