Exponential Distributions: PDF, CDF and Moments

• Exponential Distributions focus on the time interval between successes.
• A random variable X follows an exponential distribution if it takes only non-negative values

PDF and CDF

• Probability Density Function (PDF): f(x, θ) = θe^(-θx) for x > 0.
• x ~ exp(θ) signifies that x follows an exponential distribution with parameter θ.
• Cumulative Distribution Function (CDF): F(x) = P(X ≤ x)

Calculation of CDF

• F(x) is calculated by integrating the PDF from 0 to x.
• F(x) = ∫[0 to x] f(x, θ) dx
• This evaluates to F(x) = 1 - e^(-θx)

Moments

• The rth moment is given by μr' = E(X^r) = ∫[0 to ∞] x^r f(x) dx
• This involves integrating x^r * θe^(-θx) from 0 to ∞.
• Using the integral ∫[0 to ∞] e^(-ax) x^(λ-1) dx = Γ(λ) / a^λ
• λ = r+1, and a = θ.
• Γ(r+1) / θ^(r+1) = r! / θ^(r+1)

Calculations for r = 1, 2, 3, 4

• When r = 1, μ₁' = Γ(2) / θ = 1/θ.
• When r = 2, μ₂' = 2 / θ².
• When r = 3, μ₃' = 6 / θ³.
• When r = 4, μ₄' = 24 / θ⁴.

Central Moments

• Central Moment of order 3: μ₃ = μ₃' - 3μ₂'μ₁' + 2μ₁'³ = (6/θ³) - 3(2/θ²)(1/θ) + 2(1/θ)³ = 2/θ³.
• Central Moment of order 4: μ₄ = μ₄' - 4μ₃'μ₁' + 6μ₂'μ₁'² - 3μ₁'⁴ = (24/θ⁴) - 4(6/θ³)(1/θ) + 6(2/θ²)(1/θ)² - 3(1/θ)⁴ = 9/θ⁴.

Measures of Skewness and Kurtosis

• β₁ = μ₃² / μ₂³ = (2/θ³)² / (1/θ²)³ = 4, measures skewness.
• β₂ = μ₄ / μ₂² = (9/θ⁴) / (1/θ²)² = 9, measures kurtosis, = 9 indicates a leptokurtic curve.

Moment Generating Function (MGF)

• Mₓ(t) is defined as E[e^(tx)] = ∫[0 to ∞] e^(tx) f(x) dx.
• Mₓ(t) = ∫[0 to ∞] e^(tx)·θe^(-θx) dx.
• Simplifying, this becomes θ∫[0 to ∞] e^((t-θ)x) dx.
• Further evaluation gives θ[e^((t-θ)x) / (t-θ)] from 0 to ∞.
• Which simplifies to θ[0 - (1 / (t-θ))].
• Yielding (θ / (θ-t)) which can be rewritten as (1 - t/θ)^(-1).
• From the expansion, coefficients of t, t²/2!, and t³/3! on B.S.
• Then μ₁' = coefficient of t = 1/θ = mean.
• μ₂' = coefficient of t²/2! = 2/θ².
• Central Moment of order 2: μ₂ = μ₂' - μ₁'² = 2/θ² - (1/θ)² = 1/θ² = variance.
• μ₃' = coefficient of t³/3! = 6/θ³.
• Central Moment of order 3: μ₃ = μ₃' - 3μ₂'μ₁' + 2μ₁'³ = (6/θ³) - 3(2/θ²)(1/θ) + 2(1/θ)³ = 2/θ³.
• μ₄' = coefficient of t⁴/4! = 24/θ⁴.
• Central Moment of order 4: μ₄ = μ₄' - 4μ₃'μ₁' + 6μ₂'μ₁'² - 3μ₁'⁴ = (24/θ⁴) - 4(6/θ³)(1/θ) + 6(2/θ²)(1/θ)² - 3(1/θ)⁴ = 9/θ⁴.

Memorylessness

• Exponential Distribution lacks memory.
• If X has an Exponential Distribution, then for every constant a > 0, P(Y ≤ x | X > a) = P(X ≤ x) where Y = X - a.
• PDF of Exponential Distribution: f(x) = θe^(-θx) for x > 0.

Proof

• LHS considers P(Y ≤ x | X > a) = P(Y ≤ x ∩ X > a) / P(X > a).
• Which simplifies to P(X - a ≤ x ∩ X > a) / P(X > a).
• Further simplifies to P(x ≤ x + a ∩ X > a) / P(X > a).
• Which gives P(x ≤ x + a | a < X ≤ x + a) / P(X > a).
• P(a < X ≤ x + a) integrates the PDF from a to x + a: ∫[a to x+a] θe^(-θx) dx.
• This resolves to -[e^(-θx)] from a to x + a giving -[e^(-θ(x+a)) - e^(-θa)].
• Which simplifies to e^(-θa) - e^(-θ(x+a)).
• P(X > a) integrates the PDF from a to ∞: ∫[a to ∞] θe^(-θx) dx,
• Which is -[e^(-θx)] from a to ∞, resulting in e^(-θa).
• Substituting back and simplifying, P(Y ≤ x | X > a) = [e^(-θa) (1 - e^(-θx))] / e^(-θa) = 1 - e^(-θx).
• P(X ≤ x) integrates the PDF from 0 to x: ∫[0 to x] θe^(-θx) dx.
• Evaluation yields -[e^(-θx)] from 0 to x, giving 1 - e^(-θx).
• Since P(Y ≤ x | X > a) = P(X ≤ x), the Exponential Distribution lacks memory.