TheoremComplete

Common Distributions - Applications

Advanced distribution theory enables sophisticated modeling in science, engineering, and finance. We explore specialized distributions and their applications.

The Beta Distribution

Definition

XBeta(α,β)X \sim \text{Beta}(\alpha, \beta) on [0,1][0,1] has PDF: f(x)=1B(α,β)xα1(1x)β1,0<x<1f(x) = \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1}, \quad 0 < x < 1

where B(α,β)=Γ(α)Γ(β)Γ(α+β)B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} is the beta function.

Parameters: α,β>0\alpha, \beta > 0 (shape parameters)

The beta distribution is extremely flexible for modeling proportions, probabilities, and percentages.

Special Cases:

  • Beta(1,1)=Uniform(0,1)(1,1) = \text{Uniform}(0,1)
  • Beta(α,1)(\alpha, 1) concentrates near 1
  • Beta(1,β)(1, \beta) concentrates near 0
  • Beta(α,β)(\alpha, \beta) symmetric about 0.5 when α=β\alpha = \beta
Example

Bayesian Analysis: Model uncertainty about a probability pp using Beta(α,β)(\alpha, \beta) prior.

If observing kk successes in nn trials (Binomial likelihood), posterior is: pdataBeta(α+k,β+nk)p | \text{data} \sim \text{Beta}(\alpha + k, \beta + n - k)

Starting with uniform prior Beta(1,1)(1,1) and observing 7 successes in 10 trials: pdataBeta(8,4)p | \text{data} \sim \text{Beta}(8, 4)

Mean estimate: 88+4=0.667\frac{8}{8+4} = 0.667

The Weibull Distribution

Definition

XWeibull(k,λ)X \sim \text{Weibull}(k, \lambda) models lifetimes with PDF: f(x)=kλ(xλ)k1e(x/λ)k,x0f(x) = \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k}, \quad x \geq 0

Parameters: k>0k > 0 (shape), λ>0\lambda > 0 (scale)

Hazard Function: h(x)=kλ(x/λ)k1h(x) = \frac{k}{\lambda}(x/\lambda)^{k-1}

  • k<1k < 1: decreasing failure rate (infant mortality)
  • k=1k = 1: constant failure rate (Exponential(λ)(\lambda))
  • k>1k > 1: increasing failure rate (aging/wear-out)
Example

Wind speed modeling: XWeibull(2,10)X \sim \text{Weibull}(2, 10) m/s (Rayleigh distribution).

Probability wind exceeds 15 m/s: P(X>15)=e(15/10)2=e2.250.105P(X > 15) = e^{-(15/10)^2} = e^{-2.25} \approx 0.105

The t-Distribution

Definition

Student's tt-distribution with ν\nu degrees of freedom: f(x)=Γ((ν+1)/2)νπΓ(ν/2)(1+x2ν)(ν+1)/2f(x) = \frac{\Gamma((\nu+1)/2)}{\sqrt{\nu\pi}\Gamma(\nu/2)}\left(1 + \frac{x^2}{\nu}\right)^{-(\nu+1)/2}

As ν\nu \to \infty, tνN(0,1)t_{\nu} \to \mathcal{N}(0,1).

Usage: Inference about means when variance unknown and sample size small.

For sample mean Xˉ\bar{X} with sample std dev SS: T=XˉμS/ntn1T = \frac{\bar{X} - \mu}{S/\sqrt{n}} \sim t_{n-1}

Example

Sample of n=10n = 10 has mean 52 and std dev 8. Test H0:μ=50H_0: \mu = 50 vs. H1:μ50H_1: \mu \neq 50.

t=52508/10=22.530.79t = \frac{52 - 50}{8/\sqrt{10}} = \frac{2}{2.53} \approx 0.79

With 9 degrees of freedom, pp-value 0.45\approx 0.45. Do not reject H0H_0.

The F-Distribution

Definition

If Uχm2U \sim \chi^2_m and Vχn2V \sim \chi^2_n independently: F=U/mV/nFm,nF = \frac{U/m}{V/n} \sim F_{m,n}

Parameters: m,nm, n (degrees of freedom)

Usage: Compare variances, ANOVA, regression F-tests.

Example

Testing equality of variances: H0:σ12=σ22H_0: \sigma_1^2 = \sigma_2^2.

Sample variances S12=25S_1^2 = 25 (n=11), S22=16S_2^2 = 16 (n=16).

F=S12S22=2516=1.5625F10,15F = \frac{S_1^2}{S_2^2} = \frac{25}{16} = 1.5625 \sim F_{10, 15}

If critical value F0.05(10,15)2.54F_{0.05}(10,15) \approx 2.54, do not reject H0H_0.

Remark

These specialized distributions—Beta for proportions, Weibull for reliability, tt and FF for inference—extend the basic toolkit. Understanding when and how to apply each distribution is essential for advanced statistical modeling and requires both theoretical knowledge and practical experience.