基本概率分布Basic Concept of Probability Distributions 7: Uniform Distribution

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The probability density function of the uniform distribution is $$f(x; alpha, eta) = egin{cases}{1overeta-alpha} & mbox{if} alpha < x < eta\ 0 & mbox{otherwise} end{cases} $$ The cumulative distribution function of the uniform distribution is $$F(x) = egin{cases}0 & xleqalpha \ {x-alphaover eta-alpha} & alpha < x < eta\ 1 & x geq eta end{cases}$$

Proof:

$$ egin{align*} int_{-infty}^{infty}f(x; alpha, eta) dx &= int_{alpha}^{eta}{1overeta-alpha} dx\ &= {xovereta-alpha}Big|_{alpha}^{eta}\ &= {etaovereta-alpha} - {alphaovereta-alpha}\ &= 1 end{align*} $$ And $$ egin{align*} F(x; alpha, eta) &= int_{-infty}^{x}f(x; alpha, eta) dx\ &= int_{-infty}^{x}{1overeta-alpha} dx\ &= {xovereta-alpha}Big|_{alpha}^{x}\ &= {x - alphaovereta-alpha} end{align*} $$

Mean

The expected value is $$mu = E[X] = {eta + alpha over 2}$$

Proof:

$$ egin{align*} E[X] &= int_{-infty}^{infty}xf(x; alpha, eta) dx\ &= int_{alpha}^{eta}{xovereta-alpha} dx\ &= {x^2over2(eta - alpha)}Big|_{alpha}^{eta}\ &= {eta^2-alpha^2over2(eta-alpha)}\ &= {eta + alpha over 2} end{align*} $$

Variance

The variance is $$sigma^2 = mbox{Var}(X) = {(eta - alpha)^2 over 12}$$

Proof:

$$ egin{align*} Eleft[X^2 ight] &= int_{-infty}^{infty}x^2f(x;alpha, eta) dx\ &= int_{alpha}^{eta}{x^2overeta-alpha} dx\ &= {x^3over 3(eta - alpha)}Big|_{alpha}^{eta}\ &= {eta^3 - alpha^3over 3(eta - alpha)}\ &= {eta^2 + alphaeta + alpha^2over 3} end{align*} $$ Hence $$ egin{align*} mbox{Var}(X) &= Eleft[X^2 ight] - E[X]^2\ &= {eta^2 + alphaeta + alpha^2over 3} - {alpha^2+2alphaeta +eta^2 over 4}\ &= {eta^2 + alpha^2 -2alphaeta over 12}\ &= {(eta - alpha) ^2 over 12} end{align*} $$

Examples

1. If $X$ is uniformly distributed over $(0, 10)$, calculate the probability that (a) $X < 3$, (b) $X > 6$, (c) $3 < X < 8$.  

Solution:

The uniform density function is $f(x) = {1over 10}$, for $xin (0, 10)$.

(a) $$P(X < 3) = int_{0}^{3}{1over10} dx = {3over10}$$(b) $$P(X > 6) = int_{6}^{10}{1over10} dx = {4over10} = {2over5}$$ (c) $$P(3 < X < 8) = int_{3}^{8}{1over10} dx = {5over10} = {1over2}$$

2. Buses arrives at a specified stop at 15-minute interval starting at 7 A.M. That is, they arrive at 7, 7:15, 7:30, 7:45, and so on. If a passenger arrives at the stop at a time that is uniformly distributed between 7 and 7:30, find the probability that he waits (a) less than 5 minutes for a bus; (b) more than 10 minutes for a bus.

Solution:

Let $X$ be the number of minutes past 7 that the passenger arrives at the stop. The uniform density function is $f(x) = {1over 30}$, for $xin (0, 30)$.

(a) The passenger would have to wait less than 5 minutes if he arrives between 7:10 and 7:15 or between 7:25 and 7:30. $$P(10 < X < 15) + P(25 < X < 30) = int_{10}^{15}{1over30} dx + int_{25}^{30}{1over30} dx = {1over3}$$ (b) The passenger would have to wait more than 10 minutes if he arrives between 7 and 7:05 or between 7:15 and 7:20. $$P(0 < X < 5) + P(15 < X < 20) = int_{0}^{5}{1over30} dx + int_{15}^{20}{1over30} dx = {1over3}$$

Reference

1. Ross, S. (2010). A First Course in Probability (8th Edition). Chapter 5. Pearson. ISBN: 978-0-13-603313-4.