基本概率分布Basic Concept of Probability Distributions 6: Exponential Distribution

PDF version

PDF & CDF

The exponential probability density function (PDF) is $$f(x; lambda) = egin{cases}lambda e^{-lambda x} & xgeq0\ 0 & x < 0 end{cases}$$ The exponential cumulative distribution function (CDF) is $$F(x; lambda) = egin{cases}1 - e^{-lambda x} & xgeq0\ 0 & x < 0 end{cases}$$

Proof:

$$ egin{align*} F(x; lambda) &= int_{0}^{x}f(x; lambda) dx\ &= int_{0}^{x}lambda e^{-lambda x} dx \ &= lambdacdotleft(-{1overlambda} ight)int_{0}^{x}e^{-lambda x} d(-lambda x)\ &= -e^{-lambda x}Big|_{0}^{x}\ &= 1 - e^{-lambda x} end{align*} $$ And $$F(infty) = 1$$

Mean

The expected value is $$mu = E[X] = {1overlambda}$$

Proof:

$$ egin{align*} Eleft[X^k ight] &= int_{0}^{infty}x^kf(x; lambda) dx\ &= int_{0}^{infty}x^klambda e^{-lambda x} dx\ &= -x^ke^{-lambda x}Big|_{0}^{infty} + int_{0}^{infty}e^{-lambda x}kx^{k-1} dxquadquadquadquad(mbox{integrating by parts})\ &= 0 + {kover lambda}int_{0}^{infty}x^{k-1}lambda e^{-lambda x} dx\ &= {koverlambda}Eleft[X^{k-1} ight] end{align*} $$ Using the integrating by parts: $$u= x^kRightarrow du = kx^{k-1} dx, dv = lambda e^{-lambda x}Rightarrow v = intlambda e^{-lambda x} dx = -e^{-lambda x}$$ $$implies int x^klambda e^{-lambda x} dx =uv - int vdu = -x^ke^{-lambda x} + int e^{-lambda x}kx^{k-1} dx$$ Hence setting $k=1$: $$E[X]= {1overlambda}$$

Variance

The variance is $$sigma^2 = mbox{Var}(X) = {1overlambda^2}$$

Proof:

$$ egin{align*} Eleft[X^2 ight] &= {2overlambda} E[X] quadquad quadquad (mbox{setting} k=2)\ &= {2overlambda^2} end{align*} $$ Hence $$ egin{align*} mbox{Var}(X) &= Eleft[X^2 ight] - E[X]^2\ &= {2overlambda^2} - {1overlambda^2}\ &= {1overlambda^2} end{align*} $$

Examples

1. Let $X$ be exponentially distributed with intensity $lambda$. Determine the expected value $mu$, the standard deviation $sigma$, and the probability $Pleft(|X-mu| geq 2sigma ight)$. Compare with Chebyshev's Inequality.

Solution:

$$mu = {1overlambda}, sigma = {1overlambda}$$ The probability that $X$ takes a value more than two standard deviations from $mu$ is $$ egin{align*} Pleft(|X - mu| geq 2sigma ight) &= Pleft(X geq {3over lambda} ight)\ &= 1-Fleft({3overlambda} ight)\ &= e^{-3}= 0.04978707 end{align*} $$ Chebyshev's Inequality gives the weaker estimation $$Pleft(|X - mu| geq 2sigma ight) leq {1over4} = 0.25$$

2. Suppose that the length of a phone call in minutes is an exponential random variable with parameter $lambda = {1over10}$. If someone arrives immediately ahead of you at a public telephone booth, find the probability that you will have to wait (a) more than 10 minutes; (b) between 10 and 20 minutes.

Solution:

Let $X$ be the length of the call made by the person in the booth. And $$f(x) = {1over10}e^{-{1over10}x}, F(x) = 1-e^{-{1over10}x}$$ (a) $$ egin{align*} P( X > 10) &= 1 - P(X leq 10)\ &= 1 - F(10)\ &= e^{-1}= 0.3678794 end{align*} $$ (b) $$ egin{align*} P(10 < X < 20) &= P(X < 20) - P(X < 10)\ &= F(20) - F(10)\ &= (1-e^{-2}) - (1 - e^{-1})\ &= e^{-1} - e^{-2} = 0.2325442 end{align*} $$

Reference

1. Ross, S. (2010). A First Course in Probability (8th Edition). Chapter 5. Pearson. ISBN: 978-0-13-603313-4.
2. Brink, D. (2010). Essentials of Statistics: Exercises. Chapter 5. ISBN: 978-87-7681-409-0.