Chapter 5. Continuous Probability Distributions

Transcription

1 Chapter 5. Continuous Probability Distributions Sections 5.4, 5.5: Exponential and Gamma Distributions Jiaping Wang Department of Mathematical Science 03/25/2013, Monday

2 Outline Exponential: PDF and CDF Exponential: Mean and Variance Gamma: PDF and CDF Gamma: Mean and Variance More Examples

3 Part 1. Exponential: PDF and CDF

4 Probability Density Function In general, the exponential density function is given by 1 f x = e x/θ, x 0 θ 0, ooooooooo Where the parameter θ is a constant (θ>0) that determines the rate at which the curve decreases. θ = 2 θ = 1/2

5 Cumulative Distribution Function The exponential CDF is given as 0, x < 0 F x = 1 e x /θ, x 0 θ = 2 θ = 1/2

6 Part 2. Mean and Variance

7 Gamma Function The gamma function Γ(α) is given as Γ α = x α 1 e x dd 0 We can show that Γ α + 1 = αγ α So Γ n = n 1 Γ n 2 = = n 1! Specially, Γ 1/2 = π

8 Mean and Variance E X = xx x dd = x θ eee x θ dd = 1 θ x eee x θ dd = 1 θ Γ 2 θ2 = θ. 0 0 E X 2 = x2 θ eee x θ dd = 1 θ Γ 3 θ3 = Then we have V(X)=E(X 2 )-E 2 (X)=2θ 2 - θ 2 = θ 2. 0

9 Example 5.9 A sugar refinery has three processing plants, all of which receive raw sugar in bulk. The amount of sugar that one plant can process in one day can be modeled as having an exponential distribution with a mean of 4 tons for each of the three plants. If the plants operate independently, find the probability that exactly two of the three plants will process more than 4 tons on a given day. Answer: The probability that any given plant will process more than 4 tons a day, with X representing the amount used, is p = P X > 4 = 1 P X 4 = 1 F 4 = 1 1 exp 4 = 4 exp 1 = 0.37 As the plants operate independently, the problem is to find the probability of two successes out of three tries with p=0.37, which is a binomial distribution, so P(Exactly two of three plants use more than 4 tons)= = 0.26.

10 Example 5.10 Consider a particular plant in Example 5.9. How much raw sugar should be stocked for that plant each day so that the chance of running out of product is only 0.05? Answer: Let a denote the amount to be stocked. Because the amount to be used X has an exponential distribution, so that P X > a = 1 P X a = 1 1 exp a = exp a 4 4 So we choose a with P(X>a)=exp(-a/4)=0.05 a=11.98 (tons).

11 Properties 1. Memoryless: P X > a + b X > a = P X>a+b P X>a = 1 1 exp a+b θ 1 1 exp a θ = exp b θ = 1 F b = P(X > b) 2. Relation with Poisson distribution: Assume a Poisson distribution with λ events per hour, so in t hours, the number of events, Y, follows a Poisson with mean λt. Now we start at time zero and ask how long do I have to wait to see the 1 st event occur? Let X denote the length of time until 1 st event occurs. P X > t = P Y = 0 oo ttt iiiiiiii 0, t = λt 0 exp λt 0! = exp λt P(X t)=1-exp(- λt) f t = 1 exp 1, t > 0 which means the θ θ interval time between two consecutive events in Poisson distribution follows the exponential distribution.

12 Part 3. Gamma: PDF

13 Probability Density Function In general, the Gamma density function is given by 1 x f x = α 1 exp ( x ), x 0 Γ α β α β 0, ooooooooo Where the parameters α and β are constants (α >0, β>0) that determines the shape of the curve.

14 Part 4. Mean and Variance

15 E X = 1 xx x dd = 0 x 1 x α eee x Γ α β α 0 dd β x α 1 eee x dd = Γ α β α β = 1 Γ α β α Γ α + 1 β α + 1 = αα Similary, we can find E(X 2 ) = α(α + 1)β 2, so V(X) = E(X 2 ) E 2 (X) = αα 2. Suppose Y = X i with X 1, X 2,, X n being independent Gamma variables with parameters α and β, then E(Y) = nαα, V(Y) = nαα 2.

16 Example 5.11 A certain electronic system has a life length of X1, which has an exponential distribution with a mean of 450 hours. The system is supported by an identical backup system that has a life length of X2. The backup system takes over immediately when the system fails. If the system operate independently, find the probability distribution and expected value for the total life length of the primary and backup systems. Answer: Let Y denote the total life length, Y= X1+X2, where X1 and X2 are Independent exponential random variable with mean β=450. So Y is a gamma Distribution with α=2 and β=450, that is, 1 yyyy y f y =, y > 0 Γ , ooooooooo Then the mean E(Y)=αβ=2(450)=900.

17 Example 5.12 Suppose that the length of time X needed to conduct a periodic maintenance check on a pathology lab s microscope (known from previous experience) follows a gamma distribution with α=3 and β=2 (minutes). Suppose that a new repairperson requires 20 minutes to check a particular microscope. Does this time required to perform a maintenance check seem our of line with prior experience? Answer: so μ=e(x)=αβ=6, σ 2 =V(X)=αβ 2 =12, the standard deviation σ=3.446, When x=20 minutes required from the repairperson, the deviation is 20-6=14 minutes, Which exceeds the mean 6 by k=14/3.446 standard deviations, so based on the Tschebysheff s inequality, we have P( X-6 14) (3.446/14) 2 =0.06, which is really small Probability, so we can say it is out of line with prior experience.

18 Part 3. More Examples

19 Additional Example 1 An insurance policy reimburses dental expense, X, up to a maximum benefit of 250. The probability density function for X is: where c is a constant. Calculate the median benefit for this policy. Answer: If P(X>a)=1/2, then a is a median. So c=250. As F(x)=1-exp(-x/250), we have 1-exp(-x/250)=1/2 x=250[ln(2)] =

20 Additional Example 2 Let X be an exponential random variable such that P(X>2) = 2P(X>4). Find the variance of X. Answer: Let the distribution function F x = 1 exp x, based on P(X>2)=2P(X>4), θ we have 1-F(2)=2(1-F(4)) 1-(1-exp(-2/θ))=2(1-(1-exp(-4/θ)) exp(-2/θ)=2exp(-4/θ) -2/θ=ln(2)-4/θ θ = 2/ln(2) V(X)=[2/ln(2)] 2.

21 Additional Example 3 If X has probability density function given by Find the mean and variance. Answer: Change it to the standard form with α=3, β=/12, so we can find E(X)=αβ=3/2, V(X)=αβ 2 =3/4.