Download The UNIVERSITY of NORTH CAROLINA at

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Probability wikipedia , lookup

Transcript
Chapter 5. Continuous Probability
Distributions
Sections 5.4, 5.5: Exponential and Gamma
Distributions
Jiaping Wang
Department of Mathematical Science
03/25/2013, Monday
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Outline
Exponential: PDF and CDF
Exponential: Mean and Variance
Gamma: PDF and CDF
Gamma: Mean and Variance
More Examples
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 1. Exponential: PDF and
CDF
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Probability Density Function
In general, the exponential density function is given by
𝑓 π‘₯ =
1 βˆ’ π‘₯ /πœƒ
𝑒
,
πœƒ
π‘₯β‰₯0
0, π‘œπ‘‘β„Žπ‘’π‘Ÿπ‘€π‘–π‘ π‘’
Where the parameter ΞΈ is a constant (ΞΈ>0) that determines the
rate at which the curve decreases.
ΞΈ=2
ΞΈ = 1/2
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Cumulative Distribution Function
The exponential CDF is given as
0, π‘₯ < 0
/
𝐹 π‘₯ =
1 βˆ’ 𝑒 βˆ’ π‘₯ πœƒ, π‘₯ β‰₯ 0
ΞΈ=2
ΞΈ = 1/2
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 2. Mean and Variance
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Gamma Function
The gamma function Ξ“(Ξ±) is given as
∞
Ξ“ 𝛼 = 0 π‘₯𝛼 βˆ’ 1𝑒 βˆ’ π‘₯𝑑π‘₯
We can show that Ξ“ 𝛼 + 1 = 𝛼Γ 𝛼
So Ξ“ 𝑛 = 𝑛 βˆ’ 1 Ξ“ 𝑛 βˆ’ 2 = β‹― = 𝑛 βˆ’ 1 !
Specially, Ξ“ 1/2 = πœ‹
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Mean and Variance
∞
𝑬 𝑿 =
∞
𝒙𝒇 𝒙 𝒅𝒙 =
βˆ’βˆž
=
𝐄
π—πŸ
=
𝟏 ∞
𝒙
𝜽 𝟎
βˆ™ 𝒆𝒙𝒑
∞ 𝐱𝟐
𝐞𝐱𝐩
𝟎 𝛉
𝐱
βˆ’
𝛉
𝒙
βˆ’
𝜽
𝟎
𝒅𝒙 =
𝐝𝐱 =
𝟏
πšͺ
𝛉
𝒙
𝒙
𝒆𝒙𝒑 βˆ’ 𝒅𝒙
𝜽
𝜽
𝟏
Ξ“
𝜽
𝟐 𝜽𝟐 = 𝜽.
πŸ‘ π›‰πŸ‘ = πŸπ›‰πŸ.
Then we have V(X)=E(X2)-E2(X)=2ΞΈ2- ΞΈ2= ΞΈ2.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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
4
𝑝 = 𝑃 𝑋 > 4 = 1 βˆ’ 𝑃 𝑋 ≀ 4 = 1 βˆ’ 𝐹 4 = 1 βˆ’ 1 βˆ’ exp βˆ’ 4 =
exp βˆ’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)= 32 0.37 2 0.63 = 0.26.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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
π‘Ž
π‘Ž
𝑃 𝑋 > π‘Ž = 1 βˆ’ 𝑃 𝑋 ≀ π‘Ž = 1 βˆ’ 1 βˆ’ exp βˆ’ 4 = exp βˆ’ 4
So we choose a with P(X>a)=exp(-a/4)=0.05οƒ  a=11.98 (tons).
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Properties
1. Memoryless: 𝑃 𝑋 > π‘Ž + 𝑏 𝑋 > π‘Ž =
exp
𝑏
βˆ’
πœƒ
𝑃 𝑋>π‘Ž+𝑏
𝑃 𝑋>π‘Ž
π‘Ž+𝑏
=
1βˆ’ 1βˆ’exp βˆ’ πœƒ
π‘Ž
1βˆ’ 1βˆ’exp βˆ’
=
πœƒ
= 1 βˆ’ 𝐹 𝑏 = 𝑃(𝑋 > 𝑏)
2. Relation with Poisson distribution: Assume a Poisson distribution with Ξ»
events per hour, so in t hours, the number of events, Y, follows a Poisson
with mean Ξ»t. Now we start at time zero and ask β€œ how long do I have to
wait to see the 1st event occur?
Let X denote the length of time until 1st event occurs.
λ𝑑 0 exp βˆ’Ξ»π‘‘
𝑃 𝑋 > 𝑑 = 𝑃 π‘Œ = 0 π‘œπ‘› π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™ 0, 𝑑 =
= exp βˆ’Ξ»π‘‘
P(X≀ t)=1-exp(- λ𝑑) οƒ  𝑓 𝑑 =
1
exp
πœƒ
1
βˆ’
πœƒ
0!
, 𝑑 > 0 which means the
interval time between two consecutive events in Poisson distribution follows the
exponential distribution.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 3. Gamma: PDF
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Probability Density Function (PDF)
In general, the Gamma density function is given by
𝑓 π‘₯ =
1
𝛼 βˆ’ 1exp(βˆ’ π‘₯ ),
π‘₯
Ξ“ 𝛼 𝛽𝛼
𝛽
π‘₯β‰₯0
0, π‘œπ‘‘β„Žπ‘’π‘Ÿπ‘€π‘–π‘ π‘’
Where the parameters Ξ± and Ξ² are constants (Ξ± >0, Ξ²>0) that
determines the shape of the curve.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 4. Mean and Variance
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
𝑬 𝑿 =
∞
𝒙𝒇 𝒙 𝒅𝒙 =
βˆ’βˆž
𝟏
Ξ“ 𝜢 𝜷𝜢
∞
π’™βˆ™πŸ
𝒙
βˆ’πŸ
𝜢
𝒙
𝒆𝒙𝒑 βˆ’ 𝒅𝒙=
𝟎 Ξ“ 𝜢 𝜷𝜢
𝜷
∞ 𝜢
𝒙
𝟏
𝜢+𝟏
𝒙
𝒆𝒙𝒑
βˆ’
𝒅𝒙
=
Ξ“
𝜢
+
𝟏
𝜷
𝟎
𝜷
Ξ“ 𝜢 𝜷𝜢
= 𝜢𝜷
Similary , we can find 𝑬(π‘ΏπŸ) = 𝜢(𝜢 + 𝟏)𝜷𝟐, so
𝑽(𝑿) = 𝑬(π‘ΏπŸ) βˆ’ π‘¬πŸ(𝑿) = 𝜢𝜷𝟐.
Suppose 𝒀 = βˆ‘π‘Ώπ’Š with π‘ΏπŸ, π‘ΏπŸ, … , 𝑿𝒏 being independent
Gamma variables with parameters Ξ± and Ξ², then
𝑬(𝒀) = π’πœΆπœ·, 𝑽(𝒀) = π’πœΆπœ·πŸ.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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 Ξ²=450. So Y is a gamma
Distribution with Ξ±=2 and Ξ²=450, that is,
𝑓 𝑦 =
1
Ξ“ 2 450
𝑦
, 𝑦>0
2 𝑦𝑒π‘₯𝑝 βˆ’
450
0, π‘œπ‘‘β„Žπ‘’π‘Ÿπ‘€π‘–π‘ π‘’
Then the mean E(Y)=Ξ±Ξ²=2(450)=900.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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 Ξ±=3 and Ξ²=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 ΞΌ=E(X)=Ξ±Ξ²=6, Οƒ2=V(X)=Ξ±Ξ²2=12, the standard deviation Οƒ=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|β‰₯14)≀(3.446/14)2=0.06, which is really small
Probability, so we can say it is out of line with prior experience.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 3. More Examples
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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 οƒ  x=250[ln(2)] = 173.29
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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 π‘₯ = 1 βˆ’ exp βˆ’
π‘₯
πœƒ
, based on P(X>2)=2P(X>4),
we have 1-F(2)=2(1-F(4))οƒ 1-(1-exp(-2/ΞΈ))=2(1-(1-exp(-4/ΞΈ))οƒ 
exp(-2/ΞΈ)=2exp(-4/ΞΈ)οƒ  -2/ΞΈ=ln(2)-4/ΞΈ οƒ  ΞΈ = 2/ln(2) οƒ  V(X)=[2/ln(2)]2.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Additional Example 3
If X has probability density function given by
Find the mean and variance.
Answer: Change it to the standard form with Ξ±=3, Ξ²=/12, so we can find
E(X)=Ξ±Ξ²=3/2, V(X)=Ξ±Ξ²2=3/4.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL