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Chapter 5. Continuous Probability
Distributions
Sections 5.4, 5.5: Exponential and Gamma
Distributions
Jiaping Wang
Department of Mathematical Science
03/25/2013, Monday
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Outline
Exponential: PDF and CDF
Exponential: Mean and Variance
Gamma: PDF and CDF
Gamma: Mean and Variance
More Examples
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Part 1. Exponential: PDF and
CDF
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Probability Density Function
In general, the exponential density function is given by
𝑓 𝑥 =
1 − 𝑥 /𝜃
𝑒
,
𝜃
𝑥≥0
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Where the parameter θ is a constant (θ>0) that determines the
rate at which the curve decreases.
θ=2
θ = 1/2
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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
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Part 2. Mean and Variance
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Gamma Function
The gamma function Γ(α) is given as
∞
Γ 𝛼 = 0 𝑥𝛼 − 1𝑒 − 𝑥𝑑𝑥
We can show that Γ 𝛼 + 1 = 𝛼Γ 𝛼
So Γ 𝑛 = 𝑛 − 1 Γ 𝑛 − 2 = ⋯ = 𝑛 − 1 !
Specially, Γ 1/2 = 𝜋
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Mean and Variance
∞
𝑬 𝑿 =
∞
𝒙𝒇 𝒙 𝒅𝒙 =
−∞
=
𝐄
𝐗𝟐
=
𝟏 ∞
𝒙
𝜽 𝟎
∙ 𝒆𝒙𝒑
∞ 𝐱𝟐
𝐞𝐱𝐩
𝟎 𝛉
𝐱
−
𝛉
𝒙
−
𝜽
𝟎
𝒅𝒙 =
𝐝𝐱 =
𝟏
𝚪
𝛉
𝒙
𝒙
𝒆𝒙𝒑 − 𝒅𝒙
𝜽
𝜽
𝟏
Γ
𝜽
𝟐 𝜽𝟐 = 𝜽.
𝟑 𝛉𝟑 = 𝟐𝛉𝟐.
Then we have V(X)=E(X2)-E2(X)=2θ2- θ2= θ2.
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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 − 𝑃 𝑋 ≤ 4 = 1 − 𝐹 4 = 1 − 1 − exp − 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)= 32 0.37 2 0.63 = 0.26.
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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
𝑎
𝑎
𝑃 𝑋 > 𝑎 = 1 − 𝑃 𝑋 ≤ 𝑎 = 1 − 1 − exp − 4 = exp − 4
So we choose a with P(X>a)=exp(-a/4)=0.05 a=11.98 (tons).
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Properties
1. Memoryless: 𝑃 𝑋 > 𝑎 + 𝑏 𝑋 > 𝑎 =
exp
𝑏
−
𝜃
𝑃 𝑋>𝑎+𝑏
𝑃 𝑋>𝑎
𝑎+𝑏
=
1− 1−exp − 𝜃
𝑎
1− 1−exp −
=
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 1st event occur?
Let X denote the length of time until 1st event occurs.
λ𝑡 0 exp −λ𝑡
𝑃 𝑋 > 𝑡 = 𝑃 𝑌 = 0 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 0, 𝑡 =
= exp −λ𝑡
P(X≤ t)=1-exp(- λ𝑡)  𝑓 𝑡 =
1
exp
𝜃
1
−
𝜃
0!
, 𝑡 > 0 which means the
interval time between two consecutive events in Poisson distribution follows the
exponential distribution.
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Part 3. Gamma: PDF
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Probability Density Function (PDF)
In general, the Gamma density function is given by
𝑓 𝑥 =
1
𝛼 − 1exp(− 𝑥 ),
𝑥
Γ 𝛼 𝛽𝛼
𝛽
𝑥≥0
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Where the parameters α and β are constants (α >0, β>0) that
determines the shape of the curve.
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Part 4. Mean and Variance
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𝑬 𝑿 =
∞
𝒙𝒇 𝒙 𝒅𝒙 =
−∞
𝟏
Γ 𝜶 𝜷𝜶
∞
𝒙∙𝟏
𝒙
−𝟏
𝜶
𝒙
𝒆𝒙𝒑 − 𝒅𝒙=
𝟎 Γ 𝜶 𝜷𝜶
𝜷
∞ 𝜶
𝒙
𝟏
𝜶+𝟏
𝒙
𝒆𝒙𝒑
−
𝒅𝒙
=
Γ
𝜶
+
𝟏
𝜷
𝟎
𝜷
Γ 𝜶 𝜷𝜶
= 𝜶𝜷
Similary , we can find 𝑬(𝑿𝟐) = 𝜶(𝜶 + 𝟏)𝜷𝟐, so
𝑽(𝑿) = 𝑬(𝑿𝟐) − 𝑬𝟐(𝑿) = 𝜶𝜷𝟐.
Suppose 𝒀 = ∑𝑿𝒊 with 𝑿𝟏, 𝑿𝟐, … , 𝑿𝒏 being independent
Gamma variables with parameters α and β, then
𝑬(𝒀) = 𝒏𝜶𝜷, 𝑽(𝒀) = 𝒏𝜶𝜷𝟐.
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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,
𝑓 𝑦 =
1
Γ 2 450
𝑦
, 𝑦>0
2 𝑦𝑒𝑥𝑝 −
450
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Then the mean E(Y)=αβ=2(450)=900.
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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, σ2=V(X)=αβ2=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.
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Part 3. More Examples
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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)] = 173.29
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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
𝜃
, based on P(X>2)=2P(X>4),
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.
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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)=αβ2=3/4.
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