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Weibull Distribution

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Page 1: Beta n Weibul Distri

Weibull Distribution

DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is

f (x ;α, β) =

{αβα xα−1e−(x/β)α

x ≥ 0

0 x < 0

Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.

Page 2: Beta n Weibul Distri

Weibull Distribution

DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is

f (x ;α, β) =

{αβα xα−1e−(x/β)α

x ≥ 0

0 x < 0

Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.

Page 3: Beta n Weibul Distri

Weibull Distribution

DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is

f (x ;α, β) =

{αβα xα−1e−(x/β)α

x ≥ 0

0 x < 0

Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.

2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.

Page 4: Beta n Weibul Distri

Weibull Distribution

DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is

f (x ;α, β) =

{αβα xα−1e−(x/β)α

x ≥ 0

0 x < 0

Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.

Page 5: Beta n Weibul Distri

Weibull Distribution

Remark:3. When α = 1, the pdf becomes

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

which is the pdf for an exponential distribution with parameterλ = 1

β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.

Page 6: Beta n Weibul Distri

Weibull Distribution

Remark:

3. When α = 1, the pdf becomes

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

which is the pdf for an exponential distribution with parameterλ = 1

β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.

Page 7: Beta n Weibul Distri

Weibull Distribution

Remark:3. When α = 1, the pdf becomes

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

which is the pdf for an exponential distribution with parameterλ = 1

β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.

4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.

Page 8: Beta n Weibul Distri

Weibull Distribution

Remark:3. When α = 1, the pdf becomes

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

which is the pdf for an exponential distribution with parameterλ = 1

β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.

Page 9: Beta n Weibul Distri

Weibull Distribution

Page 10: Beta n Weibul Distri

Weibull Distribution

Page 11: Beta n Weibul Distri

Weibull Distribution

Page 12: Beta n Weibul Distri

Weibull Distribution

Page 13: Beta n Weibul Distri

Weibull Distribution

Proposition

Let X be a random variable such that X ∼WEI(α, β). Then

E (X ) = βΓ

(1 +

1

α

)and V (X ) = β2

(1 +

2

α

)−[

Γ

(1 +

1

α

)]2}

The cdf of X is

F (x ;α, β) =

{1− e−(x/β)α

x ≥ 0

0 x < 0

Page 14: Beta n Weibul Distri

Weibull Distribution

Proposition

Let X be a random variable such that X ∼WEI(α, β). Then

E (X ) = βΓ

(1 +

1

α

)and V (X ) = β2

(1 +

2

α

)−[

Γ

(1 +

1

α

)]2}

The cdf of X is

F (x ;α, β) =

{1− e−(x/β)α

x ≥ 0

0 x < 0

Page 15: Beta n Weibul Distri

Weibull Distribution

Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed random variable, X ∼WEB(400, 2/3).

a. Find P(X > 410).

b. Find P(X > 410 | X > 390).

c. Find E (X ) and V (X ).

d. Find the 95th percentile.

Page 16: Beta n Weibul Distri

Weibull Distribution

Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed random variable, X ∼WEB(400, 2/3).

a. Find P(X > 410).

b. Find P(X > 410 | X > 390).

c. Find E (X ) and V (X ).

d. Find the 95th percentile.

Page 17: Beta n Weibul Distri

Weibull Distribution

Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed random variable, X ∼WEB(400, 2/3).

a. Find P(X > 410).

b. Find P(X > 410 | X > 390).

c. Find E (X ) and V (X ).

d. Find the 95th percentile.

Page 18: Beta n Weibul Distri

Weibull Distribution

Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed random variable, X ∼WEB(400, 2/3).

a. Find P(X > 410).

b. Find P(X > 410 | X > 390).

c. Find E (X ) and V (X ).

d. Find the 95th percentile.

Page 19: Beta n Weibul Distri

Weibull Distribution

Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed random variable, X ∼WEB(400, 2/3).

a. Find P(X > 410).

b. Find P(X > 410 | X > 390).

c. Find E (X ) and V (X ).

d. Find the 95th percentile.

Page 20: Beta n Weibul Distri

Weibull Distribution

In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.Example (Problem 74):Let X = the time (in 10−1 weeks) from shipment of a

defective product until the customer returns the

product. Suppose that the minimum return time is γ = 3.5 andthat the excess X − 3.5 over the minimum has a Weibulldistribution with parameters α = 2 and β = 1.5.

a. What is the cdf of X?

b. What are the expected return time and variance of returntime?

c. Compute P(X > 5).

d. Compute P(5 ≤ X ≤ 8).

Page 21: Beta n Weibul Distri

Weibull Distribution

In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.

Example (Problem 74):Let X = the time (in 10−1 weeks) from shipment of a

defective product until the customer returns the

product. Suppose that the minimum return time is γ = 3.5 andthat the excess X − 3.5 over the minimum has a Weibulldistribution with parameters α = 2 and β = 1.5.

a. What is the cdf of X?

b. What are the expected return time and variance of returntime?

c. Compute P(X > 5).

d. Compute P(5 ≤ X ≤ 8).

Page 22: Beta n Weibul Distri

Weibull Distribution

In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.Example (Problem 74):Let X = the time (in 10−1 weeks) from shipment of a

defective product until the customer returns the

product. Suppose that the minimum return time is γ = 3.5 andthat the excess X − 3.5 over the minimum has a Weibulldistribution with parameters α = 2 and β = 1.5.

a. What is the cdf of X?

b. What are the expected return time and variance of returntime?

c. Compute P(X > 5).

d. Compute P(5 ≤ X ≤ 8).

Page 23: Beta n Weibul Distri

Weibull Distribution

In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.Example (Problem 74):Let X = the time (in 10−1 weeks) from shipment of a

defective product until the customer returns the

product. Suppose that the minimum return time is γ = 3.5 andthat the excess X − 3.5 over the minimum has a Weibulldistribution with parameters α = 2 and β = 1.5.

a. What is the cdf of X?

b. What are the expected return time and variance of returntime?

c. Compute P(X > 5).

d. Compute P(5 ≤ X ≤ 8).

Page 24: Beta n Weibul Distri

Weibull Distribution

In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.Example (Problem 74):Let X = the time (in 10−1 weeks) from shipment of a

defective product until the customer returns the

product. Suppose that the minimum return time is γ = 3.5 andthat the excess X − 3.5 over the minimum has a Weibulldistribution with parameters α = 2 and β = 1.5.

a. What is the cdf of X?

b. What are the expected return time and variance of returntime?

c. Compute P(X > 5).

d. Compute P(5 ≤ X ≤ 8).

Page 25: Beta n Weibul Distri

Weibull Distribution

In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.Example (Problem 74):Let X = the time (in 10−1 weeks) from shipment of a

defective product until the customer returns the

product. Suppose that the minimum return time is γ = 3.5 andthat the excess X − 3.5 over the minimum has a Weibulldistribution with parameters α = 2 and β = 1.5.

a. What is the cdf of X?

b. What are the expected return time and variance of returntime?

c. Compute P(X > 5).

d. Compute P(5 ≤ X ≤ 8).

Page 26: Beta n Weibul Distri

Lognormal Distribution

DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is

f (x ;µ, σ) =

{1√

2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0

0 x < 0

Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.

µ 6= E (X ) and σ2 6= V (X )

Page 27: Beta n Weibul Distri

Lognormal Distribution

DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is

f (x ;µ, σ) =

{1√

2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0

0 x < 0

Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.

µ 6= E (X ) and σ2 6= V (X )

Page 28: Beta n Weibul Distri

Lognormal Distribution

DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is

f (x ;µ, σ) =

{1√

2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0

0 x < 0

Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.

2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.

µ 6= E (X ) and σ2 6= V (X )

Page 29: Beta n Weibul Distri

Lognormal Distribution

DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is

f (x ;µ, σ) =

{1√

2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0

0 x < 0

Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.

µ 6= E (X ) and σ2 6= V (X )

Page 30: Beta n Weibul Distri

Lognormal Distribution

Page 31: Beta n Weibul Distri

Lognormal Distribution

Page 32: Beta n Weibul Distri

Lognormal Distribution

Proposition

If X ∼ LOGN(µ, σ2), then

E (X ) = eµ+σ2/2 and V (X ) = e2µ+σ2 · (eσ2 − 1)

The cdf of X is

F (x ;µ, σ) = P(X ≤ x) = P[ln(X ) ≤ ln(x)]

= P

(Z ≤ ln(x)− µ

σ

)= Φ

(ln(x)− µ

σ

)x ≤ 0

where Φ(z) is the cdf of the standard normal rv Z .

Page 33: Beta n Weibul Distri

Lognormal Distribution

Proposition

If X ∼ LOGN(µ, σ2), then

E (X ) = eµ+σ2/2 and V (X ) = e2µ+σ2 · (eσ2 − 1)

The cdf of X is

F (x ;µ, σ) = P(X ≤ x) = P[ln(X ) ≤ ln(x)]

= P

(Z ≤ ln(x)− µ

σ

)= Φ

(ln(x)− µ

σ

)x ≤ 0

where Φ(z) is the cdf of the standard normal rv Z .

Page 34: Beta n Weibul Distri

Lognormal Distribution

Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.

a. What is the probability that the output current is more thantwice the input current?

b. What are the expected value and variance of the ratio ofoutput to input current?

c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?

Page 35: Beta n Weibul Distri

Lognormal Distribution

Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.

a. What is the probability that the output current is more thantwice the input current?

b. What are the expected value and variance of the ratio ofoutput to input current?

c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?

Page 36: Beta n Weibul Distri

Lognormal Distribution

Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.

a. What is the probability that the output current is more thantwice the input current?

b. What are the expected value and variance of the ratio ofoutput to input current?

c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?

Page 37: Beta n Weibul Distri

Lognormal Distribution

Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.

a. What is the probability that the output current is more thantwice the input current?

b. What are the expected value and variance of the ratio ofoutput to input current?

c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?

Page 38: Beta n Weibul Distri

Lognormal Distribution

Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.

a. What is the probability that the output current is more thantwice the input current?

b. What are the expected value and variance of the ratio ofoutput to input current?

c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?

Page 39: Beta n Weibul Distri

Beta Distribution

DefinitionA random variable X is said to have a beta distribution withparameters α, β(both positive), A, and B if the pdf of X is

f (x ;α, β,A,B) =

1B−A ·

Γ(α+β)Γ(α)·Γ(β) ·

(x−AB−A

)α−1·(

B−xB−A

)β−1A ≤ x ≤ B

0 otherwise

The case A = 0,B = 1 gives the standard beta distribution.

Remark: We use X ∼ BETA(α, β,A,B) to denote that rv X has abeta distribution with parameters α, β, A, and B.

Page 40: Beta n Weibul Distri

Beta Distribution

DefinitionA random variable X is said to have a beta distribution withparameters α, β(both positive), A, and B if the pdf of X is

f (x ;α, β,A,B) =

1B−A ·

Γ(α+β)Γ(α)·Γ(β) ·

(x−AB−A

)α−1·(

B−xB−A

)β−1A ≤ x ≤ B

0 otherwise

The case A = 0,B = 1 gives the standard beta distribution.

Remark: We use X ∼ BETA(α, β,A,B) to denote that rv X has abeta distribution with parameters α, β, A, and B.

Page 41: Beta n Weibul Distri

Beta Distribution

DefinitionA random variable X is said to have a beta distribution withparameters α, β(both positive), A, and B if the pdf of X is

f (x ;α, β,A,B) =

1B−A ·

Γ(α+β)Γ(α)·Γ(β) ·

(x−AB−A

)α−1·(

B−xB−A

)β−1A ≤ x ≤ B

0 otherwise

The case A = 0,B = 1 gives the standard beta distribution.

Remark: We use X ∼ BETA(α, β,A,B) to denote that rv X has abeta distribution with parameters α, β, A, and B.

Page 42: Beta n Weibul Distri

Beta Distribution

Proposition

If X ∼ BETA(α, β,A,B), then

E (X ) = A + (B − A) · α

α + βand V (X ) =

(B − A)2αβ

(α + β)2(α + β + 1)

Page 43: Beta n Weibul Distri

Beta Distribution

Proposition

If X ∼ BETA(α, β,A,B), then

E (X ) = A + (B − A) · α

α + βand V (X ) =

(B − A)2αβ

(α + β)2(α + β + 1)

Page 44: Beta n Weibul Distri

Beta Distribution

Page 45: Beta n Weibul Distri

Beta Distribution

Page 46: Beta n Weibul Distri

Beta Distribution

Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integer-valued].

a. Let X represent a randomly selected American credit score.What are the mean value and standard deviation of thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?

b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?

Page 47: Beta n Weibul Distri

Beta Distribution

Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integer-valued].

a. Let X represent a randomly selected American credit score.What are the mean value and standard deviation of thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?

b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?

Page 48: Beta n Weibul Distri

Beta Distribution

Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integer-valued].

a. Let X represent a randomly selected American credit score.What are the mean value and standard deviation of thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?

b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?

Page 49: Beta n Weibul Distri

Beta Distribution

Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integer-valued].

a. Let X represent a randomly selected American credit score.What are the mean value and standard deviation of thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?

b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?