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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.

f (x ;α, β) =

x ≥ 0

0 x < 0

f (x ;α, β) =

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.

f (x ;α, β) =

x ≥ 0

0 x < 0

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.

Remark:

3. When α = 1, the pdf becomes

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

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.

f (x ;β) =

{1β e−x/β x ≥ 0

0 x < 0

Proposition

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

E (X ) = βΓ

)and V (X ) = β2

The cdf of X is

F (x ;α, β) =

{1− e−(x/β)α

x ≥ 0

0 x < 0

Proposition

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

E (X ) = βΓ

)and V (X ) = β2

The cdf of X is

F (x ;α, β) =

{1− e−(x/β)α

x ≥ 0

0 x < 0

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.

a. Find P(X > 410).

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

a. Find P(X > 410).

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

a. Find P(X > 410).

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

a. Find P(X > 410).

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

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).

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

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) =

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 )

f (x ;µ, σ) =

2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0

0 x < 0

µ 6= E (X ) and σ2 6= V (X )

f (x ;µ, σ) =

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 )

f (x ;µ, σ) =

2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0

0 x < 0

µ 6= E (X ) and σ2 6= V (X )

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)]

(Z ≤ ln(x)− µ

(ln(x)− µ

)x ≤ 0

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

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)]

(Z ≤ ln(x)− µ

(ln(x)− µ

)x ≤ 0

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

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?

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.

Beta Distribution

f (x ;α, β,A,B) =

1B−A ·

(x−AB−A

)α−1·(

B−xB−A

)β−1A ≤ x ≤ B

0 otherwise

Beta Distribution

f (x ;α, β,A,B) =

1B−A ·

(x−AB−A

)α−1·(

B−xB−A

)β−1A ≤ x ≤ B

0 otherwise

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)

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)

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)?

Beta Distribution

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