Bivariate Birnbaum-Saunders Distribution

Department of Mathematics & Statistics Indian Institute of Technology Kanpur January 2nd. 2013

Outline 1 Collaborators 2 3 Birnbaum-Saunders Distribution: Introduction & Properties 4 5

Collaborators 1 N. Balakrishnan 2 R.C. Gupta 3 Ahad Jamalizadeh 4 N. Kannan 5 V. Leiva 6 H.K. T. Ng, 7 A. Sanhueza.

Outline 1 Collaborators 2 3 Birnbaum-Saunders Distribution: Introduction & Properties 4 5

Transformation from Normal Random Variable Different random variables have been derived from normal random variables. Log-normal random variable: If X is a normal random variable then Y = e X has log-normal random variable. F Y (y) = Φ(ln y) The PDF of Log-normal random distribution is; f (x; µ, σ) = 1 (ln x µ)2 e 2σ 2 2πσx

Log-normal Random Variable The shape of the log-normal PDF for different σ.

Log-normal distribution: Closeness The shape of the PDF of log-normal distribution is always unimodal. The PDF of the log-normal distribution is right skewed. The PDF of log-normal distribution has been used very successfully to model right skewed data. It is observed that the PDF of log-normal distribution is very similar to the shape of the PDF of well known Weibull distribution. Quite a bit of work has been done in discriminating between these two distribution functions.

Inverse Gaussian random variable Inverse Gaussian random variable has the following PDF: f (x; µ, λ) = ( ) λ 1/2 ) λ(x µ)2 2πx 3 exp ( 2µ 2 x Inverse Gaussian has the CDF; { ( )} { λ x F (x; µ, σ) = Φ x µ 1 + e 2λ/µ Φ λ ( )} x x µ + 1

Inverse Gaussian Distribution: PDF The shape of the Inverse Gaussian PDF for different λ.

Skew normal distribution: Introduction So far skewed distribution has been obtained for non-negative random variable. Now we provide a method to introduce skewness to a random variable which may have a support on the entire real line also. Consider two independent standard normal independent random variables, say X and Y. Then because of symmetry P(Y < X ) = 1 2

Skew normal distribution: Introduction Now suppose α > 0 and consider P(Y < αx ). In this case also since Y αx is a normal random variable with mean 0, then clearly P(Y < αx ) = 1/2 On the other hand P(Y < αx ) = φ(x)φ(αx)dx = 1 2. Therefore, 2φ(x)Φ(αx) = 1.

Skew normal distribution Skew normal distribution has the following PDF: f (x; α) = 2φ(x)Φ(αx); < x < It is a skewed distribution on the whole real line

Skew Normal Random Variable The shape of the skew normal PDF for different α.

Outline 1 Collaborators 2 3 Birnbaum-Saunders Distribution: Introduction & Properties 4 5

Birnbaum-Saunders distribution: PDF and CDF Birnbaum-Saunders Distribution has the following CDF: [ { ( ) 1 x 1/2 ( ) }] β 1/2 F (x; α, β) = Φ α β x It is a skewed distribution on the positive real line. The PDF of Birnbaum-Saunders distribution is. [ (β ) 1 1/2 ( ) ] β 3/2 [ f (x) = 2 + exp 1 ( x 2παβ x x 2α 2 β + β )] x 2

Birnbaum-Saunders Random Variable The shape of the Birnbaum-Saunders PDF for different α.

Birnbaum-Saunders distribution Birnbaum-Saunders distribution has been developed to model failures due to crack. It is a assumed that the j-th cycle leads to an increase in crack X j amount. It is further assumed that n j=1 X j is approximately normally distributed with mean nµ and variance nσ 2. Then the probability that the crack does not exceed a critical length ω is ( ) ( ω nµ ω Φ σ = Φ n σ n µ ) n σ

Birnbaum-Saunders distribution It is further assumed that the failure occurs when the crack length exceeds ω. If T denotes the lifetime (in number of cycles) until failure, then the CDF of T is approximately ( ω P(T t) 1 Φ σ t µ ) t σ ( µ t = Φ σ ω ) σ. t It is exactly the same form as defined before.

Birnbaum-Saunders distribution Fatigue failure is due to repeated applications of a common cyclic stress pattern. Under the influence of this cyclic stress a dominant crack in the material grows until it reaches a critical size w is reached, at that point fatigue failure occurs. The crack extension in each cycle are random variables and they are statistically independent. The total extension of the crack is approximately normally distributed.

Birnbaum-Saunders distribution The following observations are useful: If T a Birnbaum-Saunders distribution, say, BS(α, β) then consider the following transformation [ (T X = 1 ) 1/2 ( ) ] T 1/2 2 β β Equivalently T = β (1 + 2X 2 + 2X ( 1 + X 2) ) 1/2 Then X is normally distributed with mean zero and variance α 2 /4.

Birnbaum-Saunders distribution The above transformation becomes very helpful: It can be used very easily to generate samples from Birnbaum-Saunders distribution. It helps to derive different moments of the Birnbaum-Saunders distribution. It helps to derive some other properties of the Birnbaum-Saunders distribution.

Birnbaum-Saunders distribution: Basic Properties Here α is the shape parameter and β is the scale parameter. α governs the shape of PDF and hazard function. For all values of α the PDF is unimodal. Mode cannot be obtained in explicit form, it has to be obtained by solving a non-linear equation in α. Clearly, the median is at β, for all α.

Birnbaum-Saunders distribution: Inverse Another interesting observation which is useful for estimation purposes: If T is a Birnbaum-Saunders distribution, i.e. BS(α, β), then T 1 is also a Birnbaum-Saunders with parameters α and β 1 The above observation is very useful. Immediately we obtain E(T 1 ) = β 1 (1 + 1 2 α2 ), Var(T ) = α 2 β 2 (1 + 5 4 α2 ).

Birnbaum-Saunders distribution: Hazard Function The shape of the hazard function of Birnbaum-Saunders distribution is unimodal. The turning point of the hazard function can be obtained by solving a non-linear equation involving α. The turning point of the hazard function of the Birnbaum-Saunders distribution can be approximated very well for α > 0.25, by c(α) = 1 ( 0.4604 + 1.8417α) 2. The approximation works very well for α > 0.6.

Nice representation Now we will present one nice representation of the Birnbaum-Saunders distribution. We have already defined the Inverse Gaussian random variable which has the following PDF: f (x; µ, σ) = ( 1 2σ 2 πx 3 We will denote this as IG(µ, σ 2 ). ) 1/2 exp ( (x µ)2 2σ 2 µ 2 x Suppose X 1 IG(µ, σ 2 ), and X2 1 IG(µ 1, σ 2 mu 2 ), then consider the new random variable for 0 p 1: X 1 w.p 1 p X = X 2 w.p. p )

Nice representation Then the PDF of X can be expressed as follows: f X (x) = pf X1 (x) + (1 p)f X2 (x) Note that the PDF of X 1 is a Birnbaum-Saunders PDF, and the PDF of X 2 can be easily obtained. Interestingly, when p = 1/2, the PDF of X becomes the PDF of a Birnbaum-Saunders PDF.

Outline 1 Collaborators 2 3 Birnbaum-Saunders Distribution: Introduction & Properties 4 5

Bivariate Birnbaum-Saunders distribution Remember the univariate Birnbaum-Saunders distribution has been defined as follows: P(T 1 t 1 ) = Φ [ ( 1 t α β )] β t The bivariate bivariate Birnbaum-Saunders distribution can be defined analogously as follows:

Bivariate Birnbaum-Saunders distribution Let the joint distribution function of (T 1, T 2 ) be defined as follows F (t 1, t 2 ) = P(T 1 t 1, T 2 t 2 ), then (T 1, T 2 ) is said to have bivariate Birnbaum-Saunders distribution with parameters α 1, α 2, β 2, β 2, ρ, if [ ( 1 t1 F (t 1, t 2 ) = Φ 2 α 1 β 1 β1 t 1 ), 1 α 2 ( t2 β 2 β2 Here Φ 2 (u, v) is the CDF of a standard bivariate normal vector (Z 1, Z 2 ) with the correlation coefficient ρ. Let s denote this by BVBS(α 1, β 1, α 2, β 2, ρ). t 2 ) ; ρ ]

Bivariate Birnbaum-Saunders distribution has several interesting properties. The joint PDF of (T 1, T 2 ) can be easily obtained in terms of the PDF of bivariate normal distribution. The joint PDF can take different shapes, but it is unimodal, skewed. Need not be symmetric. The correlation coefficient between T 1 and T 2 can be both positive and negative.

Birnbaum-Saunders Random Variable Contour plot of the Bivariate Birnbaum-Saunders PDF.

BVBS: Properties If (T 1, T 2 ) BVBS(α 1, β 1, α 2, β 2 ), then we have the following results: T 1 BS(α 1, β 1 ) and T 2 BS(α 2, β 2 ). (T 1 1, T 1 2 ) BVBS(α 1, β 1 1, α 2, β 1 2, ρ). (T 1 1, T 2 ) BVBS(α 1, β 1 1, α 2, β 2, ρ). (T 1, T 1 2 ) BVBS(α 1, β 1, α 2, β 1 2, ρ).

: Generation It is very easy to generate Bivariate Birnbaum-Saunders random variables using normal random numbers generator: Generate first U 1 and U 2 from N(0,1) Generate 1 + ρ + 1 ρ 1 + ρ 1 ρ Z 1 = U 1 + U 2 2 2 1 + ρ 1 ρ 1 + ρ + 1 ρ Z 2 = U 1 + U 2 2 2 Make the transformation: (1 T i = β i 1 ) 2 2 2 α iz i + 2 α iz i + 1, i = 1, 2.

: Inference The MLEs of the unknown parameters can be obtained to solve a five dimensional optimization problem. The following observation is useful: [( T 1 β 1 β1 T 1 ), ( T 2 β 2 β2 T 2 )] N 2 {(0, 0), Σ} where ( ) α 2 Σ = 1 α 1 α 2 ρ α 1 α 2 ρ α2 2.

: Inference 1 For given β 1 and β 2, the MLEs of α 1, α 2 and ρ can be obtained in explicit forms. 2 The MLEs of β 1 and β 2 can be obtained by maximizing the profile log-likelihood function. 3 Five dimensional optimization problem can be reduced to a two dimensional optimization problem.

Outline 1 Collaborators 2 3 Birnbaum-Saunders Distribution: Introduction & Properties 4 5

Copula To every bivariate distribution function F Y1,Y 2 (, ) with continuous marginals F Y1 ( ) and F Y2 ( ), corresponds a unique function C : [0, 1] [0, 1] [0, 1], called a copula such that for (y 1, y 2 ) (, ) (, ) F Y1,Y 2 (y 1, y 2 ) = C(F Y1 (y 1 ), F Y2 (y 2 )) and C(u, v) = F Y1,Y 2 (F 1 Y 1 (u), F 1 Y 2 (v))

Bivariate Gaussian Copula C G (u, v) = where Φ 1 (u) Φ 1 (v) φ 2 (x, y; ρ)dxdy = Φ 2 (Φ 1 (u), Φ 1 (v); ρ), { } 1 φ 2 (u, v; ρ) = 2π 1 ρ exp 1 2 2(1 ρ 2 ) (u2 + v 2 2ρuv)

The bivariate Birnbaum-Saunders distribution can be written as follows; F T1,T 2 (t 1, t 2 ) = C G (F T1 (t 1 ; α 1, β 1 ), F T2 (t 2 ; α 2, β 2 ); ρ). If (T 1, T 2 ) BVBS(α 1, β 1, α 2, β 2, ρ), then different properties can be established using copula structure.

Different Properties 1 It is TP 2 for ρ > 0 and RR 2 for ρ < 0. 2 The conditional failure rate of T 1 given T 2 = t 2 is a decreasing (increasing) function of t 2 for ρ > 0(ρ < 0). 3 The conditional failure rate of T 1 given T 2 > t 2 is a decreasing (increasing) function of t 2 for ρ > 0(ρ < 0). 4 T 1 is stochastically increasing in T 2 and T 2 is stochastically increasing in T 1, if ρ > 0

Thank You