An alternative for Laplace Birnbaum-Saunders distribution

Lemma A.1

On (β,∞),

1. if 0 < α ≤ 4, then r₂(x) is strictly decreasing;

2. if α > 4, then r₂(x) is unimodal.

Proof. It is easily seen that the sign of r2′(x)depends on

which is less than (α − 4)/2. Therefore, we have r2′(x)≤0if α ≤ 4, and is positive for otherwise. Furthermore, the root of r2′(x)=0 is x0=β(α/2−1)(2/α).Note that x₀> β if and only if α > 4. Thus, r₂(x) is strictly decreasing when α ≤ 4. On the other hand, the limits

imply that the root x₀ is the maximum of r₂(x). Hence the lemma.

Lemma A.2

If α ≤ 1, then r₁(x) is a strictly decreasing function with the following limits

Proof . It is easily seen that the sign of r1′(x)depends on the following expression

The third term in the last expression is negative when α ≤ 1. For the remaining terms of that expression, we have xα/2[(α+2)xα+3βα(α−2)]and

Therefore, r₁(x) is strictly decreasing. The limits of r₁(x) in the theorem can be easily obtained.

Lemma A.3

If 1 < α ≤ 4/3, then r₁(x) is unimodal.

Proof. We have

and

which is nonnegative when α ≥ 4/3, and is negative for otherwise. Therefore, if 1 < α < 4/3, there exists at least one root of r1′(x)=0in (0, β).

It can be easily seen that if r1′(x)=0then

Rearranging the last equation, we get

Now let z = (x/β)^α/2. Then we have the following cubic polynomial, say q, to be solved:

It is expected that the limiting behavior of q(z) stays the same with the r1′(x). In fact,

and

However, there is only one solution z^∗ in (0, 1) to q′(z)=0,that is,

Since q′′(z*)>0, z*is a minimum point. This implies that the root of q(z)=0  or  r1′(x)=0in (0, z^∗) is unique. From the limiting behavior of r1′(x)given in (A.1) and (A.2), this root is a maximum of r₁(x).

Lemma A.4

If 4/3 < α < α^∗, where α^∗ = 1.4670632673094746, then r₁(x) is N-shaped.

Proof. Let q(z) with minimum point z^∗ intersect the z-axis at two points. This continues to happen until q(z) = 0, that is, the z-axis becomes a tangent to the minimum point. We will first determine the value of z^∗ at which q(z^∗) = 0. Since we have

the equation q(z) = 0 with α = α^∗ is reduced to z⁴+12z³+6z²−4z−3 = 0. The only one real root of this equation is found as z^∗ = 0.6289640109259343 using Mathematica (Wolfram, 2012). It is the minimum point of q(z) that is tangent to the positive z-axis. Then α^∗ = 1.4670632673094746. Therefore, when 4/3 < α < α^∗, the function q(z) has two real roots in (0, β). From the sign behavior of r1′(x)using (A) and (A), it can be concluded that the curve r₁(x) is N-shaped.

Lemma A.5

If α > α ^∗ , then r ₁(x) is strictly increasing.

Proof. If α > α^∗ then there is no real root of q(z) = 0 in (0, 1). Further, q(z) > 0, that is r1′(x)>0,which implies that r₁(x) is strictly increasing.

Proof of Theorem 3.6. Using the stochastic representation given in (3.4), we have

The first and second integrals in (A.3) are computed by using (3.9) and (3.10), respectively. Hence, the theorem.

Proof of Theorem 3.7. Using the stochastic representation given in (3.4), we have

We take the expectation of both sides. The right hand side is found as

by using formula (5.1.8.2) in [21]. Hence the theorem.

Proof of Theorem 5.1. The first derivative of l(α) is given by

It is easily seen that limα→0l′(α)=∞and we have

If we arrange the last equation, we get

which is clearly negative. Since l′(α) is continuous, it has a root in (0,∞). Further,

is always negative. That is, l′(α) is a decreasing function on (0,∞). This proves the existence and uniqueness of the MLE of α in (0,∞).

Proof of Theorem 5.2. The log-likelihood function l(β) is continuous but not differentiable at x_i, i = 1, 2, . . . , n. This indicates that if limβ→x+l′(β)limβ→x−l′(β)<0,then l(β) reaches a local extremum at x.

Now we will look at the limit behavior of l′(β) on the boundary of its parameter space (0,∞). We have

Clearly, limβ→0l′(β)=∞:On the other hand, we have

Since

where

is continuous on [x_(r), x_(r+1)), r = 1, 2, . . . , n, it has at least one root in such intervals. Then we have l′(β)=0 for  x(r)≤β<x(r+1)and this implies

The solution of (A.4) maximizes l(β) since

which is always negative at the solution of (A.4).

Note that l′(β) < 0 if and only if γ(β)<0.Further, since

the function γ(β) is decreasing. Further, the solution of

is the solution of β=γ(β)/γ′(β),and it is the minimum of l′(β) in [x(r),x(r+1)),say β₀.

We claim that l′(β)<0for β > β₀. That is, there are no other roots for l′(β) = 0 beyond β₀. To prove this, we may assume the contrary that there exists a β^∗> β₀ such that l′(β^∗) ≥ 0. Then we have

which is a contradiction, since β*>γ(β0)/γ′(β0).

Finally, since l′(β)does not exist at x_(r), we must show that it has a jump discontinuity there.

We have

Thus, x_(r) cannot be a root of l′(β). This completes the proof.

Proof of Theorem 5.3. Let θ = (α, β). To show the existence of a local maximum according to Theorem 2.1 in [19], one may easily show that when θ tends to the boundary of the parameter space, the likelihood in (5.4) tends to zero.

To show the uniqueness of the MLE of θ according to Corollary 2.5 in [19], we should look at the second derivatives of the log-likelihood function at the solutions of the estimating equations l_α = 0 and l_β = 0. The matrix of the second derivatives of the log-likelihood function at the solutions of the estimating equations is given by

where

and

The determinant of D²l is given by

Note that detD²l is positive by Cauchy-Schwarz inequality. Thus, the uniqueness of the MLE’s is obtained.