The log transformation

3.13.1 Expectation and variance

In the case of Poisson random variables just one link function is commonly used. This is the natural logarithm and takes the form

η_i= log(θ_i),

for i = 1, 2. Clearly since θi ∈ [0, ∞) ⇒ ηⁱ ∈ (−∞, ∞) and so ηⁱ is unbounded. We link η₁ and η₂ into a Bayes linear structure. The prior and posterior expectations and variances of θ₁, θ₂ having made within-group updates are given in Equations 3.11 and

3.12.

We can find the prior expectations and variances of each ηi. They are E₀(η_i) = ψ(a_i) − log bi, Var₀(η_i) = ψ₁(a_i),

where ψ(·) is the digamma function and ψ1(·) is the trigamma function.

Proof. In order to find the expectation and variance of η_i in terms of the parameters of the marginal gamma distributions consider the gamma function

Γ (y) = Z _∞

0

z^y−1e^−zdz.

Differentiating with respect to y gives d

dyΓ (y) = Z _∞

0 log(z) × z^y−1e^−zdz.

Each subsequent derivative simply multiplies the right hand side by a further log(z) inside the integration. Therefore, since b_iθ_i∼gamma(ai, 1), if z = b_iθ_i then

1 Γ (a_i)

dⁿ

daⁿ_i Γ (a_i) = E₀[(log b_iθ_i)ⁿ].

Thus the expectation of each η_i can be found as E0[ηi] = E0[log θi]

= E0[log biθi] − log bⁱ

= 1

Γ (a_i) d

da_iΓ (a_i) − log bi

= ψ(ai) − log bⁱ,

where ψ(x) = _dx^d log[Γ (x)] is the digamma function. The variance is then found from d²

da²_iΓ (a_i) = d

da_iΓ (a_i)ψ(a_i)

= Γ (a_i)ψ₁(a_i) + Γ (a_i)ψ(a_i)²,

where ψ₁(x) = _dx^dψ(x) is the trigamma function. Thus Var0(ηi) = Var0(log biθi)

= E0[(log biθi)²] − E⁰[log biθi]²

= ψ₁(a_i) + ψ(a_i)²− ψ(ai)²

= ψ₁(a_i).

The expectation and variance of ηi having observed xi successes in group i are E₀(η_i) = ψ(A_i) − log Bi, Var₀(η_i) = ψ₁(A_i),

which are the same form as in the prior but with a_i and b_i replaced by A_i = a_i+ x_i and Bi = bi+ 1.

We propagate these changes in belief through to the other group using Equations 3.15 and 3.16. This gives E₁(η₁; x₂), Var₁(η₁; x₂), E₁(η₂; x₁) and Var₁(η₂; x₁).

We consider the sufficient condition for a unique commutative solution using Bayes linear kinematics. In this case it is

ψ₁(a_i+ x_i) < ψ₁(a_i),

for some i. Thus, as long as we make a non-zero observation in either group A_i > a_iand ψ₁(A_i) < ψ₁(a_i) as the trigamma function is monotonically decreasing on R⁺. Thus, as long as we make a non-zero observation there will always be a unique commutative solution.

If x₁= x₂= 0 then clearly the sufficient condition does not hold. However, if we refer back to the conditions from Theorem 5 of Goldstein & Shaw (2004) as given in Section 3.4.2, then, using conditions (ii) and (iii), there is a unique commutative solution if

Var⁻¹₁ (η₁; x₁) + Var⁻¹₁ (η₁; x₂) − Var⁻¹0 (η₁) > 0, or Var⁻¹₁ (η₂; x₁) + Var⁻¹₁ (η₂; x₂) − Var⁻¹0 (η₂) > 0.

If x1= x2= 0 then Var1(η1; x1) = Var0(η1). Thus, taking condition (ii), Var⁻¹₁ (η₁; x₁) + Var⁻¹₁ (η₁; x₂) − Var⁻¹0 (η₁) = Var⁻¹₁ (η₁; x₂) > 0.

We can show condition (iii) holds using the same reasoning. Therefore, in the case of x1 = x2 = 0, there is still a unique, commutative Bayes linear kinematic solution.

This solution is given in Equations 3.18 and 3.19 and provides posterior expectations E₍₂₎(ηi; x1, x2) and variances Var₍₂₎(ηi; x1, x2).

Assuming θ₁ and θ₂ still have marginal gamma distributions (see Section 3.7.2), we find the parameter values of these distributions by solving

E₍₂₎(ηi; x1, x2) = ψ(a^∗_i) − log(b^∗i), Var₍₂₎(ηi; x1, x2) = log(b^∗_i),

for a^∗_i and b^∗_i. Then θ_i; x₁, x₂∼gamma(a^∗i, b^∗_i) and the posterior moments of θ_i are E₍₂₎(θi; x1, x2) = a^∗_i

b^∗_i , Var₍₂₎(θi; x1, x2) = a^∗_i b^∗2_i . 3.13.2 Example: piston ring failures

We wish to provide a Bayes linear Bayes solution to the problem of related numbers of piston ring failures presented in Section 2.6.4. For comparability we use the same prior specifications for the prior gamma distributions of θ1 and θ2 used in the copula methodology. That is a₁ = a₂ = 30 and b₁ = b₂ = 1. We specify a prior correlation between η₁ and η₂ of 0.25. This leads to prior specifications for each η_i of

E0(η1) = E0(η2) = 3.384, Var0(η1) = Var0(η2) = 0.0339.

When we observe x₁= 44 piston ring failures in group 1 and x₂ = 33 failures in group 2 then the resulting posterior expectations and variances of η₁, η₂ are

E₁(η₁) = 3.631, E₁(η₂) = 3.442, Var₁(η₁) = 0.0132, Var₁(η₂) = 0.0160.

Thus there has been a fairly large reduction in the uncertainty on observation of the data. We can propagate these changes through to the other group using Bayes linear kinematics. A unique commutative solution exists as we have observed some failures.

It is

E₍₂₎(η1; x1, x2) = 3.633, E₍₂₎(η2; x1, x2) = 3.471, Var₍₂₎(η₁; x₁, x₂) = 0.0131, Var₍₂₎(η₂; x₁, x₂) = 0.0157.

We use these values to calculate the posterior parameter values for θ₁ and θ₂ under the continued assumption of gamma marginal distributions. They are a^∗₁= 77.01, b^∗₁ = 2.02, a^∗₂= 64.17 and b^∗₂ = 1.98. From these we calculate the posterior expectations and

variances of the Poisson parameters.

E₍₂₎(θ1; x1, x2) = 38.0918, E₍₂₎(θ2; x1, x2) = 32.4102, Var₍₂₎(θ1; x1, x2) = 18.8423, Var₍₂₎(θ2; x1, x2) = 16.3699.

3.13.3 Mode and curvature

Just as with the beta-binomial model we can utilise a guide relationship in the gamma-Poisson setup. This time we shall define our guide relationship as

ηi ≈ µⁱ = log(θi).

Thus, rather than using the mean and variance of µ_i directly, as we did in the previous section, we use a related quantity, η_i, with mean and variance given by the mode of µ_i and curvature at the mode of the log-density of µ_i as we did in the beta-binomial model.

The density of θ_i is

f_θi(θ_i) = b^a_iⁱθ_i^ai−1e^−biθi Γ (a_i) ,

and, in terms of µ_i, θ_i = e^µi. To obtain the density of µ_i, first we must differentiate the density of θ_i to find the Jacobian. The derivative is

dθ_i

dµ_i = e^µi = θi, and so dθi = θidµi. Hence the density of µi is

f_µi(µ_i) = b^a_iⁱθ_i^aie^−biθi Γ (a_i) . If we take logs,

li(µi) = ailog bi+ ailog θi− bⁱθi− log Γ (aⁱ).

Differentiating this and setting the derivative equal to zero gives us the mode of µ_i. The derivative is

d dµi

[l_i(µ_i)] = ai

θi − bi



θ_i = a_i− biθ_i. (3.30) Setting this equal to zero we find the mode. Let mi be the mode of µi. Then let

m^∗_i = e^mi.

This gives a_i− bim^∗_i = 0 so m^∗_i = a_i/b_i, and the mode of µ_i is

mi = log ai

b_i

 .

To find the curvature we must differentiate Equation 3.30 a further time. The second derivative is given by

d²

dµ²_i[l_i(µ_i)] = −biθ_i.

At the mode θ_i = a_i/b_i so, substituting this into the above equation,

 d²l_i(µ_i) Therefore the required prior variance is

Var₀(η_i) = − d²l_i(µ_i)

Hence, our two prior moments can be expressed solely in terms of the prior gamma parameters. They are

Having made the conjugate updates, the moments take the same form but using Ai = a_i+ x_iand B_i= b_i+ 1 in place of a_i and b_i. Defining a Bayes linear structure for η₁, η₂, allows the updates to be propagated to η_j, j 6= i via Equation 3.15.

From Equation 3.17 there is a unique commutative solution to the problem using Bayes linear kinematics if

Var₁(η_i) < Var₀(η_i)

for i = 1 or 2 or both. This condition will clearly hold whenever we make a non-zero observation in either of the groups. Thus a unique commutative solution will virtually always exist. In fact we showed in the previous section that a commutative solution shall exist even if both observations are zero.

Having ascertained that a unique commutative Bayes linear kinematic solution exists it is given by Equations 3.18 and 3.19. Note that, once an adjusted mean and precision for η_i are found, the parameter values of the posterior gamma distributions are found, from posterior mean ¯m_i and variance v_i, as

a^∗_i = 1

v_i, b^∗_i = 1 v_ie^{− ¯mi}.

These can then be used to find posterior means and variances of θ₁, θ₂.

3.13.4 Example: piston ring failures

If we take the same prior values as in the previous example for the gamma distribution parameters and prior correlation then the resulting prior expectations and variances for η₁, η₂ are

E₀(η₁) = E₀(η₂) = 3.401, Var₀(η₁) = Var₀(η₂) = 0.033.

When we observe 46 failures in compressor 1 and 33 failures in compressor 2 then, having made full Bayesian updates within each group, the expectations and variances become

E₁(η₁) = 3.638, E₁(η₂) = 3.500, Var₁(η₁) = 0.0132, Var₁(η₂) = 0.0159.

We can now use Bayes linear kinematics to update our beliefs in both groups as a result of these mean and variance changes. This gives E₁(η_i; x_j) and Var₁(η_i; x_j) where j 6= i for each i. A commutative solution is available, using the sufficient condition, as we observe some piston ring failures. It is

E₍₂₎(η₁; x₁, x₂) = 3.639, E₍₂₎(η₂; x₁, x₂) = 3.478, Var₍₂₎(η₁; x₁, x₂) = 0.0130, Var₍₂₎(η₂; x₁, x₂) = 0.0156.

Solving for the posterior parameter values gives a^∗₁ = 77.02, b^∗₁ = 2.02, a^∗₂ = 64.18 and b^∗₂ = 1.98. Converting back to the expected numbers of piston ring failures results in posterior expectations and variances of

E₍₂₎(θ₁; x₁, x₂) = 38.0683, E₍₂₎(θ₂; x₁, x₂) = 32.3884, Var₍₂₎(θ₁; x₁, x₂) = 18.8170, Var₍₂₎(θ₂; x₁, x₂) = 16.3450.