deriv_2nd_arma11 {simts} | R Documentation |
Analytic D matrix for ARMA(1,1) process
Description
Obtain the second derivative of the ARMA(1,1) process.
Usage
deriv_2nd_arma11(phi, theta, sigma2, tau)
Arguments
phi |
A double corresponding to the phi coefficient of an ARMA(1,1) process.
|
theta |
A double corresponding to the theta coefficient of an ARMA(1,1) process.
|
sigma2 |
A double corresponding to the error term of an ARMA(1,1) process.
|
tau |
A vec containing the scales e.g. 2τ
|
Value
A matrix
with:
The first column containing the second partial derivative with respect to ϕ
;
The second column containing the second partial derivative with respect to θ
;
The third column contains the second partial derivative with respect to σ2
.
The fourth column contains the partial derivative with respect to ϕ
and θ
.
The fiveth column contains the partial derivative with respect to σ2
and ϕ
.
The sixth column contains the partial derivative with respect to σ2
and θ
.
Process Haar WV Second Derivative
Taking the second derivative with respect to ϕ
yields:
∂ϕ2∂2νj2(ϕ,θ,σ2)=(ϕ−1)5(ϕ+1)3τj22σ2⎝⎛(ϕ−1)2((ϕ+1)2(θ2ϕ+θϕ2+θ+ϕ)τj2(ϕ2τj−1)ϕ2τj−2+(ϕ2−1)(θ2(−ϕ)+θ(ϕ2+4ϕ+1)−ϕ)τj(ϕ2τj−2)ϕ2τj−2−2(θ−1)2(ϕτj−4ϕ2τj+3))−12(ϕ+1)2(−21(θ+1)2(ϕ2−1)τj−(θ+ϕ)(θϕ+1)(ϕτj−4ϕ2τj+3))+6(ϕ+1)(ϕ−1)(21(θ+1)2(ϕ2−1)τj+(θ+ϕ)(θϕ+1)(ϕτj−4ϕ2τj+3)+(ϕ+1)(−(θ+ϕ)(θϕ+1)τj(ϕ2τj−2)ϕ2τj−1−θ(θ+ϕ)(ϕτj−4ϕ2τj+3)−(θϕ+1)(ϕτj−4ϕ2τj+3)−(θ+1)2ϕτj))⎠⎞
Taking the second derivative with respect to θ
yields:
∂θ2∂2νj2(ϕ,θ,σ2)=(ϕ−1)3(ϕ+1)τj22σ2((ϕ2−1)τj+2ϕ(ϕτj−4ϕ2τj+3))
Taking the second derivative with respect to σ2
yields:
∂σ4∂2νj2(ϕ,θ,σ2)=0
Taking the derivative with respect to σ2
and θ
yields:
∂θ∂∂σ2∂νj2(ϕ,θ,σ2)=(ϕ−1)3(ϕ+1)τj22((θ+1)(ϕ2−1)τj+(2θϕ+ϕ2+1)(ϕτj−4ϕ2τj+3))
Taking the derivative with respect to σ2
and ϕ
yields:
∂ϕ∂∂σ2∂νj2(ϕ,θ,σ2)=(ϕ−1)4(ϕ+1)2τj22⎝⎛−(ϕ−1)(ϕ+1)⎝⎛−(θ+ϕ)(θϕ+1)τj(ϕ2τj−2)ϕ2τj−1−θ(θ+ϕ)(ϕτj−4ϕ2τj+3)−(θϕ+1)(ϕτj−4ϕ2τj+3)−(θ+1)2ϕτj⎠⎞+(ϕ−1)(−21(θ+1)2(ϕ2−1)τj−(θ+ϕ)(θϕ+1)(ϕτj−4ϕ2τj+3))+3(ϕ+1)(−21(θ+1)2(ϕ2−1)τj−(θ+ϕ)(θϕ+1)(ϕτj−4ϕ2τj+3))⎠⎞
Taking the derivative with respect to ϕ
and θ
yields:
∂θ∂∂ϕ∂νj2(ϕ,θ,σ2)=−(ϕ−1)4(ϕ+1)2τj22σ2⎝⎛τj⎝⎛2(θ+1)(ϕ−1)(ϕ+1)2+2(ϕ2−1)(2θϕ+ϕ2+1)ϕ2τj−1−(ϕ2−1)(2θϕ+ϕ2+1)ϕτj−1⎠⎞+2(θ(ϕ(3ϕ+2)+1)+ϕ(ϕ2+ϕ+3)+1)(ϕτj−4ϕ2τj+3)⎠⎞
Author(s)
James Joseph Balamuta (JJB)
[Package
simts version 0.2.2
Index]