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Separable function (Rev #1)

Separable function

Idea

Details

Multiplicatively separable

(1)F(Θ,x)= i=1:nf i(θ i,x i) F(\Theta,\mathbf{x})=\prod_{i=1:n} f_i(\theta_i,x_i)

Then

(2)F(Θ,x)θ j=f j(θ j,x j)θ j i=1:n,ijf i(θ i,x i) \frac{\partial F(\Theta,\mathbf{x})}{\partial \theta_j} =\frac{\partial f_j(\theta_j,x_j)}{\partial \theta_j} \prod_{i=1:n,i \ne j} f_i(\theta_i,x_i)
(3) 2F(Θ,x)θ j 2= 2f j(θ j,x j)θ j 2 i=1:n,ijf i(θ i,x i) \frac{\partial^2 F(\Theta,\mathbf{x})}{\partial \theta_j^2} =\frac{\partial^2 f_j(\theta_j,x_j)}{\partial \theta_j^2} \prod_{i=1:n,i \ne j} f_i(\theta_i,x_i)
(4) 2F(Θ,x)θ jθ k=f j(θ j,x j)θ jf k(θ k,x k)θ k i=1:n,ij,kf i(θ i,x i) \frac{\partial^2 F(\Theta,\mathbf{x})}{\partial \theta_j \partial \theta_k} =\frac{\partial f_j(\theta_j,x_j)}{\partial \theta_j} \frac{\partial f_k(\theta_k,x_k)}{\partial \theta_k} \prod_{i=1:n,i \ne j,k} f_i(\theta_i,x_i)

Special case:

(5)F(Θ,x)=exp i=1:nf i(θ i,x i) F(\Theta,\mathbf{x})=\exp \sum_{i=1:n} f_i(\theta_i,x_i)

Then

(6)F(Θ,x)θ j=exp(f j(θ j,x j)θ j+ i=1:n,ijf i(θ i,x i)) \frac{\partial F(\Theta,\mathbf{x})}{\partial \theta_j} =\exp (\frac{\partial f_j(\theta_j,x_j)}{\partial \theta_j} +\sum_{i=1:n,i \ne j} f_i(\theta_i,x_i))
(7) 2F(Θ,x)θ j 2=exp( 2f j(θ j,x j)θ j 2+ i=1:n,ijf i(θ i,x i)) \frac{\partial^2 F(\Theta,\mathbf{x})}{\partial \theta_j^2} =\exp (\frac{\partial^2 f_j(\theta_j,x_j)}{\partial \theta_j^2} +\sum_{i=1:n,i \ne j} f_i(\theta_i,x_i) )
(8) 2F(Θ,x)θ jθ k=exp(f j(θ j,x j)θ jf k(θ k,x k)θ k+ i=1:n,ij,kf i(θ i,x i)) \frac{\partial^2 F(\Theta,\mathbf{x})}{\partial \theta_j \partial \theta_k} =\exp (\frac{\partial f_j(\theta_j,x_j)}{\partial \theta_j} \frac{\partial f_k(\theta_k,x_k)}{\partial \theta_k} +\sum_{i=1:n,i \ne j,k} f_i(\theta_i,x_i) )