Appendix B:

A computation analogous to the previous one can be performed in the case of a Poisson process. In this case, we must compute the gradient of:

$$J(x)=\int^{x=+\infty}_{x=-\infty}[(h\otimes x)-y{\rm Log}(h\otimes x)]{\rm d}u.$$ (B1)

Considering for the variable "x'', an increase "$\rho$'' in the general direction "s'', we have:

$$J(x+\rho s) \int^{u=+\infty}_{u=-\infty}[\{ h\otimes (x+\rho s) \}$$ =

$$-y {\rm Log}\{h\otimes (x+\rho s)\}]{\rm d}u$$ (B2)

$$J(x+\rho s)= \int^{u=+\infty}_{u=-\infty}\Biggl[\{(h\otimes x)$$

$$+\rho(h\otimes s)\}-y {\rm Log}\left\{(h\otimes x)(1+\rho\frac{(h\otimes s)}{(h\otimes x)}\right\} \Biggr]{\rm d}u$$ (B3)

$$J(x+\rho s)= \int^{u=+\infty}_{u=-\infty}\Biggl[(h\otimes x)+\rho(h\otimes s) -y {\rm Log}(h\otimes h)$$

$$-y{\rm Log}\left\{1+\rho\frac{(h\otimes s)}{(h\otimes x)}\right\}\Biggr]{\rm d}u$$ (B4)

with a first order series expansion:

$$J(x+\rho s)= \int^{u=+\infty}_{u=-\infty}\Biggl[(h\otimes x)+\rho(h\otimes s)$$

$$-y {\rm Log}(h\otimes x)-y\rho\frac{(h\otimes s)}{(h\otimes x)}\Biggr]{\rm d}u$$ (B5)

then

$$J(x+\rho s)= \int^{u=+\infty}_{u=-\infty}\bigl[(h\otimes x) -y {\rm Log}(h\otimes x)\bigr]{\rm d}u$$

$$+\rho\int^{u=+\infty}_{u=-\infty}\left[(h\otimes s)-y\frac{(h\otimes s)}{(h\otimes x)}\right]{\rm d}u$$ (B6)

comparing this relation with: $J(x+\rho s)=J(x)+\rho J'(x,s)$ we have :

$$J'(x,s)=\int^{u=+\infty}_{u=-\infty}\left[(h\otimes s) -y\frac{(h\otimes s)}{(h\otimes x)}\right]{\rm d}u$$ (B7)

that is:

$$J'(x,s)=\int^{u=+\infty}_{u=-\infty}\left[1-\frac{y}{(h\otimes x)}\right](h\otimes s){\rm d}u$$ (B8)

with the integral form of the convolution product, we have:

$$J'(x,s)= \int^{u=+\infty}_{u=-\infty}\left[1-\frac{y}{(h\otimes x)}\right]$$

$$\int^{t=+\infty}_{t=-\infty}s(t)h(u-t){\rm d}t{\rm d}u$$ (B9)

or, re-ordering the integrals:

$$J'(x,s)= \int^{t=+\infty}_{t=-\infty}s(t)$$

$$\int^{u=+\infty}_{u=-\infty}\left[1-\frac{y}{(h\otimes x)} \right]h(u-t){\rm d}u{\rm d}t.$$ (B10)

Let:

$$H(u)=\left[1-\frac{y(u)}{(h(u)\otimes x(u))}\right]$$ (B11)

with the change of variables: z=u-t, we obtain:

$$J'(x,s)=\int^{t=+\infty}_{t=-\infty}s(t)\int^{z=+\infty}_{z=-\infty}H(t+z)h(z){\rm d}z {\rm d}t$$ (B12)

that is:

$$J'(x,s)=\int^{t=+\infty}_{t=-\infty}s(t)[h(-t)\otimes H(t)]{\rm d}t.$$ (B13)

Using the definition of the derivative of J(x) at x in the direction s we obtain:

$$\nabla J(x)=h(-u)\otimes \left[1-\frac{y(u)}{h(u)\otimes x(u)}\right]\cdot$$ (B14)

Acknowledgements

We are grateful to Professor Jean CEA for stimulating discussions, in particular for help in the formulation of the appendices.