Can a Training Image Be a Substitute for a Random Field Model?

Abstract

In most multiple-point simulation algorithms, all statistical features are provided by one or several training images (TI) that serve as a substitute for a random field model. However, because in practice the TI is always of finite size, the stochastic nature of multiple-point simulation is questionable. This issue is addressed by considering the case of a sequential simulation algorithm applied to a binary TI that is a genuine realization of an underlying random field. At each step, the algorithm uses templates containing the current target point as well as all previously simulated points. The simulation is validated by checking that all statistical features of the random field (supported by the simulation domain) are retrieved as an average over a large number of outcomes. The results are as follows. It is demonstrated that multiple-point simulation performs well whenever the TI is a complete (infinitely large) realization of a stationary, ergodic random field. As soon as the TI is restricted to a limited domain, the statistical features cannot be obtained exactly, but integral range techniques make it possible to predict how much the TI should be extended to approximate them up to a prespecified precision. Moreover, one can take advantage of extending the TI to reduce the number of disruptions in the execution of the algorithm, which arise when no conditioning template can be found in the TI.

Appendices

Appendix A: Proof of Algorithm 1

The proof is established by induction. Let S ₀ and S ₁ be the current level sets of the outcome, and suppose that $\operatorname{Prob} \{ Z(S_{0})=0, Z(S_{1})=1 \} = \beta> 0$. Put $p = \lim_{\lambda\rightarrow\infty} \frac{N_{\lambda}}{D_{\lambda}}$ with

$$N_\lambda= \frac{\# [ \eta^I_{T_1} \cap Sq(\lambda) ] }{\# Sq(\lambda)} \qquad D_\lambda= \frac{\# [ \eta^I_{T_0} \cap Sq(\lambda) ] + \# [ \eta^I_{T_1} \cap Sq(\lambda) ]}{ \# Sq(\lambda)}. $$

By the ergodic property (Eq. (1)), one has

$$\begin{aligned} \lim_{\lambda\rightarrow\infty} N_\lambda&= \operatorname{Prob} \bigl\{ Z(S_0) = 0, Z \bigl(S_1 \cup\{\bold x\} \bigr) = 1 \bigr\} \\ \lim_{\lambda\rightarrow\infty} D_\lambda&= \operatorname{Prob} \bigl\{ Z \bigl(S_0 \cup\{ \bold x\} \bigr) = 0, Z(S_1) = 1 \bigr \} + \operatorname{Prob} \bigl\{ Z(S_0) = 0, Z \bigl(S_1 \cup\{\bold x\} \bigr) = 1 \bigr\} \\ &= \operatorname{Prob} \bigl\{ Z(S_0) = 0, Z(S_1) = 1 \bigr\}. \end{aligned}$$

As lim _λ→∞ D _λ >0, it follows

$$p = \frac{\lim_{\lambda\rightarrow\infty} N_\lambda}{\lim _{\lambda\rightarrow\infty} D_\lambda} = \operatorname{Prob} \bigl \{ Z(\bold x) = 1 \, \vert\, Z(S_0) = 0, Z(S_1) = 1 \bigr\}. $$

Let $S'_{0}$ and $S'_{1}$ be the next level sets obtained once $\bold x$ has been allocated. Note that p can take all values on [0,1]. If p<1, step (iv) of Algorithm 1 shows that $\bold x$ can be assigned the value 0, in which case $S'_{0} = S_{0} \cup\{\bold x\}$ and $S'_{1} = S_{1}$, and one has $\operatorname{Prob} \{ Z(S'_{0}) = 0 , Z(S'_{1}) = 1 \} = (1-p) \beta> 0$. Similarly, if p>0, then $\bold x$ can be assigned the value 1, in which case $S'_{0} = S_{0}$ and $S'_{1} = S_{1} \cup\{\bold x\}$, and one has $\operatorname{Prob} \{ Z(S'_{0}) = 0 , Z(S'_{1}) =1 \} = p \beta> 0$. Consequently, one has $\operatorname{Prob} \{ Z(S'_{0}) = 0 , Z(S'_{1}) = 1 \} > 0$ whatever the allocation of $\bold x$. The induction hypothesis is thus preserved, which proves the correctness of the sequential algorithm for infinite TI’s.

Appendix B: Proof of Eqs. (3) and (4)

To calculate $\operatorname{Prob} \{ Z (K_{0}) = 0 \}$, the starting point is to express that none of the Boolean objects hits K ₀

$$\operatorname{Prob} \bigl\{ Z (K_0) = 0 \bigr\} = \operatorname{Prob} \bigl\{ \forall\bold u \in\mathbb{Z}^2 , \forall n \leq N( \bold u) , \tau_{\bold{u}} A_{\bold u,n} \cap K_0 = \varnothing\bigr\}. $$

The right-hand side is now expanded using the fact that the Boolean model is made of independent objects in independent Poisson numbers

$$\begin{aligned} \operatorname{Prob} \bigl\{ Z (K_0) = 0 \bigr\} &= \prod _{\bold u \in\mathbb{Z}^2} \operatorname{Prob} \bigl\{ \forall n \leq N(\bold u) , \tau_{\bold{u}} A_{\bold u,n} \cap K_0 = \varnothing\bigr \} \\ &= \prod_{\bold u \in\mathbb{Z}^2} \sum_{n=0}^\infty \exp(- \theta) \frac {\theta^n}{n!} \bigl(\operatorname{Prob} \{ \tau_{\bold{u}} A \cap K_0 = \varnothing\} \bigr)^n \\ &= \prod_{\bold u \in\mathbb{Z}^2} \exp\bigl( - \theta+ \theta \operatorname{Prob} \{ \tau_{\bold{u}} A \cap K_0 = \varnothing\} \bigr) \\ &= \prod_{\bold u \in\mathbb{Z}^2} \exp\bigl( - \theta \operatorname{Prob} \{ \tau_{\bold{u}} A \cap K_0 \neq \varnothing\bigr). \end{aligned}$$

Moreover, one has $\tau_{\bold{u}} A \cap K_{0} \neq\varnothing$ if and only if $A \cap\tau_{-\bold u} K_{0} \neq\varnothing$, that is, $- \bold u \in\delta_{K_{0}} A$. Accordingly,

$$\begin{aligned} \operatorname{Prob} \bigl\{ Z (K_0) = 0 \bigr\} &= \exp\biggl( - \theta\sum_{\bold u \in\mathbb{Z} ^2} \operatorname{Prob} \{ - \bold u \in \delta_{K_0} A \} \biggr) \\ &= \exp\biggl( - \theta\sum_{\bold u \in\mathbb{Z}^2} E \{ 1_{- \bold u \in\delta_{K_0} A} \} \biggr) \\ &= \exp\biggl( - \theta E \biggl\{ \sum_{\bold u \in\mathbb{Z}^2} 1_{- \bold u \in\delta_{K_0} A} \biggr\} \biggr) \\ &= \exp\bigl( - \theta E \{ \# \delta_{K_0} A \} \bigr) \end{aligned}$$

as announced in Eq. (3).

To prove Eq. (4), rewrite the probability as the expectation of an indicator function

$$\begin{aligned} \operatorname{Prob} \bigl\{ Z(K_0) = 0, Z(K_1)=1 \bigr \} &= E \biggl\{ 1_{ Z(K_0) = 0 } \, \prod_{\bold x_1 \in K_1} 1_{Z(\bold x_1) = 1} \biggr\} \\ &= E \biggl\{ 1_{ Z(K_0) = 0 } \, \prod_{\bold x_1 \in K_1} ( 1 - 1_{Z(\bold x_1) = 0} ) \biggr\}. \end{aligned}$$

By expanding, one obtains

$$\begin{aligned} \operatorname{Prob} \bigl\{ Z(K_0) = 0, Z(K_1)=1 \bigr \} &= E \biggl\{ 1_{ Z(K_0) = 0 } \sum_{L \subset K_1} (-1)^{\# L} 1_{Z(L) = 0} \biggr\} \\ &= \sum_{L \subset K_1} (-1)^{\# L} E \{ 1_{ Z(K_0 \cup L) = 0 } \} \\ &= \sum_{L \subset K_1} (-1)^{\# L} \operatorname{Prob} \bigl\{ Z(K_0 \cup L) = 0 \bigr\}. \end{aligned}$$

Equation (4) is derived by replacing $\operatorname{Prob} \{ Z(K_{0} \cup L) = 0 \}$ by its expression given in Eq. (3).