Stochastic and Deterministic Models for the Metastatic Emission Process: Formalisms and Crosslinks

Abstract

Although the detection of metastases radically changes prognosis of and treatment decisions for a cancer patient, clinically undetectable micrometastases hamper a consistent classification into localized or metastatic disease. This chapter discusses mathematical modeling efforts that could help to estimate the metastatic risk in such a situation. We focus on two approaches: (1) a stochastic framework describing metastatic emission events at random times, formalized via Poisson processes, and (2) a deterministic framework describing the micrometastatic state through a size-structured density function in a partial differential equation model. Three aspects are addressed in this chapter. First, a motivation for the Poisson process framework is presented and modeling hypotheses and mechanisms are introduced. Second, we extend the Poisson model to account for secondary metastatic emission. Third, we highlight an inherent crosslink between the stochastic and deterministic frameworks and discuss its implications. For increased accessibility the chapter is split into an informal presentation of the results using a minimum of mathematical formalism and a rigorous mathematical treatment for more theoretically interested readers.

Both authors equally contributed to this work.

A Appendix: Proofs of Results of Section 5.4

The proofs provided in this section are based on the following classical result on Poisson random measures. We refer to [45, Chap. 6, pp. 251] for further details. Also, note that this result directly yields Eqs. 2 through 5.

Proposition A.1.

Let (N _t ) _{t ≥ 0} be a PP with intensity λ and P the corresponding Poisson random measure. We have for $\phi, \psi \in {L}^1\left({\mathbb{R}}^{+},\lambda (u) du\right)\cap {L}^2\left({\mathbb{R}}^{+},\lambda (u) du\right)$

$$ \mathbb{E}\left[\int \phi (u)P(du)\right]=\int \phi (u)\lambda (u) du, $$

and

$$ {\displaystyle \begin{array}{c}\mathbb{E}\left[\int \phi \left({u}_1\right)P\left(d{u}_1\right)\int \phi \left({u}_2\right)P\left(d{u}_2\right)\right]=\int \phi \left({u}_1\right)\lambda \left({u}_1\right)d{u}_1\int \psi \left({u}_2\right)\lambda \left({u}_2\right)d{u}_2\\ {}\kern15.5em +\int \phi (u)\psi (u)\lambda (u) du.\end{array}} $$

In other words, we can write the first order moment of the Poisson random measure P in a more compact form

$$ \mathbb{E}\left[P(du)\right]=\lambda (u) du, $$

as well as its second order moment

$$ \mathbb{E}\left[P\left(d{u}_1\right)P\left(d{u}_2\right)\right]=\lambda \left({u}_1\right)\lambda \left({u}_2\right)d{u}_1d{u}_2+\delta \left({u}_1-{u}_2\right)\lambda \left({u}_1\right)d{u}_1d{u}_2. $$

Moreover, to simplify notations in the forthcoming computations, we introduce the following convolution-like notation: for functions ϕ, ψ

$$ \phi \ast \psi (t):= \underset{0}{\overset{t}{\int }}\phi \left(t-u\right)\psi (u) du. $$

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1.1 A.1 Proof of Proposition 5.1

We first need to establish the following lemma, which is proven further below.

Lemma A.1.

We have

$$ {e}_f={\lambda}_p\ast \left(f\left({X}_m\right)+{e}_{m,f}\right), $$

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where e _m, f has been introduced in Proposition 5.2.

This is not exactly the renewal equation we want. To derive the desired equation (Eq. 10) we just have to make the following remark. Taking λ _p = λ _m , Lemma A.1 gives that e _m, f satisfies

$$ {e}_{m,f}={\lambda}_m\ast \left(f\left({X}_m\right)+{e}_{m,f}\right). $$

Hence, from (Eq. 17), we have

$$ {\displaystyle \begin{array}{ccc}{e}_f& =& {\lambda}_p\ast f\left({X}_m\right)+{\lambda}_p\ast \left({\lambda}_m\ast f\left({X}_m\right)+{\lambda}_m\ast {e}_{m,f}\right)\\ {}& =& {\lambda}_p\ast f\left({X}_m\right)+{\lambda}_m\ast \left({\lambda}_p\ast f\left({X}_m\right)+{\lambda}_p\ast {e}_{m,f}\right)\\ {}& =& {\lambda}_p\ast f\left({X}_m\right)+{\lambda}_m\ast {e}_f.\end{array}} $$

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Now, let T > 0, we have from the last line of (Eq. 18) that for all t ∈ [0, T],

$$ {e}_f(t)\le \parallel f{\parallel}_{L^{\infty}\left(\left[{x}_m^0,{x}_m^{\infty}\right]\right)}\parallel {\lambda}_p{\parallel}_{L^{\infty}\left(\left[0,T\right]\right)}T+\parallel {\lambda}_m{\parallel}_{L^{\infty}\left(\left[0,T\right]\right)}\underset{0}{\overset{t}{\int }}{e}_f(u) du, $$

which gives using Gronwall’s inequality

$$ \underset{t\in \left[0,T\right]}{\sup }{e}_f(t)\le {C}_{T,f,{\lambda}_p,{\lambda}_m}\parallel f{\parallel}_{L^{\infty}\left(\left[{x}_m^0,{x}_m^{\infty}\right]\right)}<+\infty . $$

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As a result, e _f (t) &lt; +∞ for all t ≥ 0 since T is arbitrary, and also

$$ \mathbb{P}\left(\right.{\mathrm{SMO}}_f(T)<+\infty \left)\right.=1. $$

Finally, using that t ↦ SMO _f (t) is an increasing non-negative function, we have

$$ \mathbb{P}\left(\right.\forall t\in \left[0,T\right],\kern1em {\mathrm{SMO}}_f(t)<+\infty \left)\right.=1, $$

and then

$$ \mathbb{P}\left(\right.\forall t\ge 0,\kern1em {\mathrm{SMO}}_f(t)<+\infty \left)\right.=\underset{n\to +\infty }{\lim}\mathbb{P}\left(\right.\forall t\in \left[0,n\right],\kern1em {\mathrm{SMO}}_f(t)<+\infty \left)\right.=1. $$

Proof (of Lemma A.1).

Let us start with the following remark. According to the recursive definition (Eq. 11) of our PP cascade, one has

$$ {T}^{\left({n}_1,\dots, {n}_k\right)}={T}^{\left({n}_1\right)}+{\overline{T}}^{\left({n}_1,\dots, {n}_k\right)}, $$

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where all the times

$$ \left\{\right.{\overline{T}}^{\left({n}_1,\dots, {n}_k\right)},\kern1em k\ge 2,\kern1em {n}_1,\dots, {n}_k\ge 1\Big\} $$

are independent of ${\Pi}^{(1)}:= {\left({T}^{\left({n}_1\right)}\right)}_{n_1\ge 1}$.

Now, from this consideration, by taking apart the first generation of metastasis, we can rewrite SMO _f as follows:

$$ {\displaystyle \begin{array}{ccc}{\mathrm{SMO}}_f(t)& =& \sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}f\left({X}_m\left(\right.t-{T}^{\left({n}_1\right)}\right)\Big)\\ {}& & +\sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}{\mathrm{SMO}}_{n_1,f}\left(t-{T}^{\left({n}_1\right)}\right)\\ {}& := & I+J,\end{array}} $$

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with

$$ {\mathrm{SMO}}_{n_1,f}(t):= \sum \limits_{k\ge 2}\sum \limits_{n_2,\dots, {n}_k\ge 1}{\mathbf{1}}_{\left({\overline{T}}^{\left({n}_1,\dots, {n}_k\right)}\le t\right)}f\left({X}_m\left(\right.t-{\overline{T}}^{\left({n}_1,\dots, {n}_k\right)}\right)\Big), $$

which are independent of ${\Pi}^{(1)}$. Note that all the times

$$ {\overline{\Pi}}_{n_1}:= \left\{\right.{\overline{T}}^{\left({n}_1,\dots, {n}_k\right)},\kern1em k\ge 2,\kern1em {n}_2,\dots, {n}_k\ge 1\Big\}, $$

can be defined following (Eq. 11), but with λ _m as intensity for all the PPs since we consider all the times from the second generation. Therefore, ${\left({\mathrm{SMO}}_{n_1,f}\right)}_{n_1\ge 1}$ are all independent. Moreover, the shape of all the ${\mathrm{SMO}}_{n_1,f}$ is similar to SMO _f except that the PPs in the cascade have all λ _m for intensity. Hence, all the ${\mathrm{SMO}}_{n_1,f}$ have the same law as SMO _m, f .

Using Proposition A.1 with the Poisson random measure P ⁽¹⁾(du) associated to (N _t ⁽¹⁾) _{t ≥ 0} (with intensity λ _p ), it is direct to see that

$$ {\displaystyle \begin{array}{cc}\mathbb{E}\left[I\right]=\mathbb{E}\left[\underset{0}{\overset{t}{\int }}f\left({X}_m\left(t-u\right)\right){P}^{(1)}(du)\right]& =\underset{0}{\overset{t}{\int }}{\lambda}_p(u)f\left({X}_m\left(t-u\right)\right) du\\ {}& ={\lambda}_p\ast f\left({X}_m\right)(t).\end{array}} $$

For the second term, using standard properties of the conditional expectation (especially $\mathbb{E}\left[X\right]=\mathbb{E}\left[\mathbb{E}\left[X|Y\right]\right]$), one has

$$ \mathbb{E}\left[ II\right]=\mathbb{E}\left[\right.\sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1,f}\left(t-{T}^{\left({n}_1\right)}\right)\left|\right.{\Pi}^{(1)}\left]\right] $$

with

$$ {\displaystyle \begin{array}{ccc}\mathbb{E}\left[\right.{\mathrm{SMO}}_{f,{n}_1}\left(t-{T}^{\left({n}_1\right)}\right)\left|\right.{\Pi}^{(1)}\left]\right.& =& \mathbb{E}{\left[{\mathrm{SMO}}_{n_1,f}\left(t-u\right)\right]}_{\mid u={T}_{n_1}^{(1)}}\\ {}& =& \mathbb{E}{\left[{\mathrm{SMO}}_{m,f}\left(t-u\right)\right]}_{\mid u={T}^{\left({n}_1\right)}}\\ {}& =& {e}_{m,f}\left(t-{T}^{\left({n}_1\right)}\right),\end{array}} $$

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and then

$$ \sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1,f}\left(t-{T}^{\left({n}_1\right)}\right)\left|\right.{\Pi}^{(1)}\Big]=\underset{0}{\overset{t}{\int }}{e}_{m,f}\left(t-u\right){P}^{(1)}(du). $$

This, together with Proposition A.1, yields

$$ \mathbb{E}\left[ II\right]=\mathbb{E}\left[\right.\underset{0}{\overset{t}{\int }}{e}_{m,f}\left(t-u\right){P}^{(1)}(du)\left]\right.=\underset{0}{\overset{t}{\int }}{\lambda}_p(u){e}_{m,f}\left(t-u\right) du={\lambda}_p\ast {e}_{m,f}(t), $$

which concludes the proof of (Eq. 17). □

1.2 A.2 Proof of Proposition 5.2

Using the same strategy as for (Eq. 18), the proof of (Eq. 15) consists only in proving the following relation:

$$ {v}_f={\lambda}_p\ast \left(\right.f\left({X}_m\right)+{e}_{m,f}{\left)\right.}^2+{\lambda}_p\ast {v}_{m,f}, $$

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with v _m, f (t): = var[SMO _m, f (t)]. Knowing Proposition 5.1 and the formula of the variance, one can focus on the term ${e}_f^{(2)}(t):= \mathbb{E}\left[{\mathrm{SMO}}_f^2(t)\right]$. Using (Eq. 21), we have to compute three terms

$$ {e}_f^{(2)}(t)=\mathbb{E}\left[{I}^2\right]+2\mathbb{E}\left[ IJ\right]+\mathbb{E}\left[{J}^2\right]. $$

The Term $\mathbb{E}\left[{I}^2\right]$

Reminding that $I={\int}_0^tf\left({X}_s\left(\right.t-u\right)\left)\right.{P}^{(1)}(du)$, and using Proposition A.1, it is direct that

$$ \mathbb{E}\left[{I}^2\right]=\left(\right.\underset{0}{\overset{t}{\int }}{\lambda}_p(u)f\left({X}_m\left(t-u\right)\right) du\Big){}^2+\underset{0}{\overset{t}{\int }}{\lambda}_p(u){f}^2\left({X}_m\left(t-u\right)\right) du. $$

The Term $\mathbb{E}\left[ IJ\right]$

Using standard properties of the conditional expectation, and that for all n ₁ ≥ 1

$$ {e}_{m,f}(t)=\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1,f}(t)\Big], $$

we have using (Eq. 22)

$$ {\displaystyle \begin{array}{c}\mathbb{E}\left[ IJ\right]=\mathbb{E}\left[\sum \limits_{n_1^1,{n}_1^2\ge 1}{\mathbf{1}}_{\left(\right.{T}^{\left({n}_1^1\right)}\le t\Big)}{\mathbf{1}}_{\left(\right.{T}^{\left({n}_1^2\right)}\le t\Big)}f\left({X}_m\left(\right.t-{T}^{\left({n}_1^1\right)}\right)\left)\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1^2,f}\left(\right.t-{T}^{\left({n}_1^2\right)}\right)\left|\right.{\uppi}^{(1)}]\right]\\ {}\kern2.3em =\mathbb{E}\left[\sum \limits_{n_1^1,{n}_1^2\ge 1}{\mathbf{1}}_{\left(\right.{T}^{\left({n}_1^1\right)}\le t\Big)}{\mathbf{1}}_{\left(\right.{T}^{\left({n}_1^2\right)}\le t\Big)}f\left({X}_m\left(\right.t-{T}^{\left({n}_1^1\right)}\right)\left){e}_{m,f}\left(\right.t-{T}^{\left({n}_1^2\right)}\right)\right].\end{array}} $$

As result, according to Proposition A.1, we have

$$ {\displaystyle \begin{array}{c}\mathbb{E}\left[ IJ\right]=\mathbb{E}\left[\underset{0}{\overset{t}{\int }}f\Big({X}_m\left(t-u\right)\left)\right.{P}^{(1)}(du)\underset{0}{\overset{t}{\int }}{e}_{m,f}\left(t-u\right)\left)\right.{P}^{(1)}(du)\right]\\ {}=\underset{0}{\overset{t}{\int }}{\lambda}_p\left({u}_1\right)f\Big({X}_m\left(t-{u}_1\right)\left)\right.d{u}_1\underset{0}{\overset{t}{\int }}{\lambda}_p\left({u}_2\right){e}_{m,f}\left(t-{u}_2\right)d{u}_2\\ {}+\underset{0}{\overset{t}{\int }}{\lambda}_p(u)f\Big({X}_m\left(t-u\right)\left)\right.{e}_{m,f}\left(t-u\right) du.\end{array}} $$

The Term $\mathbb{E}\left[{J}^2\right]$

To compute this term we have to consider two cases

$$ {\displaystyle \begin{array}{c}{J}^2=\sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}{\mathrm{SMO}}_{n_1,f}^2\left(t-{T}^{\left({n}_1\right)}\right)\\ {}+\sum \limits_{\begin{array}{c}{n}_1^1,{n}_1^2\ge 1\\ {}{n}_1^1\ne {n}_1^2\end{array}}{\mathbf{1}}_{\left({T}^{\left({n}_1^1\right)}\le t\right)}{\mathbf{1}}_{\left({T}^{\left({n}_1^2\right)}\le t\right)}{\mathrm{SMO}}_{n_1^1,f}\left(t-{T}^{\left({n}_1^1\right)}\right){\mathrm{SMO}}_{n_1^2,f}\left(t-{T}^{\left({n}_1^2\right)}\right)\\ {}:= {J}_1+{J}_2.\end{array}} $$

Following (Eq. 22), but with SMO _n, f ² instead, we have

$$ {\displaystyle \begin{array}{c}\mathbb{E}\left[{J}_1\right]=\mathbb{E}\left[\sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left(\right.{T}^{\left({n}_1^1\right)}\le t\Big)}\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1,f}^2\left(\right.t-{T}^{\left({n}_1\right)}\Big)\left|\right.{\Pi}^{(1)}]\right]\\ {}=\mathbb{E}\left[\right.\underset{0}{\overset{t}{\int }}{e}_{m,f}^{(2)}\left(t-u\right){P}^{(1)}(du)]\\ {}=\underset{0}{\overset{t}{\int }}{\lambda}_p(u){e}_{m,f}^{(2)}\left(t-u\right) du,\end{array}} $$

where ${e}_{m,f}^{(2)}(t):= \mathbb{E}\left[{\mathrm{SMO}}_{m,f}^2(t)\right]$. Now using that ${\mathrm{SMO}}_{n_1^1,f}$ and ${\mathrm{SMO}}_{n_1^2,f}$ are independent for n ₁ ¹ ≠ n ₁ ², we have, using (Eq. 22) and Proposition A.1 one more time,

$$ {\displaystyle \begin{array}{c}\mathbb{E}\left[{J}_2\right]\\ {}=\mathbb{E}\left[\right.\sum \limits_{n_1^1\ne {n}_1^2}{\mathbf{1}}_{\left({T}^{\left({n}_1^1\right)}\le t\right)}{\mathbf{1}}_{\left({T}^{\left({n}_1^2\right)}\le t\right)}\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1^1,f}\left(t-{T}^{\left({n}_1^1\right)}\right)\left|\right.{\Pi}^{(1)}]\mathbb{E}\left[\right.{\mathrm{SMO}}_{n_1^2,f}\left(t-{T}^{\left({n}_1^2\right)}\right)\left|\right.{\Pi}^{(1)}]]\\ {}=\mathbb{E}\left[\right.\left(\right.\sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}{e}_{m,f}\left(t-{T}^{\left({n}_1\right)}\right)\Big){}^2]-\mathbb{E}\left[\right.\sum \limits_{n_1\ge 1}{\mathbf{1}}_{\left({T}^{\left({n}_1\right)}\le t\right)}{e}_{m,f}^2\left(t-{T}^{\left({n}_1\right)}\right)]\\ {}=\mathbb{E}\left[\right.\left(\right.\underset{0}{\overset{t}{\int }}{e}_{m,f}\left(t-u\right){P}^{(1)}(du)\Big){}^2]-\mathbb{E}\left[\right.\underset{0}{\overset{t}{\int }}{e}_{m,f}^2\left(t-u\right){P}^{(1)}(du)]\\ {}=\left(\right.\underset{0}{\overset{t}{\int }}{\lambda}_p(u){e}_{m,f}\left(t-u\right) du\Big){}^2.\end{array}} $$

Combining the three previous computations, we obtain

$$ {e}_f^{(2)}={\left({\lambda}_p\ast \left(f\left({X}_m\right)+{e}_{m,f}\right)\right)}^2+{\lambda}_p\ast {f}^2\left({X}_m\right)+2{\lambda}_p\ast \left(f\left({X}_m\right){e}_{m,f}\right)+{\lambda}_p\ast {e}_{m,f}^{(2)}. $$

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Considering this equation for λ _p = λ _m , we obtain as for the expectation a renewal equation for e _m, f ⁽²⁾, which yields for all t > 0

$$ {e}_{m,f}^{(2)}(t)\le {C}_1+{C}_2\underset{0}{\overset{t}{\int }}{e}_{m,f}^{(2)}(u) du, $$

and then for all T > 0,

$$ \underset{t\in \left[0,T\right]}{\sup }{e}_{m,f}^{(2)}(t)\le {C}_{1,T}+{C}_{2,T}\underset{t\in \left[0,T\right]}{\sup }{e}_{m,f}^2<+\infty, $$

using Gronwall’s inequality and Proposition 5.1. This proves that $\mathbb{E}\left[{\mathrm{SMO}}_{m,f}^2(t)\right]<+\infty$ for all t ≥ 0, and then $\mathbb{E}\left[{\mathrm{SMO}}_f^2(t)\right]<+\infty$ by going back to (Eq. 24). Now, rewriting (Eq. 24), we obtain

$$ {e}_f^{(2)}={e}_f^2+{\lambda}_p\ast {\left(f\left({X}_m\right)+{e}_{m,f}\right)}^2+{\lambda}_p\ast {v}_{m,f}, $$

which is (Eq. 23).

1.3 A.3 Proof of Theorem 5.2

Using that X _m (s) ∈ [x _m ⁰, x _m ^∞] for all $s\in {\mathbb{R}}^{+}$, the σ-finiteness and absolute continuity of μ _t (for any t ≥ 0) are direct consequences of (Eq. 19). Denoting by ${\overset{\sim }{\rho}}_t$ its Radon–Nikodým density, Proposition 5.1 and Theorem 5.1 then yield

$$ \underset{x_m^0}{\overset{x_m^{\infty }}{\int }}f(x){\mu}_t(dx)=\underset{x_m^0}{\overset{x_m^{\infty }}{\int }}f(x){\overset{\sim }{\rho}}_t(x) dx=\underset{x_m^0}{\overset{x_m^{\infty }}{\int }}f(x)\rho \left(t,x\right) dx, $$

for all $f\in C\left(\left[{x}_m^0,{x}_m^{\infty}\right]\right)\cap {L}^{\infty}\left(\left[{x}_m^0,{x}_m^{\infty}\right]\right)$, which concludes the proof.