# CTRW: solve a renewal equation

Let X(t) be a continuous time random walk, with exponentially distributed waiting times of pdf $f_T(t)= k e^{-k t}\; t\geq 0$ and a jump sizes pdf $f_J(x)$. Suppose the initial distribution $\rho_{X(0)}(x)$. I would like to prove (or correct) the following intuitive expression for the survival probability above a certain threshold $a$: \begin{align} \Pr (X(t')\geq a \; \forall t'\leq t \; | \; \rho_{X(0)}) &\equiv S_a(t \; | \; \rho_{X(0)}) \\ &= \int_a^\infty \Big(\sum_{n=0}^\infty \frac{e^{-kt}(kt)^n \mathcal{O}^n}{n!}\Big) \rho_{X(0)} (x) dx\\ &= \sum_{n=0}^\infty \frac{e^{-kt}(kt)^n }{n!}\int_a^\infty\mathcal{O}^n\rho_{X(0)} (x) dx \end{align} where $\mathcal{O}$ is an operator that I believe to be $\circledast_{f_J}\circ H_a$ the multiplication by a Heaviside function with a cut off at $a$ followed by the convolution by the jump size pdf.

The first step would be to write a renewal equation for $S_a$, here is my attempt, first for $a<x_0$ and $\rho_{X(0)}=\delta_{x_0}$: \begin{align} S_{a}(t|\delta_{x_0})&=\int_t^{\infty} f_T(t')dt'+\int_0^t \int_{a}^{\infty}S_{a}(t-t'|\delta_{x})f_J(x-x_0)f_T(t')dxdt' \end{align} The probability to stay above $a$ up to time $t$ starting from $x_0$ is the probability to stay at $x_0$ (no jump) plus the probability to make any jump to $x\geq a$ at time $t'<t$ and to survive above $a$ up to time $t$. We could then define $S_a(t \; | \; \rho_{X(0)})=\int_a^\infty S_a(t \; | \; \delta_{x})\rho_{X(0)}(x)dx$ \begin{align} \int_a^\infty S_{a}(t|\delta_{x_0})\rho_{X(0)}(x_0)dx_0 &= \int_a^\infty \rho_{X(0)}(x_0)dx_0\int_t^\infty f_T(t')dt'\\ &+ \int_a^\infty \rho_{X(0)}(x_0)\int_0^t \int_a^\infty S_a(t-t'|\delta_{x})f_J(x-x_0)f_T(t')dxdt'dx_0 \end{align}

We can easily simplify the time part of this integral equation by taking its Laplace transform with respect to $t$. And I see how to get the Poisson distribution. However, the space part is more complicated. We can write explicitly the Heaviside's functions and take the Fourier transform, but this does not lead me to the expected result. The renewal equation is probably not correctly set.

Do you see anything wrong here? A resolution using a Green function may be another way.

• Slight remark : I personally find $\mathcal{O} g(x) = \mathbb{E}(g(x + J)) = \int_{\mathbb{R}} g(x + u) \, f_J(u) du$ which is slightly different from the convolution $g*f_J(x) := \int_{\mathbb{R}} g(x - u) \, f_J(u) du$. Maybe $\mathcal{O} = opp \circ \star_{f_J } \circ H_a$ where $opp\, g(x) := g(-x)$ ? Aug 18, 2017 at 0:01
• Nice comment, my problem for getting the proof is exactly around there. Indeed, if we don't consider the boundary / the Heaviside, your representation of the operator as an expectation makes sense, except that the operator should give the position density after one jump, which would mean $mathcal{O}g(y)=\mathbb{E}(g(y-J))$ (? because y=x+J) and this is the convolution. But this is conflicting with the meaning of the renewal equation. If I write $f_J(x_0-x)$, I get the convolution in the operator but $f_J(x-x_0)$ seems to be the correct one as it means "jump from x_0 to x (and survive from x)". Aug 18, 2017 at 14:33
• Maybe is it just a problem of convention in the renewal equation then ? If you consider minus your walk instead of your walk ? Intuitively, the jump from $x_0$ to $x$ would be $f(x - x_0)$ indeed... Aug 19, 2017 at 0:05
• Yes, maybe, but the reverse walk does not have any defined initial value, which prevent us from writing any other renewal equation. If I rewrite a bit the question, for $x_0=b$, $a=0$ and $t\rightarrow \infty$, I would like to know whether $$S(b)=\lim_{n\rightarrow \infty} \int_0^{\infty} [\circledast_f\circ H]^n\delta_b \; (x)dx$$ is solution of $$S(b)=\int_0^\infty S(u)f(u-b)du$$...? Plugging one expression in the other, we can append an $H$ to the operator, but I don't quite know how to deal with the shifting value of the Dirac and the operator which is acting on it and get the convolution. Aug 20, 2017 at 16:13

Using my previous remark (namely the correction of the operator $\mathcal{O}$), an expression would just involve the "skeleton" of your continuous time random walk, namely the walk with initial law $\rho_{X_0} \equiv \rho_0$ and jump law $f_J$, or, $Y_k = R_0 + \sum_{m = 1}^k J_m$. Then, this survival probability would be $\mathbb{P}(R_0 \geqslant a, Y_1 \geqslant a, \dots, Y_{[t]}\geqslant a) = \int_a^{+\infty} \mathcal{O}^{[t]}\rho_0(x) dx$ (with the corrected operator $\mathcal{O}$). Here, $[t]$ is the integer part of $t$.
Your walk is just $X_t = Y_{N_t}$ where $(N_t)$ is an independent Poisson process of parameter $k$ (i.e. $\mathbb{P}(N_t = m) = e^{-k t} (kt)^m/m!$). Thus, what you have written is more of the form $\mathbb{P}_{\rho_0}( Z_s \geqslant a, \, \forall s \geqslant 0 )$ where $Z = Y\circ N^{(kt)}$ and $N^{(kt)}$ is an independent Poisson process of parameter $kt$ (but you need to write $e^{-kt}$ in place of $e^{-knt}$).
• Thank you for pointing out that parenthesis mistake (I will edit it), and for expressing this process in a simpler way. However, I didn't know that we could represent it as a composition of two random variables. The survival probability would rather be $\mathbb{P}(Z_s\geq,\;\forall s<t)=\mathbb{P}(Inf_{[0,t]}Y_{N_t}\geq a)$. Aug 18, 2017 at 13:45
• In general, passing from a discrete time Markov chain to a continuous time Markov process involves to compose by a Poisson process. This is almost the definition : a Markov chain $Y$ is defined by $\mathbb{E}_x(f(Y_k)) = M^k f(x)$ for a good function/vector $f$ and a good operator/stochastic matrix $M$ and a Markov process $X$ is defined in a similar way by $\mathbb{E}_x(f(X_t)) = e^{t L} f(x)$ for a good operator $L$. One can show that the only independent randomisation that allows to go from $M^k$ to $e^{t L}$ is by setting $L = M - I$ and a Poisson process. Aug 18, 2017 at 23:45
• Indeed, $\mathbb{E}_x(f(Y_{N_t })) = \sum_k \mathbb{P}(N_t = k) \mathbb{E}_x(f(Y_k)) = \sum_k \mathbb{P}(N_t = k) M^k f(x)$ and the only possibility is to have $\mathbb{P}(N_t = k)$ proportional to $(\lambda t)^k/k!$ to have $e^{\lambda t M}$ and the normalisation gives the Poisson probability and the "infinitesimal generator" $L = \lambda (M - I)$ then. You can check that a stochastic matrix gives an admissible infinitesimal generator in such a way (i.e. $M 1 = 1$ gives $L 1 = 0$ where $1$ is the vector/function identically equal to 1). Aug 18, 2017 at 23:49
• In your case, $\inf_{s \in [0, t]} Y_{N_s} = \inf_{ k \in \{ 0, \dots, [t] \} } Y_k$ since the function is piecewise continuous, hence my expression. Maybe should I edit my answer with this general fact to complete it (if you want) ? Aug 18, 2017 at 23:57