What is the stress energy tensor of this Boolean Field?

Question

Question
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Let's say I have a boolean field $E_s$ at each point $S$ of a Lorentzian manifold:

 $$d \tilde  s^2  = - c^2 dt^2 + dx^2$$ 

The mean probability is given by:

$$ P(E_s) =  \lim_{n \to \infty} \sum_{s}\frac{E_{s_1} + E_{s_2} + E_{s_3} + \dots _+ E_{s_n}}{n}  $$
The equations of motion are given by:

$$ d \tilde s = E_{ s} (c d t \hat e_t+ d x \hat  e_x) +(1-E_{ s})(c d t \hat e_t - d x \hat  e_x)$$  

Does this reduce to $1$-dimensional brownian motion in the limit $c \to \infty$? Also is there some nice way to calculate the stress energy tensor of the boolean field?

Background
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Let's say I'm $1+1$ Minkowski spacetime and want to model a random walk. Let $\Delta s$ be the $2$-displacement and $\Delta x/ \Delta t =v $ be a constant. We expect after $\Delta s' =  \kappa \Delta s$ and event will happen which will either flip $\Delta x$ or remain the same. Let us assume there are only $2$ possibilities of $\Delta s$: 

$$\Delta s_0 = (c\Delta t \hat e_t+ \Delta x \hat  e_x)$$ 

$$\Delta s_1 = (c\Delta t \hat e_t - \Delta x \hat  e_x)$$

This can be expressed as:

$$ \frac{\Delta s '}{\kappa} = E_{ i} (c\Delta t \hat e_t+ \Delta x \hat  e_x) +(1-E_{ i})(c\Delta t \hat e_t - \Delta x \hat  e_x)$$  

where $E_{ \kappa s}$ is a boolean variable and denotes at the possibility of $\Delta s_0$ or $\Delta s_1$ happening. As an example to show what I mean:

$$ \sum  \frac{\Delta s '}{\kappa} = (c\Delta t \hat e_t+ \Delta x \hat  e_x) + (c\Delta t \hat e_t+ \Delta x \hat  e_x)$$

where $E_{1} = 1$ and $E_{2} = 1$

Going to the contiuum limit $  \Delta s ' / \kappa \to 0$ and $E_i \to E_s$ where $E_s$ is a boolean field as point $s$:

$$ d \tilde s = E_{ s} (c d t \hat e_t+ d x \hat  e_x) +(1-E_{ s})(c d t \hat e_t - d x \hat  e_x)  $$  

To integrate $(2E_s -1) \frac{dx}{ds} ds $ we use this:

$$ \int_{\tilde s} d \tilde s =  \int_{\tilde s} c dt \hat e_t + (2P(E_s) -1)\int_{\tilde s}  dx \hat e_{x}$$

More specifically:
$$ \int_{\tilde s} 2E_s  \frac{dx}{ds} ds = 2 \sum_{s} \lim_{n \to \infty} \frac{E_{s_1} + E_{s_2} + E_{s_3} + \dots _+ E_{s_n}}{n} \int_{\tilde s} dx  $$

Making $P(E_s)$ the probability of a collision and $\tilde s$ the path.

Let, us consider the $2$-distance:

$$ d\tilde s^2 = c^2 dt^2 - E_s^2  dx^2 - (1-E_s)^2 d x^2 +2 E_s(1-E_s) dx^2 = c^2 d \tau^2$$ 

Since, $E_s(1-E_s) =0 $, $E_s^2 = E_s$ and $(1-E_s)^2= (1-E_s)$ we have:

$$ d\tilde s^2 = c^2 dt^2 - E_s  dx^2 - (1-E_s) d x^2  = c^2 dt^2 -  d x^2  $$