I do not know if writing the integrals is simple or complicated, but I feel like it should be simple enough.
Suppose we have
- Two points, $y_1=0$ and $y_2=1$,
- a random variable $X\sim U[0,1]$ and
a random variable $Z$, with mass $\dfrac23$ on $z_1$ and $\dfrac13$ on $z_2$.
We also have a function $$\sigma: \left\{ \frac{1}{3}, \frac{1}{3} \right\}\times [0, 1] \to \{1,2\}$$ and a function $$d: [0, 1] \times \{y_1,y_2\} \to [0,1].$$
How can I write (with summations and integrals) $$ Pr(d(X,y_{\sigma(Z,X)})\leq s). \quad s\geq 0 $$
I think it is something like $$ \sum_{\{z_1,z_2\}}\left (\int_0^1 \text{something}\right). $$
So I am confused because I think I need a pdf for $d(x,y_\sigma)$ somewhere, but we do not know what this looks like (and it will depend on the specific realizations of $Z,X$, so would I need a triple integral??
If it helps, it can be assumed that $$\sigma(z,x) = \begin{cases}1; & x\in \left[0, \dfrac13\right)\cup \left(\dfrac23,1\right]\\ 2; & \text{otherwise} \end{cases}$$ So $\sigma$ does not depend on $z$, but I prefer an answer for the general case where it does. And $d(x,y_{\sigma(z,x)})$ can be taken to be the straight-line distance between $x$ and $y_{\sigma(z,x)}$, i.e. $$d(x,y_{\sigma(z,x)})=\begin{cases}x; & \sigma(z,x)=1\\1-x; & \sigma(z,x)=2\end{cases}$$
$\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}\def\Ω{{\mit Ω}}$Denote by $μ_X$ and $μ_Z$ the probability measures associated with $X$ and $Z$, respectively, and denote the density function of $X$ by $f_X$ and that of $Z$ (point-density) by $p_Z$. If $X$ and $Z$ are independent, then $μ_{X, Z} = μ_X × μ_Z$. Now,\begin{align*} P(d(X, y_{σ(Z, X)}) \leqslant s) &= \int\limits_\Ω I_{\{d(X, y_{σ(Z, X)}) \leqslant s\}} \,\d P = \int\limits_\Ω I_{\{(x, z) \mid d(x, y_{σ(z, x)}) \leqslant s\}}(X, Z) \,\d P\\ &= \int\limits_{[0, 1] × \{z_1, z_2\}} I_{\{(x, z) \mid d(x, y_{σ(z, x)}) \leqslant s\}}(x, z) \,\d μ_{X, Z} \tag{1}\\ &= \int\limits_{[0, 1] × \{z_1, z_2\}} I_{\{(x, z) \mid d(x, y_{σ(z, x)}) \leqslant s\}}(x, z) \,\d μ_X \d μ_Z\\ &= \sum_{z \in \{z_1, z_2\}} p_Z(z) \int_{[0, 1]} I_{\{(x, z) \mid d(x, y_{σ(z, x)}) \leqslant s\}}(x, z) \,\d μ_X\\ &= \sum_{z \in \{z_1, z_2\}} p_Z(z) \int_0^1 I_{\{(x, z) \mid d(x, y_{σ(z, x)}) \leqslant s\}}(x, z) f_X(x) \,\d x\\ &= \sum_{z \in \{z_1, z_2\}} \int_0^1 I_{\{(x, z) \mid d(x, y_{σ(z, x)}) \leqslant s\}}(x, z) p_Z(z) f_X(x) \,\d x, \tag{2} \end{align*} where (2) can be slightly simplified by using $f_X(x) = 1$ for $x \in [0, 1]$. If $X$ and $Z$ are not guaranteed to be independent, then (1) would be the final expression if no more information is given.