Assume the following two facts:
$x$ is distributed uniformly on $[-1,2]$.
Two variables, $s_1$ and $s_2$ are distributed uniformly on $[x-0.1,x+0.1]$, and $s_1, s_2$ are mutually independent conditional on $x$.
Suppose that some $s_1\in[0,1]$ is observed. My question is: what is the cumulative probability of $s_2$, conditional on $s_1$?
My professor has claimed in his lecture notes (without justification) that it is the following:
$$F(s_2|s_1)=\left \{ \begin{array}{11}\quad\quad\frac{(s_2-s_1+0.2)^2}{0.08}\quad if \quad s_2\leq s_1\\ \frac 1 2 +\frac{(s_2-s_1-0.2)^2}{0.08}\quad if \quad s_2>s_1 \end{array}\right.$$
But I don't know where this comes from, and it seems wrong. Is it wrong? If so, what is the correct distribution?
So for a given $s_1$ each $x\in (s_1-0.1, s_1+0.1)$ is equiprobable, also for each x every $s_2\in (x-0.1, x+0.1)$ is equiprobable (consider $x$ and $s_2$ as intervals not as discrete values)
So for some $s_2 \in (s_1-0.2, s_1+0.2)$ probability density will be ratio between lengths of intervals $|(s_1-0.1, s_1+0.1)\cap (s_2-0.1, s_2+0.1)|=0.2-|s_2-s_1|$ and $|(s_1-0.1, s_1+0.1)|=0.2$ (correspodent x and all possible x)
Try solving this: $$ F(s_2|s_1) = \int_{s_1-0.2}^{s_2}(0.2-|t-s_1|)dt : \int_{s_1-0.2}^{s_1+0.2}(0.2-|t-s_1|)dt$$