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Assume the following two facts:

  1. $x$ is distributed uniformly on $[-1,2]$.

  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?

user152497
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  • When $s_2=s_1$ + a tiny little bit then F=1. As $s_2$ increase F decreases. Not very CDF like. $s_2$ will also be bounded so this should form part of the answer. – user121049 Jun 08 '18 at 19:30
  • I think it should be $1-$ rather than $1/2+$ for $s_2 > s_1$. Otherwise it looks good, although it needs a proper proof. – D. Thomine Jun 08 '18 at 19:38
  • @D.Thomine, are you sure? could you derive it? – user152497 Jun 09 '18 at 04:07

1 Answers1

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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: $$ F(s_2|s_1) = \int_{s_1-0.2}^{s_2}(0.2-|t-s_1|)dt : \int_{s_1-0.2}^{s_1+0.2}(0.2-|t-s_1|)dt$$

Djura Marinkov
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