I'm trying to teach my self the first 10 chapters of the book out of interest: "Leonid Koralov, Yakov G. Sinai - Theory of probability and random processes".
Problem:
Find the distribution function of a random variable $\xi$ which takes positive values and satisfies $\mathrm{P}(\xi>x+y \mid\xi>x)=\mathrm{P}(\xi>y), \forall x, y > 0$.
I figure I have to use some sort of conditional probability rule, like $P(A\mid B)=P(AB)/P(B)$, to get $P(\xi>x+y \mid \xi>x)=\frac{P(\xi>x+y \cap \xi>y)}{p(\xi>x)}=P(\xi>y)$, but I do not believe this to be right.
We know that $x+y>x, x+y>y$, since $x,y>0$. Any hints would be appreciated.
You're on the right track with the conditional probability rule. What you want to use is: $$ \mathbb P[A | B] = \frac{\mathbb P[A \cap B]}{\mathbb P[B]}. $$ In this case, since $\xi > 0$, we know $\{\xi > x\} \subset \{\xi > x+y\}$ for $x,y > 0$, so this means: $$ \mathbb P[\xi > y] = \mathbb P[\xi > x + y|\xi > x] = \frac{\mathbb P[\{\xi > x + y\} \cap \{\xi > x\}]}{\mathbb P[ \xi > x]} = \frac{\mathbb P[\xi > x+y]}{\mathbb P[\xi > x]} $$ (by the way, why do we know $\mathbb P[\xi > x] \neq 0$?) So this tells us: $$ \mathbb P[\xi > x]\mathbb P[\xi > y] = \mathbb P[\xi > x+y]. $$ Now, let's say that $r = \mathbb P[\xi > 1]$. Can you figure out $\mathbb P[\xi > x]$ for $x \in \mathbb N$? What about $x \in \mathbb Q$? What about $x \in \mathbb R$?