1
$\begingroup$

In Gambler's ruin problem, suppose you are playing a game. The total money in the game is let's say $S$.

$X_t :=$ Amount of money you have at time $t$ and $X_t \in \{0,1,2, ..., S\}$.

At any time if you win, $X_t = X_{t-1} + 1$ and if you loose, $X_t = X_{t-1} - 1$. The game ends if you either hit $0$ or $S$.

We consider an event $R = \bigcup_{n \geq0} \{X_n = 0\}$ and define Ruin probability as $$Pr = P(R|X_0 = k) $$ for $k \in [0,S]$. I have question regarding this event $R$. Instead of taking a union, shouldn't we take infimum? Because once you reach $0$, you stop. There would not be further any $m(>n)$ s.t. $X_m = 0$.

$\endgroup$
1
$\begingroup$

I believe the convention is that once you hit state $0$ or $S$, you stay in that state forever; i.e. if $X_t = S$ then $X_{t+k} = S$ for all $k \in \{1,2,3,\dots\}$.

$\endgroup$
  • $\begingroup$ Ok. but if I replace this union with $infimum$ then will their be any issue? $\endgroup$ – Dark_Knight Feb 1 '18 at 7:39
  • $\begingroup$ I would define the ruin probability as $P( \text{hit } 0 \text{ or hit } S \mid X_0 = k)$ $\endgroup$ – user365239 Feb 1 '18 at 8:07
  • $\begingroup$ In fact, in further texts, this concept of infimum is brought up to define the mean duration of the game. $\endgroup$ – Dark_Knight Feb 1 '18 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.