-1
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Easy example to start: You throw a $n$-sided dice until your lucky number shows. This is a Bernoulli process with $p=1/n$, your expected number $E$ of throws is $n$.
Now imagine you play this game $m$ times in parallel. Stop if any dice shows your lucky number, i.e. first win wins the whole game. (Feel free to discuss also the last win case :-) It's still a Bernoulli process with $p'=1-(1-1/p)^m, E'=1/p'$.
Now generalize. The game $G$ is defined by fixing some winning probability $p(i)$ for each move $i$, we only assume $E$ shall be finite. Play $m$ copies of $G$ parallel, again first win in a copy ends the game. What can one say about $E'$?
1. $E'\le E$. (trivial)
2. $E'\ge E/m$ ? (tempting :-)
3. Let $p(i)$ stem from some well-known probability distribution, say, instead of the geometric from above a Poisson or whatnot. Surely $E'$ already has been computed for many of these distributions?

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  • $\begingroup$ What is $i$? Time index or index of a parallel game? $\endgroup$ – kodlu Apr 28 at 14:35
  • $\begingroup$ In the general case, your question amounts to the following: you have a positive-integer-valued random variable $X$ (with $\mathbb{E}X=E$) and you take a random variable $Y$ which is the minimum of $m$ independent copies of $X$ (with $\mathbb{E}Y=E'$). $\endgroup$ – James Martin Apr 28 at 22:04
  • $\begingroup$ Property 2. doesn't hold in general. e.g. let $m=2$, and let $X$ take value $1$ with probability $2/3$ and value $100$ with probability $1/3$. Then $Y$ takes value $1$ with probability $8/9$ and value $100$ with probability $1/9$. You get $E=34$ and $E'=12$. $\endgroup$ – James Martin Apr 28 at 22:10
  • $\begingroup$ @kodlu: time index. I (and James) assume the games (including their $p_i$) are identical $\endgroup$ – Hauke Reddmann Apr 29 at 11:34
  • $\begingroup$ @JamesMartin: E'=12>11.333...=34/3=E/m exactly as I supposed $\endgroup$ – Hauke Reddmann Apr 29 at 11:38

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