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Suppose I have a computer that solves a certain problem in an average time T. The average time T is calculated from the probability distribution p(t), where p(t) is the probability that the program halts at time t meaning the problem has been solved.

Now suppose we run n identical computers simultaneously and independent of each other solving the same problem as I just described. I am interested in the new average time that it takes before atleast one of the computers have solved the problem.

I'm not sure how I should tackle this problem.

Higgsino
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1 Answers1

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Let $T_1,\ldots,T_n$ be the random times it takes each computer to solve the problem. What you're interested in here is the random variable $$ T_{\min}:=\min\{T_1,\ldots,T_n\}. $$

Given that, here's my hint: $$ P(T_{\min}>t)=P(T_1>t,T_2>t,\ldots,T_n>t). $$

Nick Peterson
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  • Don't understand. – Higgsino Mar 22 '17 at 15:22
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    That's an incredibly unhelpful comment. Do you not understand what I said? Do you not understand how it leads toward a solution? Please be more specific. – Nick Peterson Mar 22 '17 at 15:23
  • How do I calculate the new average time if I know P(T_min > t) ? I assume P(T_min > t) refers to the probability that it takes more then t time before atleast one computer gives the answer. BUt where do I go from there? – Higgsino Mar 22 '17 at 15:26
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    @Higgsino $$E(T_{\mathrm{min}})=\int_0^\infty P(T_{\mathrm{min}}>t),dt$$ – Did Mar 22 '17 at 15:33
  • But now T_min refers to an interval. When I caclulate an average I usually take p(t) at a specifik time and then multiply with the time and integrate to get the average. I just don't understand how your integral will give the average of T_min. – Higgsino Mar 22 '17 at 15:41
  • http://math.stackexchange.com/questions/64186/intuition-behind-using-complementary-cdf-to-compute-expectation-for-nonnegative – Nick Peterson Mar 22 '17 at 15:50