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I take a sample of 10 from a population that I suspect is non-normal (but it is continuous). I need to calculate the probability that the next sample will be larger than the maximum of the previous 10 points (given the previous 10 observations).

I'm an engineer with some extra statistical knowledge. I was taught from a frequentist perspective. Am I correct in thinking I can solve this problem with a Bayesian credible interval?

Added: I did consider a nonparametric prediction interval, but since that is a frequentist's perspective it doesn't give a probability of the next sample, but that the probability applies to the procedure used to calculate the interval (see this explanation of prediction intervals paragraph 21).

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  • $\begingroup$ Try researching 'extreme value theory' and 'extreme value distributions'. I am not an expert on these topics but it sounds like they may apply to your problem. $\endgroup$
    – Zachary Blumenfeld
    Commented Feb 22, 2016 at 1:57
  • $\begingroup$ "the next sample will be higher" - what does this mean exactly? $\endgroup$
    – Aksakal
    Commented Feb 22, 2016 at 14:37
  • $\begingroup$ @Aksakal if the sample is 3,5,6,4,7,8,2,5,6,4 then what is the probability that the next test is > 8 $\endgroup$
    – Zaralynda
    Commented Feb 22, 2016 at 14:39
  • $\begingroup$ @ZacharyBlumenfeld am looking at it now, thanks for the suggestion $\endgroup$
    – Zaralynda
    Commented Feb 22, 2016 at 14:40
  • $\begingroup$ Are you asking for this probability given the first 10 observations or before seeing the first 10? $\endgroup$
    – Juho Kokkala
    Commented Feb 22, 2016 at 14:47

1 Answer 1

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Given a sample $X_1, \dotsc, X_n$ from the same population (distribution) (not even independence is necessary, it is enough with symmetry (or exchangeability), that all permutations have the same distribution). Then the probability that the next sampled value $X_{n+1}$ from that same distribution will be the maximum is $\frac1{n+1}$. You can argument directly from symmetry: $X_{n+1} > \max_{i=1,\dotsc, n} X_i$ has a probability that is unchanged by permuting the indices, so $X_3 > \max_{i\not = 3} X_i$ and so on, so the commom probability must be $\frac1{n+1}$.

This leads to a theory of records: Let $X_1, X_2, X_3, \dotsc $ be a sequence of independent (or exchangeable ...) random variables, in this case, we must have exchangeability for the finite subsequences $X_1, \dotsc, X_n$ for all $n$. Let $$ T_n=\begin{cases} 1 ~\text{if $X_n$ is the record observation (max) up to time $n$,} \\ 0 ~\text{otherwise} \end{cases} $$ Then $P(T_n=1)=\frac{1}{n}$, and so the expected number of records up to time $n$ is $\sum_{i=1}^n \frac1{i}$ which is diverging in $n$, so there will never be a last record, for instance. That sum is (for moderate $n$) close to $\ln n$, so the expected number of records in the first thousand trials is approximately 6.9 (exact 7.49).

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    $\begingroup$ Note that (as clarified in the latest edit) the question is asking about the conditional probability$P(X_{n+1} > \max_{i=1,\ldots,n}X_i \mid X_1,\ldots,X_n)$ rather than the prior $P(X_{n+1} > \max_{i=1,\ldots,n X_i})$, so the symmetry argument does not directly apply $\endgroup$
    – Juho Kokkala
    Commented Feb 22, 2016 at 14:59
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    $\begingroup$ If then, you must tell us what is your probability model as the answer will depend (maybe even strongly) on it. Do you have a fixed, known model, in which case the answer might depend strongly on the conditioning variables, or do you estimate parameters simultaneously, in which case I guess the dependence on the conditioning variables will be weaker. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Feb 22, 2016 at 15:10

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