0

I'm not sure how to accurately ask that question so I let me go with example.

Let's say I've got 3 not-fair number generators: A, B, C. They draw numbers between 0 and 1000. I'd like to guess for each of the generators, the probability of having the highest expected value. In other words I'd like to know chances for each of them of being the best (generating the highest values on average).

I made 5 simulations for simplicity and those were results:

A: 845, 13, 345, 765, 900

B: 87, 105, 345, 10, 5

C: 950, 25, 4, 870, 930

I've got number of samples, I can calculate EV and std dev over those samples. How to calculate probability for each generator of being the one producing the highest numbers? Thanks for help!

Edit: Let's assume generators has normal distribution.

Jakub Bielan
  • 109
  • It sounds like you need to try to find an approximate distribution each group follows (normal, uniform, etc.) Usually random generation follows a uniform distribution. From there you could find the distribution of the sample maximum within each group and go from there. – Remy Feb 22 '19 at 07:58
  • Thanks Remy. Generators are only an example. For my problem, there's normal distribution. – Jakub Bielan Feb 22 '19 at 08:05
  • In other words, you would like to find $P(A \geq max(B,C))$ for $A,B,C$ values of a normal distribution. Related question for 2 variables only - https://math.stackexchange.com/questions/40224/probability-of-a-point-taken-from-a-certain-normal-distribution-will-be-greater – Gareth Ma Feb 22 '19 at 08:13
  • It sounds to me like OP has three normal distributions such that $X\sim\mathsf N(\mu_X,\sigma_X^2)$, $Y\sim\mathsf N(\mu_Y,\sigma_Y^2)$, and $Z\sim\mathsf N(\mu_Z,\sigma_Z^2)$ with sample sizes $n$ for each, and wants to find $\mathsf P\left(\text{max}{{X_1,...,X_n}}>\text{max}{{Y_1,...,Y_n,Z_1,..,Z_n}}\right)$ – Remy Feb 22 '19 at 08:18
  • @Remy Jakub asks for highest expected value though. So to me it sounds like we have n samples from each of 3 different normal distributions with unknown means and variances and the problem is to estimate the means. – nikkou Feb 22 '19 at 08:37
  • (not the means themselves but the probability of one mean being larger or equal than all others) – nikkou Feb 22 '19 at 08:47
  • Exactly what @nikkou said. – Jakub Bielan Feb 22 '19 at 09:32

0 Answers0