For a side project I am working on I need to find the probability that a given Normal rv from $N(\mu_a,\sigma_a^2)$ is greater than or equal to at least one of the values from a set of other, unique, Normal rv's $N(\mu_1,\sigma_1^2)$, $N(\mu_2,\sigma_2^2)$, $N(\mu_3,\sigma_3^2)$, ... $N(\mu_m,\sigma_m^2)$

To add some context, I'm trying to check how likely the first Normal rv is to improve (higher is better) the "best m" set of rv's if it were to be added to the set.

There are about 400 unique Normal rv's in the set I want to test, and each one of these will potentially have a different set of Normal rv's to compare against.

For performance reasons, I would rather not run a simulation for each one of these comparisons - and am looking for something that either (a) solves this directly, or (b) is a "good enough" approximation of the result.

If anyone can help me out it would be greatly appreciated!


1 Answer 1


It can be calculated in a pretty straightforward manner if independence is assumed. You want the following probability: $$P(X\geq \min_i(X_i))=1-\prod_{i=1}^mP(X < X_i)$$

Because $X-X_i$ is $\mathcal{N}(\mu_a-\mu_i, \sigma_a^2+\sigma_i^2)$, we can write the inner term as:$$P(X-X_i<0)=\Phi\left(\frac{\mu_i-\mu_a}{\sqrt{\sigma_a^2+\sigma_i^2}}\right)$$ where $\Phi$ denotes the CDF of standard normal.

  • $\begingroup$ This is fantastic, thanks for the quick reply. A question on independence.. we need guarantee each individual RV to be independent, and only that, right? There would definitely be dependence between each of the comparisons, I believe. I.e. if my first RV is greater than one of the RVs in the comparison set, it is much more likely that it is greater than the other ones as well? That type of "dependence" doesn't matter here? $\endgroup$ Feb 5, 2020 at 17:15
  • $\begingroup$ No, that doesn't count. That is conditional. $\endgroup$
    – gunes
    Feb 5, 2020 at 17:17
  • $\begingroup$ Awesome, thanks much $\endgroup$ Feb 5, 2020 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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