I am tasked with developing confidence intervals for the function x = 1 - |(a+b)/c| where a, b and c are random variables. a and b are normally distributed, but c is heavily skewed left. further there b and c are correlated (Pearson = .21). x belongs to [-inf, 1] and is skewed left because of its domain. I have 2000 "rows" of data.

I can't see an analytic solution for this equation/problem. I was thinking this would be a great opportunity to bootstrap an estimate. I've approached it two ways; 1) resample a, b, and c independently. Then, calculate x^ each time, then take summary statistics for every iteration. Iterate 10,000 or so times. 2) resample the rows of data for a, b, and c. Calculate x^ and take summary statistics. Again, iterate 10,000 or so times.

I think the correct approach is 2 because then the correlation between b and c will be present in the data. Please tell me if I'm right.

Now once if have a big collection of means of x^ can I use the standard deviation of the means of x^ to get the 95% or whatever % confidence bounds?

  • 3
    $\begingroup$ I'm voting to close this question as off-topic because it's a question for the stats site. this is the kind of thing that might come up in a quant interview; but it's purely about statistics and not the quant space. $\endgroup$ – Theodore Aug 22 '19 at 18:11
  • $\begingroup$ Is there a way I can move it? I didn't find a stats site when i signed up. $\endgroup$ – Ksnapp Aug 22 '19 at 18:29
  • $\begingroup$ yep, the site is Cross Validated. one option if you don't want to wait for enough votes to go through for this question to be migrated, is you can just delete the question from here and post it there. $\endgroup$ – Theodore Aug 22 '19 at 18:37

Confidence intervals are applied to estimates to give a sense for potential error. If a, b, and c are R.V. as you described, they're independent by nature, making confidence intervals irrelevant. Either you're misunderstanding what was asked of you or your problem is misspecified (or you just misspecified it here).

  • $\begingroup$ Thank you, I specified the problem here. Also I'm going to move the question to a difference site because it does seem off topic. $\endgroup$ – Ksnapp Aug 22 '19 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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