mathematix Posted August 23, 2013 Share Posted August 23, 2013 Hi, I feel confused when discussing this issue related to confidence intervals. Assume we have two random variables: X and Y. We construct a 95%-confidence interval of the mean of X and Y. The two intervals are denoted: CI(X) and CI(Y).What is the statistical meaning of the area/region (rectangle) defined by the cartesian product of CI(X) and CI(Y)? Suppose that X and Y are the random output of two stochastic processes ("Cost" and "Service Level" for example). The random parameter of these two processes are the customers' demand denoted D, which follows a known probability distribution. In order to simulate X and Y, someone can generate many demands' samples {d(1), d(2),..., d(n)) and for each d(i), just compute the resulting cost and service level responses. Does this simulation process lead to independent X-Y samples? I developed a simulation script to test empirically if the definition of "confidence region" defined by two CIs is meaningful. The result is positive! I took two normal random variables X and Y with a known means mX and mY. I sampled X and Y more than 1000 times. Each time i compute the two Confidence intervals CI(X) and CI(Y) of the sample means. Then, using Excel, i calculated the number of times the the true means mX and mY are inside the confidence region CI(X) - CI(Y). I repeated this many times. I always obtain a correct probability of being inside the region, respecting the following formula: ConfidenceRegionLevel(X, Y) = ConfidenceLevel(CI(X)) x ConfidenceLevel(CI(Y)) = (1 - alpha)^2 That is, i get approximately 89.5% region confidence level, when i use two a=95% CIs. Thank you in advance. Link to comment Share on other sites More sharing options...

cyberwulf Posted August 23, 2013 Share Posted August 23, 2013 While this isn't really the place for this type of question, the quick answer is that if X and Y are independent, then the probability that the CI "box" for muX and muY contains both is, indeed, P(CI(X) contains muX) * P(CI(Y) contains muY) = 0.95 * 0.95 = 0.9025. Since you've generated X and Y independently in your simulation, you are getting the expected result. If X and Y are dependent, then the coverage probability of the "box" may not be 0.9025. Link to comment Share on other sites More sharing options...

mathematix Posted August 24, 2013 Author Share Posted August 24, 2013 thank you very much. where can i ask more similar questions. Link to comment Share on other sites More sharing options...

cyberwulf Posted August 24, 2013 Share Posted August 24, 2013 I would try CrossValidated (http://stats.stackexchange.com/). Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now