• • • # Area/Region defined by confidence intervals

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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.

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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.

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thank you very much. where can i ask more similar questions.

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I would try CrossValidated (http://stats.stackexchange.com/).

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