I'd think the rectangle isn't probably very meaningful. In two dimensional case the relevant area would be ellipse. With independent variables the ellipse could perhaps be derived from the independently calculated intervals. For normally distributed estimators of the mean the general case can be calculated using Hotelling's T-squared distribution.

If a 95% confidence CI interval means that there is a probability of 0.95 that the true mean belongs to CI. Then, the region defined by two 95% confidence intervals which have the same center (the same sample mean) can be considered as a 90.25% confidence region. Confidence region seems to have the same sense as confidence intervals.

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disclaimer : i'm definitely not an expert on such matters !

anyway, and for what it's worth, i remember that there is something tricky here : CI are random variables (random intervals, as they depend on the (random) sample on which they are calculated) and the cartesian product of such random intervals yields a proper confidence region when the CI (considered as random variables) are independent ; whether this implies that X and Y must be stochastically independent has always been unclear to me (although i understand it is a suffcient condition) : what happens for instance if X and Y are not stochastically independent but CI(X ; S) and CI(Y ; S') have been obtained on independent samples S ans S' ?

as i said, this is unclear to me and any advice would be most welcome !

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For a given data set, define the rectangle defined by the product of the two CI's as R. Write the test statistics that were used to estimate the means of X and Y as Avx and Avy, respectively. When Avx and Avy are stochastically independent, R can be interpreted: the joint probability that Avx is in R AND Avy is in R is 0.95*0.95 =0.9025.

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Nico, that's where i start to be confused ... AvX and AvY are constants (the "true values")so the notion of stochastic independence does not seem to be applicable to them ; rather, it has to be applied to the confidence intervals considered as random variables (and having some probability, upon repeated experiments on random samples, to contain the values AvX or AvY) ... and from this point my confusion is close to complete and guidance would be welcome !

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Clerot - before you have seen your data, Avx and Avy are random variables. Also, the interval bounds are random variables, as you point out correctly. Thus one can very well assume independence of Avx and Avy.

But when you have observed your data, you fill inn the relevant values for the observables, and obtain empirical values for Avx and Avy (and the interval bounds). Clearly, these are two numbers, and for these it is nonsense to assume anything about (in-)dependence.

Compare with rolling a die twice. Before you did that, the numbers that will show up in the two trials are independent random variables. After you rolled the die you get 4 and 1, say. Nothing dependent or independent for these two numbers.

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 am confused.

Merci beaucoup pour votre aide. Je suis entrain de répondre à toutes les remarques et les questions que vous m'avez posées.

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(the notation is corrected now)

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I need normality assumptions because, the normality of "the empirical mean of X" which is a random variable depends on the asymmetry of the primary random variable "X". (Reference = page 8: biol09.biol.umontreal.ca/BIO2041e/Sujet_05-Int_confiance.pdf). That is, if the skewness of X is to high, the normality of m(X, S) will be affected. That is why i assumed that X is normal. Otherwise, a correct CI should be calculated using Bootstrapping or by performing the appropriate transformation to X (log, exp, Legendre, etc.) so that the transformed random variable X' would have better skewness and kurtosis.

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The big confusion is here. It is okay that a "CI" of a certain parameter "m" is considered as random entity that covers the true "m" with a probability of (1 - a). An that an instance of CI (i.e. [a, b]) cannot be assumed to contains "m" with a predefined probability simply because the event "m in [a, b]" is static and is not connected to any probability (m is either inside [a, b] or no). Being convinced of all this, i conclude that any instance of a given CI does not have any importance, and does not add to the estimation of the true parameter. In other way, saying that a queue's expected waiting time is about: 20min +/- 5min give no information about the true waiting time which may be greater than 25min or less than 15min. What you said about "ignorance does not imply uncertainty" makes me think that (a) CI are interesting only if they are kept in their analytical "variable" form, (b) that a single CI instance is useless and (c) multiple CIs can provide information about the probability of existence of the true parameter "m" inside all the CIs.

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To my knowledge, a stochastic process adds the time axis (discrete or continuous) to the randomness of a random variable. I think it should at least inherits the complexity of the Probability Theory.

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The most important advantage of the Bonferroni inequality is that it does not require the independence of the events "ei" it is based on. Thus, it can be applied to construct a rectangular area with variable bounds that contains a specific random point with a predefined probability.

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I thought about Bootstrapping Confidence Intervals when i observed that the skewness coefficients of X and Y are too high. It is among the motivations of using such method.

Dear Fabrice and Nabil,

I was out in this days so excuse me for the delayed answer.

The use of the normal distribution assumption is not so necessary. However, it should be true that the estimator of the mean (because of the Central Limit Theorem) can be considered as a normal. However, the asymptotics on Maximum Likelihood estimators (thus based on hypothesis on the random variables at the basis of the process) furnishes clear solutions for the definition of the confidence regions.

For a rapid discussion see these slides

http://www.public.iastate.edu/~maitra/stat501/lectures/InferenceForMeans-Confidence.pdf

@Antonio Irpino:

Lot of thanks for the useful link. Do you know any inequality that offers a better lower bound than the Bonferroni's. That is:

Proba(all ci true) >= newLowerBound >= 1 - (a1 + ... + an)

@Antonio: Join this discussion please.

https://www.researchgate.net/post/Rectangular_confidence_region_of_the_means_of_two_dependent_stochastic_processes

Call for Papers:

"12th International Probabilistic Workshop" held in Weimar Germany, on November 4th and 5th 2014"

https://www.researchgate.net/post/Call_for_Papers_12th_International_Probabilistic_Workshop_held_in_Weimar_Germany_on_November_4th_and_5th_2014