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Students of statistics and probability theory sometimes develop misconceptions about the normal distribution, ideas that may seem plausible but are mathematically untrue. For example, it is sometimes mistakenly thought that two linearly uncorrelated, normally distributed random variables must be statistically independent.
Further, two jointly normally distributed random variables are independent if they are uncorrelated, [4] although this does not hold for variables whose marginal distributions are normal and uncorrelated but whose joint distribution is not joint normal (see Normally distributed and uncorrelated does not imply independent).
DEFINITION:Random variable A is said to be normal, denoted A ∈ ƒ, when its sample observations follow a univariate or multivariate Gaussian distribution of some fixed mean and (co)variance; when making statements about more than one multivariate random variable, e.g. three multivariate random variables A ∈ ƒ, B ∈ ƒ, C ∈ ƒ, then ...
Pairwise independence does not imply mutual independence, as shown by the following example attributed to S. Bernstein. [3]Suppose X and Y are two independent tosses of a fair coin, where we designate 1 for heads and 0 for tails.
In probability theory and statistics, the multivariate normal distribution, multivariate Gaussian distribution, or joint normal distribution is a generalization of the one-dimensional normal distribution to higher dimensions.
Berkson's paradox occurs when this observation appears true when in reality the two properties are unrelated—or even positively correlated—because members of the population where both are absent are not equally observed. For example, a person may observe from their experience that fast food restaurants in their area which serve good ...
If two random variables are subindependent, and if their covariance exists, then they are uncorrelated. [ 1 ] Subindependence has some peculiar properties: for example, there exist random variables X and Y that are subindependent, but X and αY are not subindependent when α ≠ 1 [ 1 ] and therefore X and Y are not independent.
Note that not all finite exchangeable sequences are mixtures of i.i.d. To see this, consider sampling without replacement from a finite set until no elements are left. The resulting sequence is exchangeable, but not a mixture of i.i.d. Indeed, conditioned on all other elements in the sequence, the remaining element is known.