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The simplest case of a normal distribution is known as the standard normal distribution or unit normal distribution. This is a special case when μ = 0 {\textstyle \mu =0} and σ 2 = 1 {\textstyle \sigma ^{2}=1} , and it is described by this probability density function (or density): φ ( z ) = e − z 2 2 2 π . {\displaystyle \varphi (z ...
The normal-exponential-gamma distribution; The normal-inverse Gaussian distribution; The Pearson Type IV distribution (see Pearson distributions) The Quantile-parameterized distributions, which are highly shape-flexible and can be parameterized with data using linear least squares. The skew normal distribution
The fact that two random variables and both have a normal distribution does not imply that the pair (,) has a joint normal distribution. A simple example is one in which X has a normal distribution with expected value 0 and variance 1, and = if | | > and = if | | <, where >. There are similar counterexamples for more than two random variables.
Diagram showing the cumulative distribution function for the normal distribution with mean (μ) 0 and variance (σ 2) 1. These numerical values "68%, 95%, 99.7%" come from the cumulative distribution function of the normal distribution. The prediction interval for any standard score z corresponds numerically to (1 − (1 − Φ μ,σ 2 (z)) · 2).
Let X 1, X 2, ..., X n be independent, identically distributed normal random variables with mean μ and variance σ 2.. Then with respect to the parameter μ, one can show that ^ =, the sample mean, is a complete and sufficient statistic – it is all the information one can derive to estimate μ, and no more – and
It is possible to have variables X and Y which are individually normally distributed, but have a more complicated joint distribution. In that instance, X + Y may of course have a complicated, non-normal distribution. In some cases, this situation can be treated using copulas.
[7] [4] [8] The normal distribution is a commonly encountered absolutely continuous probability distribution. More complex experiments, such as those involving stochastic processes defined in continuous time, may demand the use of more general probability measures.
Proof: We will prove this statement using the portmanteau lemma, part A. First we want to show that (X n, c) converges in distribution to (X, c). By the portmanteau lemma this will be true if we can show that E[f(X n, c)] → E[f(X, c)] for any bounded continuous function f(x, y). So let f be such arbitrary bounded continuous function.