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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 ...
Normal numbers exist. Moreover, computable normal numbers exist. These non-probabilistic existence theorems follow from probabilistic results: (a) a number chosen at random (uniformly on (0,1)) is normal almost surely (which follows easily from the strong law of large numbers); (b) some probabilistic inequalities behind the strong law. The ...
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
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.
This is a list of some of the more commonly known problems that are NP-complete when expressed as decision problems. As there are thousands of such problems known, this list is in no way comprehensive. Many problems of this type can be found in Garey & Johnson (1979).
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).
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.
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. However, this is ...