normal-distribution

Multivariate Normal CDF in Python using scipy

由 谁都会走 提交于 2019-11-26 21:47:41
问题 In order to calculate the CDF of a multivariate normal, I followed this example (for the univariate case) but cannot interpret the output produced by scipy: from scipy.stats import norm import numpy as np mean = np.array([1,5]) covariance = np.matrix([[1, 0.3 ],[0.3, 1]]) distribution = norm(loc=mean,scale = covariance) print distribution.cdf(np.array([2,4])) The output produced is: [[ 8.41344746e-01 4.29060333e-04] [ 9.99570940e-01 1.58655254e-01]] If the joint CDF is defined as: P (X1 ≤ x1,

C++: generate gaussian distribution

由 谁说我不能喝 提交于 2019-11-26 21:37:24
问题 I would like to know if in C++ standard libraries there is any gaussian distribution number generator, or if you have any code snippet to pass. Thanks in advance. 回答1: The standard library does not. Boost.Random does, however. I'd use that if I were you. 回答2: C++ Technical Report 1 adds support for random number generation. So if you're using a relatively recent compiler (visual c++ 2008 GCC 4.3), chances are that it is available out of the box. See here for sample usage of std::tr1::normal

Creating a Gaussian Random Generator with a mean and standard deviation

由 北城余情 提交于 2019-11-26 21:12:59
问题 I am trying to create a one dimensional array and use a random number generator(Gaussian generator that generates a random number with means of 70 and a standard deviation of 10) to populate the array with at least 100 numbers between 0 and 100 inclusive. How would i go about doing this in C++ ? 回答1: In C++11 this is relatively straight forward using the random header and std::normal_distribution ( live example ): #include <iostream> #include <iomanip> #include <string> #include <map>

How to calculate the inverse of the normal cumulative distribution function in python?

由 好久不见. 提交于 2019-11-26 19:43:21
How do I calculate the inverse of the cumulative distribution function (CDF) of the normal distribution in Python? Which library should I use? Possibly scipy? NORMSINV (mentioned in a comment) is the inverse of the CDF of the standard normal distribution. Using scipy , you can compute this with the ppf method of the scipy.stats.norm object. The acronym ppf stands for percent point function , which is another name for the quantile function . In [20]: from scipy.stats import norm In [21]: norm.ppf(0.95) Out[21]: 1.6448536269514722 Check that it is the inverse of the CDF: In [34]: norm.cdf(norm

Seeing if data is normally distributed in R

由 雨燕双飞 提交于 2019-11-26 19:14:31
Can someone please help me fill in the following function in R: #data is a single vector of decimal values normally.distributed <- function(data) { if(data is normal) return(TRUE) else return(NO) } Ian Fellows Normality tests don't do what most think they do. Shapiro's test, Anderson Darling, and others are null hypothesis tests AGAINST the the assumption of normality. These should not be used to determine whether to use normal theory statistical procedures. In fact they are of virtually no value to the data analyst. Under what conditions are we interested in rejecting the null hypothesis that

Random Gaussian Variables

由 柔情痞子 提交于 2019-11-26 12:51:03
Is there a class in the standard library of .NET that gives me the functionality to create random variables that follow Gaussian distribution? Jarrett's suggestion of using a Box-Muller transform is good for a quick-and-dirty solution. A simple implementation: Random rand = new Random(); //reuse this if you are generating many double u1 = 1.0-rand.NextDouble(); //uniform(0,1] random doubles double u2 = 1.0-rand.NextDouble(); double randStdNormal = Math.Sqrt(-2.0 * Math.Log(u1)) * Math.Sin(2.0 * Math.PI * u2); //random normal(0,1) double randNormal = mean + stdDev * randStdNormal; //random

How to get a normal distribution within a range in numpy?

由 戏子无情 提交于 2019-11-26 10:45:17
问题 In machine learning task. We should get a group of random w.r.t normal distribution with bound. We can get a normal distribution number with np.random.normal() but it does\'t offer any bound parameter. I want to know how to do that? 回答1: The parametrization of truncnorm is complicated , so here is a function that translates the parametrization to something more intuitive: from scipy.stats import truncnorm def get_truncated_normal(mean=0, sd=1, low=0, upp=10): return truncnorm( (low - mean) /

Seeing if data is normally distributed in R

由 不想你离开。 提交于 2019-11-26 06:53:46
问题 Can someone please help me fill in the following function in R: #data is a single vector of decimal values normally.distributed <- function(data) { if(data is normal) return(TRUE) else return(NO) } 回答1: Normality tests don't do what most think they do. Shapiro's test, Anderson Darling, and others are null hypothesis tests AGAINST the the assumption of normality. These should not be used to determine whether to use normal theory statistical procedures. In fact they are of virtually no value to

Generate random numbers following a normal distribution in C/C++

由 孤人 提交于 2019-11-26 06:29:28
How can I easily generate random numbers following a normal distribution in C or C++? I don't want any use of Boost. I know that Knuth talks about this at length but I don't have his books at hand right now. S.Lott There are many methods to generate Gaussian-distributed numbers from a regular RNG . The Box-Muller transform is commonly used. It correctly produces values with a normal distribution. The math is easy. You generate two (uniform) random numbers, and by applying an formula to them, you get two normally distributed random numbers. Return one, and save the other for the next request

Random Gaussian Variables

由 China☆狼群 提交于 2019-11-26 03:37:24
问题 Is there a class in the standard library of .NET that gives me the functionality to create random variables that follow Gaussian distribution? 回答1: Jarrett's suggestion of using a Box-Muller transform is good for a quick-and-dirty solution. A simple implementation: Random rand = new Random(); //reuse this if you are generating many double u1 = 1.0-rand.NextDouble(); //uniform(0,1] random doubles double u2 = 1.0-rand.NextDouble(); double randStdNormal = Math.Sqrt(-2.0 * Math.Log(u1)) * Math
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