python-2.7,numpy,random,normal-distribution

numpy.random.randint Return random integers from the “discrete uniform” distribution in the “half-open” interval. numpy.random.normal Draw random samples from a normal (Gaussian) distribution. randint() and normal() do not pick a number the same way. The odds of getting any number in the chosen interval using randint() is the same, unlike numbers...

We can easily help you, if you give as a definition of L(.). Ok, let's believe the answer by Gavin Kelly. In that case, L0 <- function(x) exp(-(x^2)/2)/sqrt(2*pi) - x * (1 - pnorm(x)) L <- function(x) dnorm(x) - x * pnorm(x, lower.tail=FALSE) is a numerically better, namely accurate also...

python,statistics,scipy,normal-distribution,cdf

edit: you actually need import norm from scipy.stats. I found the answer. You need to use ppf in scipy.stats which stands for "percent point function". So let's say you have a normal distribution with stdDev = 1, and mean = 0 and you want to find the value at which...

Let's say you run the following: rnorm(5, 10, 2) What you get is the value of five points randomly drawn from a normal distribution that has a mean of 10 and a standard deviation of 2. It's a random draw, so each time you rerun this line you will get...

statistics,wolfram-mathematica,normal-distribution,cdf

1) MultinormalDistribution is now built in, so don't load MultivariateStatistics it unless you are running version 7 or older. If you do you'll see MultinormalDistribution colored red indicating a conflict. 2) this works: sig = .5; u = .5; dist = MultinormalDistribution[{0, 0}, sig IdentityMatrix[2]]; delta = CDF[dist, {xx, yy}]...

It sounds like you want to find the values that divide the area under the probability distribution function into segments of equal probability. This can be done in matlab by applying the norminv function. In your particular case: segmentBounds = norminv(linspace(0,1,10),0,1) Any two adjacent values of segmentBounds now describe the...

python,random,gaussian,normal-distribution

Thread-safe pieces of code must account for possible race conditions during execution. This introduces overhead as a result of synchronization schemes like mutexes, semaphores, etc. However, if you are writing non-reentrant code, no race conditions normally arise, which essentially means that you can write code that executes a bit faster....

r,graph,ggplot2,normal-distribution,gplots

It's really more or less the same. I prefer to use annotate for things like this. (I'd do your red line with annotate as well, personally.) Just calculate where you want things to go and add them to the plot: qplot(my_data,dist, geom="line")+xlab("x values")+ylab("Density")+ geom_point()+ ggtitle("cool graph Distribution") + geom_line(color="black", size=0.1)...

r,statistics,normal-distribution

Since both T1 and T2 rely on X1, X2, Y1, and Y2, you should first simulate those four random variables: X1 <- rnorm(1e4, mu1, sigma) X2 <- rnorm(1e4, mu1, sigma) Y1 <- rnorm(1e4, mu2, sigma) Y2 <- rnorm(1e4, mu2, sigma) Then you can run your code to get all simulated...

python,scikit-learn,gaussian,normal-distribution

Try looking into pypr. From the documentation, here is how you would find a GMM conditioned on one or more of the variables: # Now we will find the conditional distribution of x given y (con_cen, con_cov, new_p_k) = gmm.cond_dist(np.array([np.nan, y]), \ cen_lst, cov_lst, p_k) As far as i remember,...

matlab,for-loop,vectorization,normal-distribution,norm

Discussion and Code You can remove that loop with a bsxfun based vectorized version - N_g_sigma = l*sqrt(sum(bsxfun(@minus,x,g).^2,2)); N_g_normrnd = bsxfun(@plus,randn(size(N_g_sigma)).*N_g_sigma,g); N_g = bsxfun(@times,N_g_normrnd - x,1.4962*r_t); N_p_sigma = l*sqrt(sum(bsxfun(@minus,x,P_best).^2,2)); N_p_normrnd = bsxfun(@plus,randn(size(N_p_sigma)).*N_p_sigma,P_best); N_p = bsxfun(@times,N_p_normrnd - x,1.4962*r_t); It is based on a hacked version of normrnd.m as also exploited in...

c++,statistics,drawing,normal-distribution

Well, after investing this day I think I finally found a suitable solution. The idea is to compute the point on the x-axis which has the desired likelihood w.r.t. to the standard normal distribution. This can be done with a bisection-algorithm. Afterwards I can use the distance of this point...

python,loops,numpy,random,normal-distribution

There are two minor issues - the first relates to how to select the name of the files (which can be solved using pythons support for string concatenation), the second relates to np.random.normal - which only allows a size parameter when loc is a scalar. data = pl.loadtxt("20100101.txt") density =...

c++,c++11,random,normal-distribution,poisson

You can center both distributions in a point that suits your needs. But if M is small, then the Poisson distribution has a 'fat tail', that is, the probability of getting a number above M is higher compared to the normal distribution. In the normal case, you can control this...

r,normal-distribution,standard-deviation

Here's another version using ggvis: library(dplyr) library(ggvis) ## -- data generation copied from @NickK -- ## data.frame(group = letters[1:4], m = c(130, 134, 132, 105), s = c(20, 14, 12, 10)) %>% group_by(group) %>% do(data_frame(group = .$group, x = 50:200, y = dnorm(x, .$m, .$s), withinSd = abs(x - .$m)...

r,distribution,normal-distribution,beta-distribution

Use uniroot(). uniroot(function(x) dbeta(x, 1, 2)-dnorm(x, 0, 1), c(0, 1)) ## $root ## [1] 0.862456 ## ## $f.root ## [1] 5.220165e-05 ## ## $iter ## [1] 3 ## ## $estim.prec ## [1] 6.103516e-05 This solves an equation dbeta(x, ...) == dnorm(x, ...) w.r.t. x (in the inverval [0,1], as this...

python,numpy,random,normal-distribution

If you want to bootstrap you could use random.choice() on your observed series. Here I'll assume you'd like to smooth a bit more than that and you aren't concerned with generating new extreme values. Use pandas.Series.quantile() and a uniform [0,1] random number generator, as follows. Training Put your random sample...

javascript,random,normal-distribution

OK, so the central limit theorem says that the average of a sufficiently large number of uniformally distributed variables will be normal. In the statistics classes I have taken, 30 is usually used as the cutoff, so you might want to increase your simulation's "sample size". However, you can find...

sql,oracle,random,plsql,normal-distribution

If you want to generate numbers from an arbitrary normal distribution, there are only two parameters that make sense, the mean and the standard deviation. I'm not sure what "interval" you'd want to specify unless you want to produce a truncated distribution. Given a number from a standard normal distribution...

c,statistics,normal-distribution

That should do the trick. It's objective-c code but should be easily convertible into c. I use it for statistical calculations and it works just fine. - (double)getInverseCDFValue:(double)p { double a1 = -39.69683028665376; double a2 = 220.9460984245205; double a3 = -275.9285104469687; double a4 = 138.3577518672690; double a5 =-30.66479806614716; double a6...

r,statistics,distribution,normal-distribution

Here is a simple implementation. Like @DanielJohnson says you can just use the cdf form univariate normal, but it should be same as using the pmvnorm, shown below. The version using pnorm is much faster. ## Choose the matrix dimensions yticks <- xticks <- seq(-3, 3, length=100) side <- diff(yticks[1:2])...

java,statistics,normal-distribution

First of all, the question cannot be answered as is because in a continous distribution like the normal distribution, the probability of an specific point is always zero. You need to ask yourself what it is exactly you want to know in terms of an interval. For example, cern.jet.stat.Probability.normal(double) will...

c++,boost,random,normal-distribution

A few tricky bits: the inclusion order is important (Using Boost.Random to generate multiprecision integers from a seed) and you should disable expression templates for the cpp_int parameter to independent_bits. Live On Coliru #include <boost/multiprecision/random.hpp> #include <boost/random.hpp> #include <boost/multiprecision/cpp_int.hpp> #include <boost/multiprecision/cpp_dec_float.hpp> #include <boost/multiprecision/number.hpp> int main() { namespace mp = boost::multiprecision;...

python,numpy,statistics,normal-distribution

If I got it right, you want to have an output vector with the following properties: boolean vector same number of elements as in the input vector probability of each element being True depends on its value w.r.t. threshold the number of Trues is the same as if we used...

probability,matlab,normal-distribution,standard-deviation

Ideally, there are no such maximum and minimum. A normal (Gaussian) pdf has infinite support, so it can produce any value, no matter how high or low, with positive probability. Of course, exceeding a value x is less probable as x grows; but the probability is never 0. In reality,...

r,gaussian,normal-distribution

With qnorm: qnorm(.025) # [1] -1.959964 qnorm(.5) # [1] 0 qnorm(.975) # [1] 1.959964 ...

c++,boost,noise,normal-distribution

Here's my take on it: #include <boost/random/normal_distribution.hpp> #include <boost/random.hpp> int main() { boost::mt19937 gen(42); // seed it once boost::normal_distribution<double> nd(0.0, 1.0); boost::variate_generator<boost::mt19937&, boost::normal_distribution<double> > randNormal(gen, nd); std::vector<double> data(100000, 0.0), nsyData; nsyData.reserve(data.size()); double sd = 415*randNormal(); std::transform(data.begin(), data.end(), std::back_inserter(nsyData),...

There is no issue. You got that answer because its correct. Perhaps you need to show more digits in your cell? Norm.Dist(x,m,s,0) gives you the value of the probability density function of the normal distribution function, evaluated at x, with mean=m and standard deviation = s. For the values of...

c,probability,normal-distribution,quantitative-finance,probability-theory

The standard normal cumulative distribution function is exactly (1/2)*(1 + erf(z/sqrt(2))) where erf is the Gaussian error function, which is found in many C programming libraries. Check the development environment you are using -- chances are good that erf is already in one of its libraries.

r,distribution,normal-distribution,binning

If your range of data is from -2:2 with 15 intervals and the sample size is 77 I would suggest the following to get the expected heights of the 15 intervals: rn <- dnorm(seq(-2,2, length = 15))/sum(dnorm(seq(-2,2, length = 15)))*77 [1] 1.226486 2.084993 3.266586 4.716619 6.276462 7.697443 8.700123 9.062576 8.700123...

python,random,simulation,normal-distribution

You can generate the random numbers using random.gauss. For example I'll create a list of 10 random numbers, with a mean of 10 and standard deviation of 1 >>> import random >>> nums = [random.gauss(10, 1) for _ in range(10)] >>> nums [11.959067391283675, 9.736968009359552, 9.034607376861388, 9.431664007796622, 11.522646114577977, 9.777134678502273, 10.954304068858296, 9.641278997034552,...

excel,distribution,gaussian,normal-distribution

The results from NORM.DIST are correct... if you directly implement the Gaussian function in your sheet using: =(1/($F$8*SQRT(2*PI())))EXP( -((M3-$F$7)^2)/(2$F$8^2)) which is an implementation of the standard Gaussian function e.g. f(x) on: http://mathworld.wolfram.com/GaussianFunction.html then the results exactly match Excel's NORM.DIST built in function. When you say the values "should be" in...

python,pandas,statistics,normal-distribution

Caveat here is that I'm not a stats expert but basically scipy has a number of tests you can conduct on your data to test whether it could be considered to be a normalused Gaussian distribution. Here I create 2 series one is simple a linear range and the other...

matlab,random,multidimensional-array,normal-distribution

If by definition you refer to the density of the multivariate normal distribution: it contains neither the Cholesky decomposition nor the matrix square root of Σ, but its inverse and the scalar square root of its determinant. But for numerically generating random numbers from this distribution, the density is not...