probability

I am new with poisson point process. I did one simluation (matlab) as below. My intensity lambda = 50 ; clear all; lambda=50; npoints = poissrnd(lambda); pproc = rand(npoints, 2); plot(pproc(:, 1), pproc(:, 2), '.'); Then I have plot, However, the link http://connor-johnson.com/2014/02/25/spatial-point-processes/ showed me that when intensity lamuda = 0.2, smaller than 1 , he got The link also showed the code in Python.Please check it. Here is my question, why intensity is smaller than 1 , he still can plot something here? If I let my code's lamda = 0.2, there will be no value to plot. I think

I have applied gaussmix function in voicebox MATLAB tools to calculate GMM. However, the code gives me error when I run it for 512 GMM components. No_of_Clusters = 512; No_of_Iterations = 10; [m_ubm1,v_ubm1,w_ubm1]=gaussmix(feature,[],No_of_Iterations,No_of_Clusters); Error using * Inner matrix dimensions must agree. Error in gaussmix (line 256) pk=px*wt; % pk(k,1) effective number of data points for each mixture (could be zero due to underflow) I need 1024 or 2048 Mixtures for Universal Background Model (UBM) construction. Could anyone give me matlab code to calculate GMM for big number of

I know how to create an histogram in Python, but I would like that it is the probability density distribution. Let's start with my example. I have an array d , with a size of 500000 elements. With the following code I am building a simple histogram telling me how many elements of my array d are between every bin. max_val=log10(max(d)) min_val=log10(min(d)) logspace = np.logspace(min_val, max_val, 50) H=hist(select,bins=logspace,histtype='step') The problem is that this plot is not what I want. I would like to have the probability distribution function of my array d . Instead of having the

I have a question regarding the hist() function with matplotlib. I am writing a code to plot a histogram of data who's value varies from 0 to 1. For example: values = [0.21, 0.51, 0.41, 0.21, 0.81, 0.99] bins = np.arange(0, 1.1, 0.1) a, b, c = plt.hist(values, bins=bins, normed=0) plt.show() The code above generates a correct histogram (I could not post an image since I do not have enough reputation). In terms of frequencies, it looks like: [0 0 2 0 1 1 0 0 1 1] I would like to convert this output to a discrete probability mass function, i.e. for the above example, I would like to get a

Is there a package in Perl that allows you to compute the height of probability distribution at each given point. For example this can be done in R this way: > dnorm(0, mean=4,sd=10) > 0.03682701 Namely the probability of point x=0 falls into a normal distribution, with mean=4 and sd=10, is 0.0368. I looked at Statistics::Distribution but it doesn't give that very function to do it. Why not something along these lines (I am writing in R, but it could be done in perl with Statistics::Distribution): dn <- function(x=0 # value ,mean=0 # mean ,sd=1 # sd ,sc=10000 ## scale the precision ) { res <-

I have two documents. Doc1 is in the below format: TOPIC: 0 5892.0 site 0.0371690427699 Internet 0.0261371350984 online 0.0229124236253 web 0.0218940936864 say 0.0159538357094 TOPIC: 1 12366.0 web 0.150331554262 site 0.0517548115801 say 0.0451237263464 Internet 0.0153647096879 online 0.0135856380398 ...and so on till Topic 99 in the same pattern. And Doc2 is in the format: 0 0.566667 0 0.0333333 0 0 0 0.133333 .......... and so on... There are totally 100 values each value for each topic. Now, I have to find the weighted average probability for each word, that is: P(w) = alpha.P(w1)+ alpha.P

I'm implementing the forward algorithm for HMMs to calculate the probability of a given HMM emitting a given observation sequence. I'd like my algorithm to be robust to underflow. I can't work in log-space because the forward algorithm requires the multiplication AND addition of probabilities. What is the best way to avoid underflow? I've read some sources about this but the best suggestion I get is scaling the probabilities at each time step Section 6 Here . By the end of the algorithm you won't be left with the exact probability you want (of the observation sequence). Also, unless I'm