normal-random | Generates normally distributed random variates
kandi X-RAY | normal-random Summary
kandi X-RAY | normal-random Summary
Normal Random Variables === [NPM version][npm-image]][npm-url] [Build Status][travis-image]][travis-url] [Coverage Status][codecov-image]][codecov-url] [Dependencies][dependencies-image]][dependencies-url].
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QUESTION
As a complete beginner to C++, I would like to generate a random number from a normal distribution.
With the following code (derived from this post), I am able to do so:
...ANSWER
Answered 2020-Mar-17 at 11:15Use a seed to initialize your generator
. Here I am using a time-based seed.
QUESTION
I have a copula representing the dependence between two variables X and Y. I want to compute the following formula: E(X|Y≤1%). It is the expected value of X conditional on Y being lower than 1%. I see that a somewhat similar question was asked there but the R code provided does not give the value I am looking for. Below are some details about the copula and marginal distribution.
...ANSWER
Answered 2019-Jun-12 at 12:22You have to evaluate this double integral: integral of x*pdf(x,y), -oo < x < +oo, -oo < y < 1%
, and divide it by Pr(Y < 1%)
. This is done below. I also perform an approximation by simulations to have a check.
QUESTION
I have been given a task of translating the simulations inside of the Excel plug-in @Risk to Python. The functionalities closely line up with numpy's random number simulation given a distribution type and mu, sigma, or high and low values. An example for what I am doing is here.
In the linked example, mu=2 and sigma=1. Using numpy I get the same distribution as @Risk.
...ANSWER
Answered 2018-Jul-19 at 15:48Take another look at the @RISK documentation that you linked to and the docstring for numpy.random.lognormal
. The @RISK function whose parameters match those of numpy.random.lognormal
is RiskLognorm2. The parameters for numpy.random.lognormal
and RiskLognorm2
are the mean and standard deviation of the underlying normal distribution. In other words, they describe the distribution of the logarithm of the data.
The @RISK documentation explains that the parameters for RiskLognorm
are the mean and standard distribution of the log-normal distribution itself. It gives the formulas for translating between the two methods of parametrizing the distribution.
If you are sure that the parameters in the @RISK code are correct, then you will have to translate those parameters to the form used by numpy.random.lognormal
. Given the values mean
and stddev
as the parameters used by RiskLognorm
, you can get the parameters mu
and sigma
of numpy.random.lognormal
as follows:
QUESTION
I am curious as to how I can add a normal-randomized 300 dimension vector (elements' type = tf.float32) whenever a word unknown to the pre-trained vocabulary is encountered. I am using pre-trained GloVe word embeddings, but in some cases, I realize I encounter unknown words, and I want to create a normal-randomized word vector for this new found unknown word.
The problem is that with my current set up, I use tf.contrib.lookup.index_table_from_tensor to convert from words to integers based on the known vocabulary. This function can create new tokens and hash them for some predefined number of out of vocabulary words, but my embed
will not contain an embedding for this new unknown hash value. I am uncertain if I can simply append a randomized embedding to the end of the embed
list.
I also would like to do this in an efficient way, so pre-built tensorflow function or method involving tensorflow functions would probably be the most efficient. I define pre-known special tokens such as an end of sentence token and a default unknown as the empty string ("at index 0), but this is limited in its power to learn for various different unknown words. I currently use tf.nn.embedding_lookup() as the final embedding step.
I would like to be able to add new random 300d vectors for each unknown word in the training data, and I would also like to add pre-made random word vectors for any unknown tokens not seen in training that is possibly encountered during testing. What is the most efficient way of doing this?
...ANSWER
Answered 2017-Aug-19 at 09:32I never tried it but I can try to provide a possible way using the same machineries of your code, but I will think of it more later.
The index_table_from_tensor
method accepts a num_oov_buckets
parameter that shuffles all your oov words into a predefined number of buckets.
If you set this parameter to a certain 'enough large' value, you will see your data spreads among these buckets (each bucket has an ID > ID of the last in-vocabulary word).
So,
- if (at each lookup) you set (i.e.
assign
) the last rows (those corresponding to the buckets) of yourembedding_init
Variable to a random value - if you make
num_oov_buckets
enough large that collisions will be minimized
you can obtain a behavior that is (an approximation of) what you are asking in a very efficient way.
The random behavior can be justified by a theory similar to the hash table ones: if the number of buckets is enough large, the hashing method of the strings will assign each oov word to a different bucket with high probability (i.e. minimizing collisions to the same buckets). Since, you are assigning a different random number to each different bucket, you can obtain a (almost) different mapping of each oov word.
QUESTION
How do we go about numerically solving equations of the sort below using R?
Please note, this can be shown to be convex and there is a separate thread on this. https://stats.stackexchange.com/questions/158042/convexity-of-function-of-pdf-and-cdf-of-standard-normal-random-variable
This question has been posted on the Mathematics Forum to get Closed Form or other Theoretical Approaches, but it seems numerically solutions are the way to go? https://math.stackexchange.com/questions/2689251/solving-equations-with-standard-normal-cdf-and-pdf-optimization
...ANSWER
Answered 2018-Mar-14 at 11:28You can use the build in optimize
function to directly optimize the original function:
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