does random number in the real world follow uniform distribution - random

I am confused about the concept of uniform distribution and random number. Does random number follow uniform distribution or does random number not follow any distribution?

Traditionally, random is just that, random. There is nothing that guarantees that after 100 random numbers, at least one of them is non-zero. You'd probably think something is broken, and while you'd probably be right, it is just as possible as any other combination of numbers in the same given range.
Uniform distribution will ensure that statistically speaking, your values will be spread out across a given range. In that case if you got 100 random uniformly distributed numbers and they were all zero, something is definitely broken.

Any number generator guaranteeing a uniform distribution is not random. That said, the more numbers you generate, the more likely it is to resemble a uniform distribution.

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How to get a representative random number from a set of pseudo random numbers?

Let's say I got three pseudo random numbers from different pseudo random number generators.
Since the generators would reflect only a part of the real random number generating process, I believe that one way to get a number closer to real random might be to somehow get a "center" of the three pseudo random numbers.
An easy way to get that "center" would be to take average, median or mode (if any) of them.
I am wondering if there's a more sophisticated way due to the fact that they should represent random numbers.
Well, there is an approach, called entropy extractor, which allows to get (good) random numbers from not quite random source(s).
If you have three independent but somewhat low quality (biased) RNGs, you could combine them together into uniform source.
Suppose you have three generators giving you a single byte each, then uniform output would be
t = X*Y + Z
where addition and multiplication are done over GF(28) finite field.
Some code (Python)
def RNG1():
return ... # single random byte
def RNG2():
return ... # single random byte
def RNG3():
return ... # single random byte
from pyfinite import ffield
def muRNG():
X = RNG1()
Y = RNG2()
Z = RNG3()
GF = ffield.FField(8)
return GF.Add(GF.Multiply(X, Y), Z)
Paper where this idea was stated
Trying to use some form of "centering" turns out to be a bad idea if your goal is to have a better representation of the randomness.
First, a thought experiment. If you think three values gives more randomness, wouldn't more be even better? It turns out that if you take either the average or median of n Uniform(0,1) values, as n→∞ these both converge to 0.5, a point. It also happens to be the case that replacing distributions with a "representative" constant is generally a bad idea if you want to understand stochastic systems. As an extreme example, consider queues. As the arrival rate of customers/entities approaches the rate at which they can be served, stochastic queues get progressively larger on average. However, if the arrival and service distributions are constant, queues remain at zero length until the arrival rate exceeds the service rate, at which point they go to infinity. When the rates are equal, the stochastic queue would have infinite queues, while the deterministic queue would remain at its initial length (usually assumed to be zero). Infinity and zero are about as wildly different as you can get, illustrating that replacing distributions in a queueing model with their means would give you no understanding of how queues actually work.
Next, empirical evidence. Below histograms of the medians and averages constructed from 10,000 samples of three uniforms. As you can see, they have different distribution shapes but are clearly no longer uniform. Values bunch in the middle and are progressively rarer towards the endpoints of the range (0,1).
The uniform distribution has maximum entropy for continuous distributions on a closed interval, so both of these alternatives, being non-uniform, are clearly lower entropy, i.e., more predictable.
To get good random numbers, it's advisable to get some bits of entropy. Depending on whether they are used for security purposes or not, you could just get the time from the system clock as a seed for a random number generator, or use more sophisticated means. The project PWGen download | SourceForge.net is open-sourced, and monitors Windows events as a source of random bits of entropy.
You can find more info on how to random numbers in C++ from this SO ? too: Random number generation in C++11: how to generate, how does it work? [closed]. It turns out C++'s random numbers aren't always all that random: Everything You Never Wanted to Know about C++'s random_device; so looking for a good way to seed, i.e. by passing the time in mS to srand() and calling rand() might be a quick and dirty way to go.

Random number from many other random numbers, is it more random?

We want to generate a uniform random number from the interval [0, 1].
Let's first generate k random booleans (for example by rand()<0.5) and decide according to these on what subinterval [m*2^{-k}, (m+1)*2^{-k}] the number will fall. Then we use one rand() to get the final output as m*2^{-k} + rand()*2^{-k}.
Let's assume we have arbitrary precision.
Will a random number generated this way be 'more random' than the usual rand()?
PS. I guess the subinterval picking amounts to just choosing the binary representation of the output 0. b_1 b_2 b_3... one digit b_i at a time and the final step is adding the representation of rand() to the end of the output.
It depends on the definition of "more random". If you use more random generators, it means more random state, and it means that cycle length will be greater. But cycle length is just one property of random generators. Cycle length of 2^64 usually OK for almost any purpose (the only exception I know is that if you need a lot of different, long sequences, like for some kind of simulation).
However, if you combine two bad random generators, they don't necessarily become better, you have to analyze it. But there are generators, which do work this way. For example, KISS is an example for this: it combines 3, not-too-good generators, and the result is a good generator.
For card shuffling, you'll need a cryptographic RNG. Even a very good, but not cryptographic RNG is inadequate for this purpose. For example, Mersenne Twister, which is a good RNG, is not suitable for secure card shuffling! It is because observing output numbers, it is possible to figure out its internal state, so shuffle result can be predicted.
This can help, but only if you use a different pseudorandom generator for the first and last bits. (It doesn't have to be a different pseudorandom algorithm, just a different seed.)
If you use the same generator, then you will still only be able to construct 2^n different shuffles, where n is the number of bits in the random generator's state.
If you have two generators, each with n bits of state, then you can produce up to a total of 2^(2n) different shuffles.
Tinkering with a random number generator, as you are doing by using only one bit of random space and then calling iteratively, usually weakens its random properties. All RNGs fail some statistical tests for randomness, but you are more likely to get find that a noticeable cycle crops up if you start making many calls and combining them.

Generating Gaussian Random Numbers without a Uniform Random Number Generator

I know many uniform random number generators(RNGs) based on some algorithms, physical systems and so on. Eventually, all these lead to uniformly distributed random numbers. It's interesting and important to know whether there is Gaussian RNGs, i.e. the algorithm or something else creates Gaussian random numbers. Much precisely I want to say that I don't want to use transformations such as Box–Muller or Marsaglia polar method to get Gaussian from Uniform RNGs. I am interested if there is some paper, algorithm or even idea to create Gaussian random numbers without any of use Uniform RNGs. It's just to say we pretend that we don't know there exist Uniform random number generators.
As already noted in answers/comments, by virtue of CLT some sum of any iid random number could be made into some reasonable looking gaussian. If incoming stream is uniform, this is basically Bates distribution. Ami Tavory answer is pretty much amounts to using Bates in disguise. You could look at closely related Irwin-Hall distribution, and at n=12 or higher they look a lot like gaussian.
There is one method which is used in practice and does not rely on transformation of the U(0,1) - Wallace method (Wallace, C. S. 1996. "Fast Pseudorandom Generators for Normal and Exponential Variates." ACM Transactions on Mathematical Software.), or gaussian pool method. I would advice to read description here and see if it fits your purpose
As others have noted, it's a bit unclear what is your motivation for this, and therefore I'm not sure if the following answers your question.
Nevertheless, it is possible to generate (an approximation of) this without the specific formulas transforming uniform RNGs that you mention.
As with any RNG, we have to have some source of randomness (or pseudo-randomness). I'm assuming, therefore, that there is some limitless sequence of binary bits which are independently equally likely to be 0 or 1 (note that it's possible to counter that this is a uniform discrete binary RNG, so I'm unsure if this answers your question).
Choose some large fixed n. For each invocation of the RNG, generate n such bits, sum them as x, and return
(2 x - 1) / √n
By the de Moivre–Laplace theorem this is normal with mean 0 and variance 1.

How to choose seeds for two random number generators so that sequences would be completely uncorrelated?

I want to use rejection sampling to generate random numbers from a given distribution. I want to be quite general so that I don't want to relay on things like Box-Muller transformation which can generate only normal distributed random numbers. I am using linear congruential generator to generate a random sequence between 0 and 1 with uniform distribution. To use rejection sampling, I need to generate two sequences of random numbers so that I would be able to generate uniform points inside a 2d region. This can be done using two random sequences (one for x coordinate and other for y coordinate). I searched on Internet but nowhere I saw how to make sure that these two sequences are really uncorrelated. Is there any way to choose seeds for these such that these sequences are uncorrelated? If I randomly give seeds then the final distribution of these numbers is not quite like what I am looking for.
Thank you

Pseudorandom Number Generation with Specific Non-Uniform Distributions

I'm writing a program that simulates various random walks (with differing distributions). At each timestep, I need randomly generated, two dimensional step distances and angles from the distribution of the random walk. I'm hoping someone can check my understanding of how to generate these random numbers.
As I understand it I can use Inverse Transform Sampling as follows:
If f(x) is the pdf of our random walk that has a non-uniform distribution, and y is a random number from a uniform distribution.
Then if we let f(x) = y and solve to find x then we have a random number from the non-uniform distribution.
Is this a feasible solution?
Not quite. The function that needs to be inverted is not f(x), the pdf, but F(x)=P(X<=x)=int_{-inf}^{x}f(t)dt, the cdf. The good thing is that F is monotone, so actually has a unique inverse (unlike f).
There are multiple other ways of generating random numbers according to a given distribution. For example, if the cdf F is difficult to compute or to invert, rejection sampling can be a good option if f is easy to compute.
You are close, but not quite. Every probability density function (pdf) has a corresponding cumulative density function (cdf). An important property about CDF(x) is that they are always between 0 and 1. Because it is relatively easy to draw a random number between 0 and 1, we can use that to work our way backwards to the distribution. So changing the word pdf to CDF in your question makes the statement correct.
As an aside for this to make sense computationally you need to find an easy to calculate inverse of the CDF. One way to do this is to fit a polynomial approximation to the CDF and find the inverse of that function. There are more advanced techniques for simulating probability distributions with messy distributions. See this book chapter for the details.

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