Generationf of uniform random

2 次查看（过去 30 天）
htet wai
htet wai 2017-11-10
评论： htet wai 2017-11-11
How can I generate the uniform random array with the constraint of second value not changing more than 20% of the first value
for example (0.5 0.45 0.54 0.63 0.52...........) The difference between first random to second random should be within 20% of first value and the overall random should in uniform manner.
Thank You.
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2017-11-10
The values after the first are all to be within 20% of the first? Or the values after the first each have to be within 20% of the immediately proceeding value?
htet wai
htet wai 2017-11-10
I want to generate 1000 random where random(i+1) is greater or less than x% (0-20%) of random(i). but all the 1000 random should be within uniformly distributed.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2017-11-10
V = zeros(1,100);
V(1) = rand;
for K = 2 : 100; V(K) = V(K-1) + (rand() * 2 - 1) * V(K-1) * .2; end
  5 个评论
显示 3更早的评论
Walter Roberson
Walter Roberson 2017-11-11
Suppose we remove the restriction that the maximum value can be 1. Suppose we simplify the model by alternately adding and subtracting the full 20%, a simple up/down randomness. Then each time we subtract 20% the previous value would be multiplied by 4/5, and each time we add 20% the previous value would be multiplied by 6/5. In this simplified model, those occur in pairs, so each pair would give 4/5 * 6/5 = 24/25 times the result of the previous pair. As you continue this, the end result is going to be (24/25)^n which for large n, the result is clearly going to tend to 0.
htet wai
htet wai 2017-11-11
Thank you for your answer and I understand your point.
Is it possible to generate the uniform random where the changes between each random term is +/-0.2 (instead of 20% of the first value)?

请先登录，再进行评论。

更多回答（1 个）

Kaushik Lakshminarasimhan
编辑：Kaushik Lakshminarasimhan 2017-11-10
This should work:
N=10; % number of random numbers
success = false;
while ~success
x = rand(N,1);
[minval,minindx] = min(abs(x(2:end) - x(1)));
if minval < 0.2*x(1)
x([2 minindx+1]) = x([minindx+1 2]);
success = true;
end
end
disp(x);
Note that as Star Strider pointed out the sequence will no longer be random.

类别

在 Help Center 和 File Exchange 中查找有关 Random Number Generation 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by