如何生成代表离散均匀分布之和的数字 [英] How to generate a number representing the sum of a discrete uniform distribution

问题描述

第1步:

比方说，我想生成值为-1或1的离散均匀随机数.因此，换句话说，我想生成具有以下分布的数:

Let's say that I want to generate discrete uniform random numbers taking the value -1 or 1. So in other words I want to generate numbers having the following distribution:

P(X = -1) = 0.5
P(X =  1) = 0.5

要生成一个由100个数字组成的数组，我可以编写以下代码:

To generate an array of 100 of those numbers I can write this code:

n   = 100
DV  = [-1,1];          % Discrete value
RI  = unidrnd(2,n,1);  % Random uniform index
DUD = DV(RI);          % Discrete uniform distribution

我的DUD数组如下:[-1,1,1,1,-1,-1,1,-1,...]

My DUD array looks like: [-1,1,1,1,-1,-1,1,-1,...]

第2步:

现在我想生成10个等于sum(DUD)的数字，因此10个数字的分布对应于遵循离散均匀分布的100个数字的总和.

Now I would like to generate 10 numbers equal to sum(DUD), so 10 numbers having a distribution corresponding to the sum of 100 numbers following a discrete uniform distribution.

我当然可以做到:

for ii = 1:10
    n   = 100;
    DV  = [-1,1];          % Discrete value
    RI  = unidrnd(2,n,1);  % Random index
    DUD = DV(RI);          % Discrete uniform distribution
    SDUD(ii) = sum(DUD);
end

使用

SDUD =

   2   2  -6  -2  -4   2   4   4   0   2 

是否有数学/矩阵技巧可以做到这一点?而无需使用for循环.

Is there a mathematical/matlab trick to do that ? without using a for loop.

SDUD的直方图(具有10000个值且n = 100)如下所示:

The histogram of SDUD (with 10000 values and n=100) looks like this:

奖金:

如果可以修改原始离散值，那就太好了.因此，离散值可以是[0,1,2]，而不是[-1,1]，每个都出现p = 1/number_of_discrete_value，因此在此示例中为1/3.

It will be great if the original discrete values could be modified. So instead of [-1,1] the discrete value could be, for example, [0,1,2], each with an occurence p = 1/number_of_discrete_value, so 1/3 in this example.

推荐答案

对于两个值

从本质上讲，这是二项式分布(请参阅Matlab的

For two values

That's essentially a binomial distribution (see Matlab's binornd), only scaled and shifted because the underlying values are given by DV instead of being 0 and 1:

n = 100;
DV = [-1 1];
p = .5; % probability of DV(2)
M = 10;
SDUD = (DV(2)-DV(1))*binornd(n, p, M, 1)+DV(1)*n;

对于任意数量的值

您拥有的是多项式分布(请参阅Matlab的

For an arbitrary number of values

What you have is a multinomial distribution (see Matlab's mnrnd):

n = 100;
DV = [-2 -1 0 1 2];
p = [.1 .2 .3 .3 .1]; % probability of each value. Sum 1, same size as DV
M = 10;
SDUD = sum(bsxfun(@times, DV, mnrnd(n, p, M)), 2);

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