如何生成表示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:

假设我想生成取值 -1 或 1 的离散均匀随机数.换句话说，我想生成具有以下分布的数字:

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

要生成包含 100 个这些数字的数组，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 数组如下所示:[-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 个数字之和.

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.

当然可以:

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

有

SDUD =

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

是否有数学/matlab 技巧可以做到这一点? 不使用 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 个值和 n=100)如下所示:

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

奖励:

如果可以修改原始离散值，那就太好了.因此，离散值可以代替 [-1,1]，例如 [0,1,2]，每个出现 p = 1/number_of_discrete_value，因此在本例中为 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

这本质上是一个二项式分布(参见 Matlab 的 binornd)，仅缩放和移位，因为基础值由 DV 给出 而不是 0 和 1:

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 多项分布(参见 Matlab 的 mnrnd):

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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