How do I estimation of the means of a random variable?

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Zaref Li
Zaref Li 2021-11-9
回答： Sulaymon Eshkabilov 2021-11-9
Hi everyone. I'm trying to understand this code. Are these operations done for the Gaussian random variable? How do I change the Gaussain random variable? Should I write normrnd() instead of rand()?
% lower and upper bounds of the Uniform random variable X:
a= 0;
b = 2;
% True values of the mean and variance of X:
m = (b-a) / 2;
v = (b-a)^2 / 12;
N = 10; % Number of observations
m_h = zeros(1,N); % Preallocation for speed
% Estimation of the mean and variance:
for i = 1 :10
X = b * rand(N,1);
m_h(i) = sum (X) / N;
end
v_h = v/N;
% Mean value of the estimates:
m_h_mean = sum (m_h,2)/N;
% Demonstrate the results:
disp([ 'Mean value of the estimates is: ',num2str(m_h_mean)])
disp([' True mean value of X is: ',num2str(m)])
stem(m_h)
hold
M = m*ones(1,length(m_h)+2);
plot(0:11,M)
M_h = m_h_mean*ones(1,length(m_h)+2);
plot(0:11,M_h, ' --')
axis([0 10.5 0 2])
  1 个评论
Star Strider
Star Strider 2021-11-9
The rand function creates uniformly-distribured random numbers on the interval [0,1] while randn creates normally distributed random numbers , so multiplying it changes σ and adding to it changes μ.
Choose the function that returns the desired result.
.

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采纳的回答

Sulaymon Eshkabilov
% lower and upper bounds of the Uniform random variable X:
a= 0;
b = 2;
% True values of the mean and variance of X:
m = (b-a) / 2;
v = (b-a)^2 / 12;
N = 10; % Number of observations
m_h = zeros(1,N); % Preallocation for speed
mu0=30; % Single Mean value of population
% mu0=[30:20:100]; % A few different mean values of population
sigma0=3.5; % STD
SETs = 1; % How many sets of Population to generate
% SETs = 5; % Five sets of Population to generate in each
% iteration
% Estimation of the mean and variance:
for i = 1 :10
X = normrnd(mu0, sigma0, N, SETs);
m_h(i) = mean(X);
end
% OR without [for .. end] loop, use this one:
X2 = normrnd(mu0, sigma0, N, N);
m_h2 = mean(X2,2);
...

更多回答（1 个）

Sulaymon Eshkabilov
Ran in:
In order to generate a normally distributed random numbers, you can employ
clc; clearvars; close all
Npop = 7.5e2; % Population size
mu0=30; % Mean value of population
sigma0=3.5; % STD
SETs = 3; % How many sets of Population to generate
P0 = normrnd(mu0, sigma0, Npop, SETs);
histfit(P0(:,1), 30, 'normal'), title('One Set of Population')

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