PDF function does not give same result as normpdf

조회 수: 2 (최근 30일)
George Ansari
George Ansari 2017년 8월 21일
답변: Star Strider 2017년 8월 21일
Hello all, I'm using the following function to create the PD of a RV
function [ X, f ] = Normdist( mu, sigma, min_x, max_x, n )
X = zeros(n,1);
x = min_x;
dx = (max_x - min_x)/n;
for k = 1:n
X(k) = x;
f(k) = 1/sqrt(2*pi*sigma)*exp(-(x-mu)^2/(2*sigma));
x = x+dx;
end
end
I call it as follows:
[LRV, LPDF] = Normdist(0, 2.5, -10, 10, 7);
LRV = [-10; -7,142; -4,285; -1,428; 1,428; 4,285; 7,14]
but when I call:
A = normpdf(linspace(-10,10,7),0,2.5)
I get:
A = [5,353; 0,004; 0,065; 0,159; 0,065; 0,004; 5,353]
what is wrong with the function? George.
  댓글 수: 1
George Ansari
George Ansari 2017년 8월 21일
Sorry I mean LPDF = [5,200; 9,340; 0,006; 0,167; 0,167; 0,006; 9,340]

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Star Strider
Star Strider 2017년 8월 21일
You need to calculate ‘x’ differently (so that it creates the same interval that linspace does), and square ‘sigma’ in the denominator of the exp argument:
X = zeros(n,1);
f = zeros(n,1);
x = min_x;
dx = (max_x - min_x)/(n - 1);
for k = 1:n
X(k) = x;
f(k) = 1/(sqrt(2*pi)*sigma)*exp(-(x-mu)^2/(2*sigma^2));
x = x+dx;
end

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