PDF function does not give same result as normpdf

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George Ansari
George Ansari 2017 年 8 月 21 日
回答済み: Star Strider 2017 年 8 月 21 日
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.
  1 件のコメント
George Ansari
George Ansari 2017 年 8 月 21 日
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 2017 年 8 月 21 日
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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