PDF function does not give same result as normpdf

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George Ansari
George Ansari el 21 de Ag. de 2017
Respondida: Star Strider el 21 de Ag. de 2017
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 comentario
George Ansari
George Ansari el 21 de Ag. de 2017
Sorry I mean LPDF = [5,200; 9,340; 0,006; 0,167; 0,167; 0,006; 9,340]

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Respuesta aceptada

Star Strider
Star Strider el 21 de Ag. de 2017
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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