# how to plot exponential pdf over a distributed data ?

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Amr Hashem on 24 Oct 2015
Commented: Star Strider on 26 Oct 2015
in order to find the best fit model
I want to produce this figure (a data & best fit over it):
so I try:
v=1:25;
n=[10;7;6;21;14;18;23;33;28;31;34;30;32;48;43;46;39;47;40;45;51;47;45;45;23];
plot(v,n,'*b')
ex=expfit(n);
E=exppdf(n,ex);
hold on
plot(v,E,'-r')
but it produce this figure:
how to modify the code to get the first figure?
Amr Hashem on 24 Oct 2015
why there aren't any interaction?

Star Strider on 24 Oct 2015
Fourteen hours ago I was asleep here in GMT-7 land.
I believe you are mistaking the exponential distribution for the exponential function.
This produces a good fit to your data:
v=1:25;
n=[10;7;6;21;14;18;23;33;28;31;34;30;32;48;43;46;39;47;40;45;51;47;45;45;23];
plot(v,n,'bp')
expfcn = @(b,x) b(1) + b(2).*exp(b(3).*x); % Three-Parameter Exponential Function
B0 = [50; -50; -0.5]; % Initial Parameter Estimates
B = nlinfit(v(:), n(:), expfcn, B0); % Estimate Parameters
E = expfcn(B,v); % Simulate Function
hold on
plot(v,E,'-r')
I used the nlinfit function because I know you have the Statistics Toolbox.
Amr Hashem on 26 Oct 2015
Edited: Amr Hashem on 26 Oct 2015
but when I try it on:
expfcn = @(b,x) b(2).*exp(b(3).*x);
x intercept at -7413 and the place of fitted curve changed I use
expfcn = @(b,x) b(2).*exp(b(3).*x);
as this the function of exponential i want
so I adjust the code to :
v=1:25;
n=[10;7;6;21;14;18;23;33;28;31;34;30;32;48;43;46;39;47;40;45;51;47;45;45;23];
plot(v,n,'*b')
expfcn = @(b,x) ( b(2).*exp(b(3).*x)); % Three-Parameter Exponential Function
B0 = [50; -50; -0.5]; % Initial Parameter Estimates
B = nlinfit(v(:), n(:), expfcn, B0); % Estimate Parameters
vrng = linspace(-35, max(v)); % Create Matching Vector
E = expfcn(B,vrng); % Simulate Function With ‘vrng’
hold on
plot(vrng,E,'-r')
this is work, but I didn't know how to find the intercept I write it manually ?
Star Strider on 26 Oct 2015
The x-intercept, such as it is, fzero calculates as -20970.520.
You could probably find the intercept manually, but your code is incorrect when estimating the parameters.
My code here is the correct method:
v=1:25;
n=[10;7;6;21;14;18;23;33;28;31;34;30;32;48;43;46;39;47;40;45;51;47;45;45;23];
plot(v,n,'*b')
expfcn = @(b,x) b(1).*exp(b(2).*x); % Two-Parameter Exponential Function
B0 = [50; 0]; % Initial Parameter Estimates
B = nlinfit(v(:), n(:), expfcn, B0); % Estimate Parameters
Y0 = fzero(@(x)expfcn(B,x), 1); % Find ‘v’ At E=0
vrng = linspace(Y0, max(v)); % Create Matching Vector
E = expfcn(B,vrng); % Simulate Function
hold on
plot(vrng,E,'-r')
fprintf(1, '\n\tE(%.3f) = 0\n', Y0)
You are only fitting two parameters, so you can only specify two in the nlinfit argument list, and you have to refer to them correctly in the ‘expfcn’ objective function. Otherwise, nlinfit will attempt to fit all three of them, even though your function is using only two, and will likely return inaccurate results for the two it does estimate.