how to plot exponential pdf over a distributed data ?

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]

why there aren't any interaction?

Answer 1

[Star Strider]

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]

Thanks

but I want to plot the " exponential distribution "

as I want to create the best curve fitting

so i want to try:

exponential distribution

poisson distribution

weibull distribution

that is explain why I use "exppdf"

waiting for your advices

[Star Strider]

You can fit it to whatever distribution you want (and are appropriate to your data). You simply cannot plot it as the plot you posted. All probablilty distributions to the best of my knowledge have a maximum amplitude of 1. What was plotted in the figure is an exponential function, not an exponential distribution.

The plot of the exponential distribution of your data are:

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];
ex=expfit(n);
E=exppdf(v,ex);
hold on
plot(E,'-r')

[Amr Hashem]

Thanks , for your advice

that's work:

 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(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')

Now I have a question how to prolong the fitted curve to reach the x-axis?

[Star Strider]

My pleasure.

That requires an fzero call to find it, and to create an appropriate vector for ‘v’ that includes it.

The code changes slightly:

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) + 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
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 With ‘vrng’
hold on
plot(vrng,E,'-r')
fprintf(1, '\n\tE(%.3f) = 0\n', Y0)

The x-intercept is about 0.432.

[Amr Hashem]

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]

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.