how to plot exponential pdf over a distributed data ?

Question

Amr Hashem 2015-10-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/250745-how-to-plot-exponential-pdf-over-a-distributed-data

评论： Star Strider 2015-10-26

采纳的回答： Star Strider

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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?

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Amr Hashem 2015-10-24

why there aren't any interaction?

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Answer 1

Star Strider 2015-10-24

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https://ww2.mathworks.cn/matlabcentral/answers/250745-how-to-plot-exponential-pdf-over-a-distributed-data#answer_197224

在 MATLAB Online 中打开

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.

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Amr Hashem 2015-10-26

编辑：Amr Hashem 2015-10-26

在 MATLAB Online 中打开

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 2015-10-26

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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.

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