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Say I have data as $x_1,x_2,..x_n$ and I am fitting a model to these data for example, $Y(t)=Y_0 e^{-\lambda t}$. I am trying to estimate the parameter $\lambda$ by fitting to these data.
In these estimation procedures I don't quite understand the meaning of using the logarithm of the data for parameter estimation?

Is there an advantage in using log data when the data vary over a large range, for example from $10^7$ to $1$

  1. When using data without transformation,

     model = @(x,t) data(1)*exp(x(1).*t);
    
    [lambda,error]=lsqcurvefit(model,initial,t,data);
    
  2. If using log data I will call the function as,

     model = @(x,t) log10(data(1))*exp(x(1).*t);
    
    [lambda,error]=lsqcurvefit(model,initial,t,log10(data));  
    

So, the model itself does not change and only the logarithm of the data is taken.

Is there a relationship between the $\lambda$ value that I obtain from the above 2 methods?

Also, which one is actually the correct $\lambda$ value?

When using the log data, should the entire model also be transformed, so that I am fitting to $log(Y)=log(Y_0)-\lambda t$ or can I simply use log of data while the model remains unchanged?

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The model $$Y=Y_0 e^{-\lambda t}$$ is nonlinear but, taking logarithms, $$z=\log(Y)=\log(Y_0)-\lambda t=\alpha+\beta t$$ is linear. So, a first linear regression based on the transformed equation gives $(\alpha,\beta)$. So, you have the estimates $Y_0=e^\alpha$ and $\lambda=-\beta$ and now you can start the nonlinear regression for the true model.

Do not skip the second step since what is measured is $Y$ and not $\log(Y)$.

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  • $\begingroup$ Thank you for the answer. What is the meaning of "now you can start the nonlinear regression for the true model"? Isn't finding $\alpha$ and $\beta$ enough?So, is it wrong to use $log(x_1,x_2,...x_n)$ to estimate the parameters of $Y=Y_0 e^{-\lambda t}$?Is the true parameter value of $\lambda$ obtained when the data are used without transformation ? $\endgroup$ – sam_rox Apr 27 at 13:47
  • $\begingroup$ @sam_rox. Taking the logarithm of $Y$ makes the model linearized $z=\alpha+\beta t$; then this gives estimates of $Y_0$ and $\lambda$. With these, you start the nonlinear regression to get the true values of $Y_0$ and $\lambda$ since what is measured is $Y$ and not $\log(Y)$. $\endgroup$ – Claude Leibovici Apr 27 at 13:51
  • $\begingroup$ Can't I use a non-linear least squares optimiser and straight away fit $Y=Y_0 e^{-\lambda t}$ to data without transforming and estimate the parameter? What is the advantage of using the log data $\endgroup$ – sam_rox Apr 28 at 0:31
  • $\begingroup$ @sam_rox. For sure, you can since the problem is very simple. The advantage of using the log data is that it immediately gives very good estimates. Just make the problem a bit more difficult as $Y=Y_0 e^{-\lamda t}+k$; good values for the guesses could be crucial. $\endgroup$ – Claude Leibovici Apr 28 at 2:24

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