I have an independent variable $\delta$ and dependent set of statistical classified data $\widehat{\Lambda }(\delta)$. The model I want to fit to this data is: $$ \Lambda (\delta )=Ae^{-k\delta}*f(k) $$ There are many cases but for this example we assume: $$ f(k)=\frac{2}{k^2\sigma ^2}\big(e^{\frac{k^2\sigma ^2}{2}\Delta T}-1\big) $$ Note:$\Delta T$ and $\sigma$ are known variables.
In the white paper I've got this equation there was this equation given for the sum of the squared errors, that needs to be minimized: $$ r(A,k)=\sum_{\delta=1}^{\delta_{max}}\bigg(\log\big(\widehat{\Lambda}(\delta)\big)+k\delta-\log\big(A\big)-\log\big(f(k)\big)\bigg)^2 $$ Now since $f(k)$ is not dependent on $\delta$ I would substitute: $$ u=A*f(k) $$ Which gives you this equation as the model: $$ \Lambda(\delta)=ue^{-k\delta} $$ The function of the squared errors that needs to be minimized for this quasilinear model is : $$ r(A,k)=\sum_{\delta=1}^{\delta_{max}}\bigg(\log\big(\widehat{\Lambda}(\delta)\big)+k\delta-\log\big(u\big)\bigg)^2 $$ Now we have a linear equation we need to do our regression for, which is way easier than any non-linear approach. And with $k$ and $\log(u)$ in hand we can easily solve for $A$
I did some back tests with scipy's optimization functions and I got the same results when I optimized the parameters $A$ and $k$ with the initial model and the Levenberg-Marquardt optimization function, and when I tried fit $\log(u)$ and $k$ with the linear least squares fitting approach by hand and then solved for $A$.
But I want to ask you, the experts, if this is really mathematically justified.
Many thanks in advance.