# Fitting multiple functions with shared parameters to data

consider a set of real-valued continuous functions $$f_i(t;a)$$ for $1\leq i\leq n$ with some shared non-negative parameters $a=(a_1,\ldots,a_d)\in\mathbb{R}^d$. Furthermore, assume that there exists for each $1\leq i\leq n$ a set of non-negative data $$y^{(i)}:=\{y_1^{(i)},\ldots,y_m^{(i)}\}\in\mathbb{R}^d$$ Now, I want to estimate the parameters $a$. Usually, it should suffice to numerically compute $$\min_{a>0}\sum_{i=1}^n\sum_{j=1}^m\left(y^{(i)}_j-f_i(t_j;a)\right)^2$$ where the $t_j$ in $f_i(t_j;a)$ corresponds to the time point, at which $y_j^{(i)}$ was recorded.

The problem is, that for different $i$ the data ranges in which all elements of $y^{(i)}$ lie, vary in orders of magnitude, e.g. all elements of $y^{(1)}$ are in the interval $[0,50]$, while the elements of $y^{(2)}$ are in the interval $[0,1500]$. This causes that $f_2$ shows a very good fit to $y^{(2)}$, while data with a much smaller data range is fit very poorly.

My first idea was to consider the relative distance between data and function at each point and fit the data with respect to

$$\min_{a>0}\sum_{i=1}^n\sum_{j=1}^m\frac{\left(y^{(i)}_j-f_i(t_j;a)\right)^2}{y^{(i)}_j+1}$$ where the $+1$ causes the denominator to be non-zero. But since a lot of data points $y^{(i)}_j$ are zero, there is a big penalty for the distance between function and data point in this case, but only a small if $y^{(i)}_j$ is large.

Does anybody have a good suggestion for this issue or maybe can refer to literature on this problem?

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Do you know about weighted least squares? –  joriki Oct 12 '12 at 11:36
@FrederikZiebell What's the structure of $f$ (i.e. is it linear in some arguments? bounded maybe?), and what kinds of errors do you expect in the $y^{(i)}_j$? If you say most of the $y^{(i)}_j$ are zero, do you mean exactly zero or approximately zero? Without knowing these things, it's hard to come up with a good way to estimate $a$. –  fgp Oct 12 '12 at 13:03
@fgp Thanks for your reply. All $f_i$ are bounded and non-linear in their argument. Perhaps I should mention, that I know the errors of all the $y_j^{(i)}$ and yes I mean exactly zero (with error zero). Maybe it is worth mentioning that if $y_j^{(i)}$ is zero, then so are all $y_k^{(i)}$ for all $k>j$ and the corresponding $f_i$ is a function of exponential decay. –  Frederik Ziebell Oct 12 '12 at 17:27
@FrederikZiebell Isn exactly zero and exponentially decaying a contradiction? –  fgp Oct 12 '12 at 17:35
@fgp: Yes, but the size $n$ of the sample behind each $y_k^{(i)}$ is with 5 very small. So it is safe to assume that exponential decay is a good approximation of reality in this case. –  Frederik Ziebell Oct 13 '12 at 7:50