Suppose we have a set of experimental data $\{(x_i, Y_i, S_i)\}_{i=1}^N$ where the $x_i$'s are our measurement points, the $Y_i$'s are the mean value of the response $y$ over $m$ experiments at $x_i$, and $S_i$ are the standard errors at those $x_i$'s. In other words, each $Y_i$ is the average of $m$ experiments to measure $y$ at $x_i$.

We wish to do a nonlinear fit under the model $y = f(x;\theta)$ for some parameters $\theta$.

I want to estimate $\theta$ via maximum likelihood. I need a likelihood, not just an estimator for $\theta$. Just doing nonlinear least squares is not what I am looking for.

Yes, I know I can find $\hat \theta = \text{arg min}_\theta \sum_{i=1}^N \frac{(Y_i - f(x_i;\theta))^2}{S_i^2}$.

My questions are quite simple.

  1. What is the proper likelihood function $\mathcal{L}$ to maximize? Assuming a normal distribution for the error in measuring each $y_i$, we could write

$\mathcal{L}(\theta) = \prod_{i=1}^N \frac{1}{\sqrt{2 \pi} S_i} \text{e}^{-(Y_i - f(x_i;\theta))^2/(2 S_i^2)}$.

But is $S_i$, the standard error, correct here? Or should we be using the standard deviations, i.e., replacing $S_i$ by $\sqrt{m} S_i$ in the equation above?

For the nonlinear least squares fit, this makes no difference. But for getting an actual likelihood, this is important.

  1. Do the $m$ experiments done at each $x_i$ have any effect upon the resultant standard error we obtain for $\hat \theta$? Is it still correct to write:

$\delta \hat \theta = \sqrt{ \frac{-\mathcal{H}}{N}}$

where $\mathcal{H}$ is the Hessian of $\mathcal{L}$ at the optimal $\theta$?


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