# Maximum Likelihood with Experimental Data: Standard Errors and Standard Deviations

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 $$\theta$$?