Maximum-likelihood estimator for data points with errors

Question

Suppose there are N measurements of a random variable x which has Gaussian p.d.f. with unknown mean $\mu$ and variance $\sigma^2$. Classical textbook solution for estimation $\mu$ and $\sigma$ is to maximize likelihood function:

\begin{equation} L(\mu, \sigma) = \Pi_{i=1}^N\frac{1}{\sqrt{2\pi\sigma^2}}\,e^{-\frac{(x_i - \mu)^2}{2\sigma^2}} \end{equation}

This method is explicitly based on the assumption that all data points $x_i$ are known without error.

However I want to construct ML estimator for $\mu$ and $\sigma$ in case data points are measured with errors $x_i \pm \sigma_i$ and errors are known. Measurements are independent and each data point has it's own error $\sigma_i$. Assume also that errors are not correlated and they are known to be Gaussian.

I guess the likelihood function in this case will be the product of convolutions of the p.d.f. I'm trying to estimate with Gaussian distribution of individual measurement:

\begin{equation} L(\mu, \sigma) = \Pi_{i=1}^N\,\, \int \mathrm{d}x \frac{1}{\sqrt{2\pi\sigma_i^2}}\,e^{-\frac{(x_i - x)^2}{2\sigma_i^2}} \,\, \frac{1}{\sqrt{2\pi\sigma^2}}\,e^{-\frac{(x - \mu)^2}{2\sigma^2}} \end{equation}

This expression is based on my intuition, but it has two desired properties:

1. When errors tends to zero, exponent turns into delta function and expression reduces to standard MLE
2. When errors are huge, the first exponent can be considered as constant and integral goes only over second exponent. So the integral equals to constant and as expected we lose any sensitivity to parameters.

Summing up I have the following questions:

1. Is the above expression right? If not, what is the proper way to incorporate data errors into MLE estimation of $\mu$ and $\sigma$?
2. Can you provide refs to literature where I can read about this?

Answer 1

As you have pointed out correctly, for data without errors Likelihood is $$ L(\theta) = \Pi_{i=1}^{N} f(x_i|\theta) = \Pi_{i=1}^{N}\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - \mu)^2}{\sigma^2}}$$

and this leads to the MLE estimate as $\hat{\mu}_{MLE} = \frac{\sum_{i=1}^{N}x_i}{N}$. In case each data point is corrupted with an error $\sigma_i$, we can modify the Likelihood as follows:

$$ L(\theta) = \Pi_{i=1}^{N} f(x_i|\theta) = \Pi_{i=1}^{N}\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i \pm \sigma_i - \mu)^2}{\sigma^2}} = \Pi_{i=1}^{N}\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(y_i - \mu)^2}{\sigma^2}}$$

, where $y_i = x_i \pm \sigma_i $.

Using the formula above, $$\hat{\mu}_{MLE,err} = \frac{\sum_{i=1}^{N}y_i}{N} = \frac{\sum_{i=1}^{N}x_i \pm \sigma_{i}}{N} = \hat{\mu}_{MLE} + \frac{\sum_{i=1}^{N} \pm \sigma_{i}}{N}$$.

If we know the errors exactly, or the error generating process, one can estimate the 2nd term.