Normally distributed data?

[James.L]

Homework Statement

Suppose I have a set of measurements of a quantity Q, where the resolution R is the same for all data d_i. However, R is unknown, and I wish to find it.

In my book they do this by writing the likelihood L (or rather, ln(L)) as the Gaussian, so

$$\ln L = \sum\limits_i { - \ln R_i \sqrt {2\pi } } - \sum\limits_i {\frac{{\left( {d_i - Q } \right)^2 }}{{2R_i }}}$$

Now they differentiate wrt. the mean Q and the deviation R_i = R, yielding two equations. Solving these yields

$$R^2 = \frac{1}{N}\sum\limits_i {\left( {d_i - Q} \right)^2 }$$

This is the standard result we are "used" to. But does this mean that every single time I use this formula on a set of data, then I am implicitly assuming that the data is normally distributed?

Cheers.

 

[LightHero]

Homework Equations \ln L = \sum\limits_i { - \ln R_i \sqrt {2\pi } } - \sum\limits_i {\frac{{\left( {d_i - Q } \right)^2 }}{{2R_i }}} R^2 = \frac{1}{N}\sum\limits_i {\left( {d_i - Q} \right)^2 } The Attempt at a Solution No, you are not implicitly assuming that the data is normally distributed. The formula you are using is simply the maximum likelihood estimator for the resolution R, which is a measure of the uncertainty in the data. Therefore, the formula is only applicable if the data is actually normally distributed, since it assumes that the likelihood is a Gaussian. However, the formula itself does not imply that the data must be normally distributed - it simply provides an estimator for the resolution, assuming that the data is normally distributed.