Understanding an empirical distribution defined by a Dirac measure There is a definition in this paper (equation 2.3) which I cannot seem to grasp.
I'm writing it here in a simpler form:
Let $\delta_a$ be the Dirac measure which assigns probability 1 to the point $a$. Denote measure $\mu \in \mathcal{P}_{ac}(R^d)$ where $\mathcal{P}_{ac}(R^d)$ is the family of all Lebesgue absolutely continuous probability measures on $R^d$. Let $\boldsymbol{X}$ be a random vector, and its empirical observations are $\boldsymbol{X}_1, \boldsymbol{X}_2, ..., \boldsymbol{X}_n\  \stackrel{i.i.d.}{\sim}\ \mu \in \mathcal{P}_{ac}(R^d)$.
Now, in the paper they define the empirical distribution of $\boldsymbol{X}$ as: $$\mu_n^{\boldsymbol{X}} := \frac{1}{n} \sum_{i=1}^n \delta_{\boldsymbol{X}_i},$$
and this is the part I don't get. How does the above sum represent an empirical distribution? is it actually like saying the observations are uniformly distributed because for each observation the value of the above sum will be $1/n$?


is it actually like saying the observations are uniformly distributed because for each observation the value of the above sum will be $1/$?

Not really. The distribution would be uniform if the observed values would be uniformly distributed. The empirical distribution would depend on observed data, so if you observed the values $(1, 1, 3)$, the probability for $1$ would be $2/3$ rather than $1/3$ since there’s sum involved.
