# Understanding Empirical Data Distribution

I've been trying to understand this paper and am having trouble understanding this part:

"We can approximate $$p(x,y)=p(x)p(y|x)$$ using the empirical data distribution $$p(x,y) =\frac{1}{N} \sum_{n=1}^N \delta_{x_n}(x) \delta_{y_n}(y)"$$

In another part of the paper they say $$p(y|y_n) =\delta_{y_n}(y)$$.

I have some background in probability but none in statistics; I was able to figure out what an Empirical CDF is, but not a pdf like here, so I'm not sure exactly what the authors are doing. Does the $$\delta$$ refer to the Dirac delta distribution?

The empirical data distribution is a probability distribution which allocates probability $$1/N$$ to point in the training dataset and 0 otherwise. More formally, it is supported on $$N$$ points $$(x_i,y_i)$$ of the training set each having probability mass $1/N$$and so all other points have mass 0. Yes, $$\delta_{x_n}(x)$$ and $$\delta_{y_n}(y)$$ are indicator functions which are $$1$$ when $$x=x_n$$ and $$0$$ otherwise; similarly for $$\delta_{y_n}(y)$$. $$P(x,y)$$ is the joint probability mass and you can check that it is 0 for any point not in the training set and $$1/N$$ for a point in the training set. • So$\delta_{x_n}(x)$=$\delta(x-x_n)$right? But if$\delta$is the dirac delta function, then wouldn't it be infinity when$x = x_n$? – Adam Commented Jun 8, 2019 at 15:32 • No it is the indicator function. It is equal to 1 when$x=x_n\$, otherwise 0. Similarly for y Commented Jun 9, 2019 at 17:55