# Density of uniform distribution over two disjoint squares

A probability distribution $$P$$ over $$X \times \{0, 1\}$$. $$P$$ can be defined in term of its marginal distribution over $$X$$ , which we will denote by $$P_X$$ and the conditional labeling distribution, which is defined by the regression function $$\mu(x) = P_{ (x,y) \sim P} [y = 1 \mid x]$$ Consider a 2-dimensional Euclidean domain, that is $$X = \mathbb R^2$$, and the following process of data generation: The marginal distribution over $$X$$ is uniform over two square areas $$[1, 2] \times [1, 2] \cup [3, 4] \times [1.5, 2.5]$$. Points in the first square $$Q_1 = [1, 2] \times [1, 2]$$ are labeled 0 (blue) and points in the second square $$Q_2 = [3, 4] \times [1.5, 2.5]$$ are labeled 1 (red). Describe the density function of $$P_X$$, and the regression function, Bayes predictor and Bayes risk of $$P$$.

In the image, I have defined the Probability density function. I am having trouble in figuring out the pdf of this function in 2 dimensional space.

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Oct 9, 2020 at 4:04
The density $$p_R$$ of the uniform distribution over a rectangle $$R$$ is given by $$p_R(x) = 0$$ if $$x \notin R$$, and $$p_R(x) = 1/\mathit{area}(R)$$ otherwise. Indeed, up to scaling the distribution must have this form, and the choice $$1/\mathit{area}(R)$$ ensures that $$p_R$$ is a distribution.
In your case, you have a mixture of two such distributions: with probability $$1/2$$ you take a uniform point in $$[1,2] \times [1,2]$$, and with probability $$1/2$$ you take a uniform point in $$[3,4] \times [1.5,2.5]$$. The density function of the mixture is the average of the density functions of the individual distributions.