choose a point with uniform distribution in a triangle $(0,1),(0,0),(1,0)$

Question

We choose a point with uniform distribution in a triangle $(0,1),(0,0),(1,0)$. Calculate $Cov(X,Y)$. In the solution they stated that: $$ f_{X,Y}(x,y)=\frac{1}{0.5}I_{\{0<y<1,y<1-x,0<x<1\}} $$ I'm a bit confused about $I$. Is it an indicator? What does it represents here? why ${\{0<y<1,y<1-x,0<x<1\}}$ and which formula did they use to calculate it? I guess it's the base.

Answer 1

Yes, that notation is an indicator function. Sometimes written with $I$, or my preference, $\mathbb 1$, it is simply $$\mathbb 1_{(x,y) \in S} = \begin{cases}1, & (x,y) \in S \\ 0, & (x,y) \not\in S \end{cases}$$ where $S$ is some set. Then, we can characterize the points inside this triangle with vertices $(0,1), (0,0), (1,0)$ as those points $(x,y)$ satisfying the simultaneous inequalities $$0 < x < 1 \\ 0 < y < 1 \\ x + y < 1.$$ In fact, this specification is a bit redundant: the edge joining $(0,0)$ and $(1,0)$ is attained with only the inequality $y > 0$; the edge joining $(0,0)$ and $(0,1)$ is attained with $x > 0$; and the last edge joining $(0,1)$ and $(1,0)$ is given by $x + y < 1$. No other inequalities are needed.

We can then use this to write the indicator in other ways; e.g., we could write it as $$\mathbb 1_{\{x > 0\}} \mathbb 1_{\{y > 0\}} \mathbb 1_{\{x+y < 1\}},$$ for example.

Note that the value of the density $f(x,y)$ in this triangular region must be $2$ because the area of the triangle is $1/2$.

As for the covariance, it is not difficult to compute $\operatorname{E}[XY]$, $\operatorname{E}[X]$, and $\operatorname{E}[Y]$. The latter two must be equal due to symmetry of $X$ and $Y$ on the joint support.