What does "over the unit square" mean?
So i was told that a joint distribution of two volumes $X$ and $Y$ (both ranging from 0 to 1) is $f(x)=c(x+y^2)$ over the unit square, 0 otherwise.
My question might be trivial, but I've never seen this expression before, what does "over the unit square" mean?
$\endgroup$ 22 Answers
$\begingroup$It means the joint distribution is nonzero only at points in the square with all side-lengths 1 lying in the first quadrant with one corner at the origin and sides lying along the positive $x$- and $y$-axes.
In other words, it consists of the points with $x$ and $y$ coordinates both between $0$ and $1$.
(Note: "unit" refers to the number $1$. A unit square has side-length 1, and the unit square is the unique such square contained in the first quadrant with one corner at the origin.)
$\endgroup$ $\begingroup$"Unit" means one. "Unit square" is a square with length 1 on each side (restating what you already said: $X$ and $Y$ range from 0 to 1). "Unit circle" is a circle with radius 1. And so forth.
Edit: The words "over" or "on" can be used to refer to the domain of a particular function. This presents an interesting case, because while many of us are familiar with this usage, going through a half-dozen books on my shelf, I can't find any that formally declare/define that usage. As one example, in the Wikipedia article on functions, the word "over" gets used in the middle of an example without prior usage:
The unique function over a set $X$ that maps each element to itself is called the identity function for $X$.
Likewise for the article on domains which starts using the word "on" in the middle of an example:
For example, the function $f$ defined by $f(x)=1/x$ has no value for $f(0)$. Thus, the set of all real numbers, $\mathbb R$, cannot be its domain. In cases like this, the function is either defined on $\mathbb R$\{0} or the "gap is plugged" by explicitly defining $f(0)$.
Another term that might have been used for the question here is support, which is the part of the domain that produces nonzero values for the function.
$\endgroup$ 5
