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We formalize as follows:
A pair $\{X,\phantom{\rule{0.166667em}{0ex}}Y\}$ of random variables considered jointly is treated as the pair of coordinate functions for a two-dimensional random vector $W=(X,\phantom{\rule{0.166667em}{0ex}}Y)$ . To each $\omega \in \Omega $ , W assigns the pair of real numbers $(t,\phantom{\rule{0.166667em}{0ex}}u)$ , where $X\left(\omega \right)=t$ and $Y\left(\omega \right)=u$ . If we represent the pair of values $\{t,\phantom{\rule{0.166667em}{0ex}}u\}$ as the point $(t,\phantom{\rule{0.166667em}{0ex}}u)$ on the plane, then $W\left(\omega \right)=(t,\phantom{\rule{0.166667em}{0ex}}u)$ , so that
is a mapping from the basic space Ω to the plane R ^{2} . Since W is a function, all mapping ideas extend. The inverse mapping ${W}^{-1}$ plays a role analogous to that of the inverse mapping ${X}^{-1}$ for a real random variable. A two-dimensional vector W is a random vector iff ${W}^{-1}\left(Q\right)$ is an event for each reasonable set (technically, each Borel set) on the plane.
A fundamental result from measure theory ensures
$W=(X,\phantom{\rule{0.166667em}{0ex}}Y)$ is a random vector iff each of the coordinate functions X and Y is a random variable.
In the selection example above, we model $X\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}$ (the number of juniors selected) and Y (the number of seniors selected) as random variables. Hence the vector-valued function
In a manner parallel to that for the single-variable case, we obtain a mapping of probability mass from the basic space to the plane. Since ${W}^{-1}\left(Q\right)$ is an event for each reasonable set Q on the plane, we may assign to Q the probability mass
Because of the preservation of set operations by inverse mappings as in the single-variable case, the mass assignment determines ${P}_{XY}$ as a probability measure on the subsets of the plane R ^{2} . The argument parallels that for the single-variable case. The result is the probability distribution induced by $W=(X,\phantom{\rule{0.166667em}{0ex}}Y)$ . To determine the probability that the vector-valued function $W=(X,\phantom{\rule{0.166667em}{0ex}}Y)$ takes on a (vector) value in region Q , we simply determine how much induced probability mass is in that region.
To determine $P(1\le X\le 3,\phantom{\rule{0.277778em}{0ex}}Y>0)$ , we determine the region for which the first coordinate value (which we call t ) is between one and three and the second coordinate value (which we call u ) is greater than zero. This corresponds to the set Q of points on the plane with $1\le t\le 3$ and $u>0$ . Gometrically, this is the strip on the plane bounded by (but not including) the horizontal axis and by thevertical lines $t=1$ and $t=3$ (included). The problem is to determine how much probability mass lies in that strip. How this is acheived depends upon the nature of thedistribution and how it is described.
As in the single-variable case, we have a distribution function.
Definition
The joint distribution function ${F}_{XY}$ for $W=(X,\phantom{\rule{0.166667em}{0ex}}Y)$ is given by
This means that ${F}_{XY}(t,\phantom{\rule{0.166667em}{0ex}}u)$ is equal to the probability mass in the region ${Q}_{tu}$ on the plane such that the first coordinate is less than or equal to t and the second coordinate is less than or equal to u . Formally, we may write
Now for a given point $(a,b)$ , the region ${Q}_{ab}$ is the set of points $(t,u)$ on the plane which are on or to the left of the vertical line through $(t,0)$ and on or below the horizontal line through $(0,u)$ (see Figure 1 for specific point $t=a,u=b$ ). We refer to such regions as semiinfinite intervals on the plane.
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