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For a pair {X, Y} having joint distribution on the plane, the approach is analogous to the single variable case. To find the probability an absolutely continuous pair takes on values in a set Q on the plane, integrate the joint density over the set. In the discrete case, identify those pairs of values which meet the defining conditions for Q and add the associated probabilities. To find the probability that g(X, Y ) takes on a a value in set M, determine the set Q of those pairs (t, u) mapped into M by the function g and then determine, as in the previous case, the probability the pair {X, Y} takes on values in Q.

Introduction

The general mapping approach for a single random variable and the discrete alternative extends to functions of more than one variable. It is convenient to consider thecase of two random variables, considered jointly. Extensions to more than two random variables are made similarly, although the details are more complicated.

The general approach extended to a pair

Consider a pair { X , Y } having joint distribution on the plane. The approach is analogous to that for a single random variable with distribution on the line.

  1. To find P ( ( X , Y ) Q ) .
    1. Mapping approach . Simply find the amount of probability mass mapped into the set Q on the plane by the random vector W = ( X , Y ) .
      • In the absolutely continuous case, calculate Q f X Y .
      • In the discrete case, identify those vector values ( t i , u j ) of ( X , Y ) which are in the set Q and add the associated probabilities.
    2. Discrete alternative . Consider each vector value ( t i , u j ) of ( X , Y ) . Select those which meet the defining conditions for Q and add the associated probabilities. This is the approach we use in the MATLAB calculations. It does not require that we describe geometricallythe region Q .
  2. To find P ( g ( X , Y ) M ) . g is real valued and M is a subset the real line.
    1. Mapping approach . Determine the set Q of all those ( t , u ) which are mapped into M by the function g . Now
      W ( ω ) = ( X ( ω ) , Y ( ω ) ) Q iff g ( ( X ( ω ) , Y ( ω ) ) M Hence
      { ω : g ( X ( ω ) , Y ( ω ) ) M } = { ω : ( X ( ω ) , Y ( ω ) ) Q }
      Since these are the same event, they must have the same probability. Once Q is identified on the plane,determine P ( ( X , Y ) Q ) in the usual manner (see part a, above).
    2. Discrete alternative . For each possible vector value ( t i , u j ) of ( X , Y ) , determine whether g ( t i , u j ) meets the defining condition for M . Select those ( t i , u j ) which do and add the associated probabilities.

We illustrate the mapping approach in the absolutely continuous case. A key element in the approach is finding the set Q on the plane such that g ( X , Y ) M iff ( X , Y ) Q . The desired probability is obtained by integrating f X Y over Q .

A graph with a Diagonal line from points (0,0) to (2,2). Another line extends from (2,0) to (2,2) creating a triangle with the diagonal line. This line is labeled t=2 and this triangle is shaded. There is one more line originating at (0,1) and ending at (1,1). This line is labeled u=1. The diagonal line is labeled u=t. Below this graph is the equation f_XY(t,u)=(6/37)(t+2u). A graph with a Diagonal line from points (0,0) to (2,2). Another line extends from (2,0) to (2,2) creating a triangle with the diagonal line. This line is labeled t=2 and this triangle is shaded. There is one more line originating at (0,1) and ending at (1,1). This line is labeled u=1. The diagonal line is labeled u=t. Below this graph is the equation f_XY(t,u)=(6/37)(t+2u).
Distribution for [link] .

A numerical example

The pair { X , Y } has joint density f X Y ( t , u ) = 6 37 ( t + 2 u ) on the region bounded by t = 0 , t = 2 , u = 0 , u = max { 1 , t } (see Figure 1). Determine P ( Y X ) = P ( X - Y 0 ) . Here g ( t , u ) = t - u and M = [ 0 , ) . Now Q = { ( t , u ) : t - u 0 } = { ( t , u ) : u t } which is the region on the plane on or below the line u = t . Examination of the figure shows that for this region, f X Y is different from zero on the triangle bounded by t = 2 , u = 0 , and u = t . The desired probability is

P ( Y X ) = 0 2 0 t 6 37 ( t + 2 u ) d u d t = 32 / 37 0 . 8649
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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