Skip to main content

Chapter 8. Random Vectors and joint Distributions

8.1. Random Vectors and Joint Distributions*

Introduction

A single, real-valued random variable is a function (mapping) from the basic space Ω to the real line. That is, to each possible outcome ω of an experiment there corresponds a real value t=X(ω). The mapping induces a probability mass distribution on the real line, which provides a means of making probability calculations. The distribution is described by a distribution function FX. In the absolutely continuous case, with no point mass concentrations, the distribution may also be described by a probability density function fX. The probability density is the linear density of the probability mass along the real line (i.e., mass per unit length). The density is thus the derivative of the distribution function. For a simple random variable, the probability distribution consists of a point mass pi at each possible value ti of the random variable. Various m-procedures and m-functions aid calculations for simple distributions. In the absolutely continuous case, a simple approximation may be set up, so that calculations for the random variable are approximated by calculations on this simple distribution.

Often we have more than one random variable. Each can be considered separately, but usually they have some probabilistic ties which must be taken into account when they are considered jointly. We treat the joint case by considering the individual random variables as coordinates of a random vector. We extend the techniques for a single random variable to the multidimensional case. To simplify exposition and to keep calculations manageable, we consider a pair of random variables as coordinates of a two-dimensional random vector. The concepts and results extend directly to any finite number of random variables considered jointly.

Random variables considered jointly; random vectors

As a starting point, consider a simple example in which the probabilistic interaction between two random quantities is evident.

Example 8.1A selection problem

Two campus jobs are open. Two juniors and three seniors apply. They seem equally qualified, so it is decided to select them by chance. Each combination of two is equally likely. Let X be the number of juniors selected (possible values 0, 1, 2) and Y be the number of seniors selected (possible values 0, 1, 2). However there are only three possible pairs of values for , or . Others have zero probability, since they are impossible. Determine the probability for each of the possible pairs.

SOLUTION

There are equally likely pairs. Only one pair can be both juniors. Six pairs can be one of each. There are ways to select pairs of seniors. Thus

(8.1)

These probabilities add to one, as they must, since this exhausts the mutually exclusive possibilities. The probability of any other combination must be zero. We also have the distributions for the random variables conisidered individually.

(8.2)

We thus have a joint distribution and two individual or marginal distributions.

We formalize as follows:

A pair of random variables considered jointly is treated as the pair of coordinate functions for a two-dimensional random vector . To each ωΩ, W assigns the pair of real numbers , where X(ω)=t and Y(ω)=u. If we represent the pair of values as the point on the plane, then , so that

(8.3)

is a mapping from the basic space Ω to the plane R2. 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(Q) is an event for each reasonable set (technically, each Borel set) on the plane.

A fundamental result from measure theory ensures

is a random vector iff each of the coordinate functions X and Y is a random variable.

In the selection example above, we model (the number of juniors selected)   and Y (the number of seniors selected) as random variables. Hence the vector-valued function

Induced distribution and the joint distribution 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(Q) is an event for each reasonable set Q on the plane, we may assign to Q the probability mass

(8.4)

Because of the preservation of set operations by inverse mappings as in the single-variable case, the mass assignment determines PXY as a probability measure on the subsets of the plane R2. The argument parallels that for the single-variable case. The result is the probability distribution induced by . To determine the probability that the vector-valued function takes on a (vector) value in region Q, we simply determine how much induced probability mass is in that region.

Example 8.2Induced distribution and probability calculations

To determine , 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≤t≤3 and u>0. Gometrically, this is the strip on the plane bounded by (but not including) the horizontal axis and by the vertical 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 the distribution and how it is described.

As in the single-variable case, we have a distribution function.

Definition

The joint distribution function FXY for is given by

(8.5)

This means that is equal to the probability mass in the region Qtu 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

(8.6)

Now for a given point (a,b), the region Qab 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.

The theoretical result quoted in the real variable case extends to ensure that a distribution on the plane is determined uniquely by consistent assignments to the semiinfinite intervals Qtu. Thus, the induced distribution is determined completely by the joint distribution function.

Distribution function for a discrete random vector

The induced distribution consists of point masses. At point in the range of there is probability mass . As in the general case, to determine we determine how much probability mass is in the region. In the discrete case (or in any case where there are point mass concentrations) one must be careful to note whether or not the boundaries are included in the region, should there be mass concentrations on the boundary.

Example 8.3Distribution function for the selection problem in Example 8.1

The probability distribution is quite simple. Mass 3/10 at (0,2), 6/10 at (1,1), and 1/10 at (2,0). This distribution is plotted in Figure 8.2. To determine (and visualize) the joint distribution function, think of moving the point on the plane. The region Qtu is a giant “sheet” with corner at . The value of is the amount of probability covered by the sheet. This value is constant over any grid cell, including the left-hand and lower boundariies, and is the value taken on at the lower left-hand corner of the cell. Thus, if is in any of the three squares on the lower left hand part of the diagram, no probability mass is covered by the sheet with corner in the cell. If is on or in the square having probability 6/10 at the lower left-hand corner, then the sheet covers that probability, and the value of . The situation in the other cells may be checked out by this procedure.

Distribution function for a mixed distribution

Example 8.4A mixed distribution

The pair produces a mixed distribution as follows (see Figure 8.3)

Point masses 1/10 at points (0,0), (1,0), (1,1), (0,1)

Mass 6/10 spread uniformly over the unit square with these vertices

The joint distribution function is zero in the second, third, and fourth quadrants.

• If the point is in the square or on the left and lower boundaries, the sheet covers the point mass at (0,0) plus 0.6 times the area covered within the square. Thus in this region

(8.7)
• If the pont is above the square (including its upper boundary) but to the left of the line t=1, the sheet covers two point masses plus the portion of the mass in the square to the left of the vertical line through . In this case

(8.8)
• If the point is to the right of the square (including its boundary) with 0≤u<1, the sheet covers two point masses and the portion of the mass in the square below the horizontal line through , to give

(8.9)
• If is above and to the right of the square (i.e., both 1≤t and 1≤u). then all probability mass is covered and in this region.

Marginal distributions

If the joint distribution for a random vector is known, then the distribution for each of the component random variables may be determined. These are known as marginal distributions. In general, the converse is not true. However, if the component random variables form an independent pair, the treatment in that case shows that the marginals determine the joint distribution.

To begin the investigation, note that

(8.10)

Thus

(8.11)

This may be interpreted with the aid of Figure 8.4. Consider the sheet for point .

If we push the point up vertically, the upper boundary of Qtu is pushed up until eventually all probability mass on or to the left of the vertical line through is included. This is the total probability that Xt. Now FX(t) describes probability mass on the line. The probability mass described by FX(t) is the same as the total joint probability mass on or to the left of the vertical line through . We may think of the mass in the half plane being projected onto the horizontal line to give the marginal distribution for X. A parallel argument holds for the marginal for Y.

(8.12)

This mass is projected onto the vertical axis to give the marginal distribution for Y.

Marginals for a joint discrete distribution

Consider a joint simple distribution.

(8.13)

Thus, all the probability mass on the vertical line through is projected onto the point ti on a horizontal line to give . Similarly, all the probability mass on a horizontal line through is projected onto the point uj on a vertical line to give .

Example 8.5Marginals for a discrete distribution

The pair produces a joint distribution that places mass 2/10 at each of the five points

(See Figure 8.5)

The marginal distribution for X has masses 2/10, 2/10, 4/10, 2/10 at points t=0,1,2,3, respectively. Similarly, the marginal distribution for Y has masses 4/10, 4/10, 2/10 at points u=0,1,2, respectively.

Example 8.6

Consider again the joint distribution in Example 8.4. The pair produces a mixed distribution as follows:

Point masses 1/10 at points (0,0), (1,0), (1,1), (0,1)

Mass 6/10 spread uniformly over the unit square with these vertices

The construction in Figure 8.6 shows the graph of the marginal distribution function FX. There is a jump in the amount of 0.2 at t=0, corresponding to the two point masses on the vertical line. Then the mass increases linearly with t, slope 0.6, until a final jump at t=1 in the amount of 0.2 produced by the two point masses on the vertical line. At t=1, the total mass is “covered” and FX(t) is constant at one for t≥1. By symmetry, the marginal distribution for Y is the same.

8.2. Random Vectors and MATLAB*

m-procedures for a pair of simple random variables

We examine, first, calculations on a pair of simple random variables X,Y, considered jointly. These are, in effect, two components of a random vector , which maps from the basic space Ω to the plane. The induced distribution is on the -plane. Values on the horizontal axis (t-axis) correspond to values of the first coordinate random variable X and values on the vertical axis (u-axis) correspond to values of Y. We extend the computational strategy used for a single random variable.

First, let us review the one-variable strategy. In this case, data consist of values ti and corresponding probabilities arranged in matrices

(8.14)

To perform calculations on Z=g(X), we we use array operations on X to form a matrix

(8.15)

which has in a position corresponding to in matrix PX.

Basic problem. Determine P(g(X)∈M), where M is some prescribed set of values.

• Use relational operations to determine the positions for which . These will be in a zero-one matrix N, with ones in the desired positions.

• Select the in the corresponding positions and sum. This is accomplished by one of the MATLAB operations to determine the inner product of N and PX

We extend these techniques and strategies to a pair of simple random variables, considered jointly.

1. The data for a pair {X,Y} of random variables are the values of X and Y, which we may put in row matrices

(8.16)

and the joint probabilities in a matrix P. We usually represent the distribution graphically by putting probability mass at the point on the plane. This joint probability may is represented by the matrix P with elements arranged corresponding to the mass points on the plane. Thus

(8.17)
2. To perform calculations, we form computational matrices t and u such that — t has element ti at each position (i.e., at each point on the ith column from the left) — u has element uj at each position (i.e., at each point on the jth row from the bottom) MATLAB array and logical operations on t,u,P perform the specified operations on ti,uj, and at each position, in a manner analogous to the operations in the single-variable case.

3. Formation of the t and u matrices is achieved by a basic setup m-procedure called jcalc. The data for this procedure are in three matrices: is the set of values for random variable X is the set of values for random variable Y, and , where . We arrange the joint probabilities as on the plane, with X-values increasing to the right and Y-values increasing upward. This is different from the usual arrangement in a matrix, in which values of the second variable increase downward. The m-procedure takes care of this inversion. The m-procedure forms the matrices t and u, utilizing the MATLAB function meshgrid, and computes the marginal distributions for X and Y. In the following example, we display the various steps utilized in the setup procedure. Ordinarily, these intermediate steps would not be displayed.

Example 8.7Setup and basic calculations

>> jdemo4                         % Call for data in file jdemo4.m
>> jcalc                          % Call for setup procedure
Enter JOINT PROBABILITIES (as on the plane)  P
Enter row matrix of VALUES of X  X
Enter row matrix of VALUES of Y  Y
Use array operations on matrices X, Y, PX, PY, t, u, and P
>> disp(P)                        % Optional call for display of P
0.0360    0.0198    0.0297    0.0209    0.0180
0.0372    0.0558    0.0837    0.0589    0.0744
0.0516    0.0774    0.1161    0.0817    0.1032
0.0264    0.0270    0.0405    0.0285    0.0132
>> PX                             % Optional call for display of PX
PX =  0.1512    0.1800    0.2700    0.1900    0.2088
>> PY                             % Optional call for display of PY
PY =  0.1356    0.4300    0.3100    0.1244
- - - - - - - - - -               % Steps performed by jcalc
>> PX = sum(P)                    % Calculation of PX as performed by jcalc
PX =  0.1512    0.1800    0.2700    0.1900    0.2088
>> PY = fliplr(sum(P'))           % Calculation of PY (note reversal)
PY = 0.1356    0.4300    0.3100    0.1244
>> [t,u] = meshgrid(X,fliplr(Y)); % Formation of t, u matrices (note reversal)
>> disp(t)                        % Display of calculating matrix t
-3     0     1     3     5    % A row of X-values for each value of Y
-3     0     1     3     5
-3     0     1     3     5
-3     0     1     3     5
>> disp(u)                        % Display of calculating matrix u
2     2     2     2     2    % A column of Y-values (increasing
1     1     1     1     1    % upward) for each value of X
0     0     0     0     0
-2    -2    -2    -2    -2


Suppose we wish to determine the probability . Using array operations on t and u, we obtain the matrix .

>> G = t.^2 - 3*u                 % Formation of G = [g(t_i,u_j)] matrix
G  = 3    -6    -5     3    19
6    -3    -2     6    22
9     0     1     9    25
15     6     7    15    31
>> M = G >=  1                    % Positions where G >= 1
M =  1     0     0     1     1
1     0     0     1     1
1     0     1     1     1
1     1     1     1     1
>> pM = M.*P                    % Selection of probabilities
pM =
0.0360         0         0    0.0209    0.0180
0.0372         0         0    0.0589    0.0744
0.0516         0    0.1161    0.0817    0.1032
0.0264    0.0270    0.0405    0.0285    0.0132
>> PM = total(pM)               % Total of selected probabilities
PM =  0.7336                    % P(g(X,Y) >= 1)


4. In Example 3 from "Random Vectors and Joint Distributions" we note that the joint distribution function FXY is constant over any grid cell, including the left-hand and lower boundaries, at the value taken on at the lower left-hand corner of the cell. These lower left-hand corner values may be obtained systematically from the joint probability matrix P by a two step operation.

• Take cumulative sums upward of the columns of P.

• Take cumulative sums of the rows of the resultant matrix.

This can be done with the MATLAB function cumsum, which takes column cumulative sums downward. By flipping the matrix and transposing, we can achieve the desired results.

Example 8.8Calculation of FXY values for Example 3 from "Random Vectors and Joint Distributions"
>> P = 0.1*[3 0 0; 0 6 0; 0 0 1];
>> FXY = flipud(cumsum(flipud(P)))  % Cumulative column sums upward
FXY =
0.3000    0.6000    0.1000
0    0.6000    0.1000
0         0    0.1000
>> FXY = cumsum(FXY')'              % Cumulative row sums
FXY =
0.3000    0.9000    1.0000
0    0.6000    0.7000
0         0    0.1000


Comparison with Example 3 from "Random Vectors and Joint Distributions" shows agreement with values obtained by hand.
The two step procedure has been incorprated into an m-procedure jddbn. As an example, return to the distribution in Example Example 8.7

Example 8.9Joint distribution function for Example 8.7
>> jddbn
Enter joint probability matrix (as on the plane)  P
To view joint distribution function, call for FXY
>> disp(FXY)
0.1512    0.3312    0.6012    0.7912    1.0000
0.1152    0.2754    0.5157    0.6848    0.8756
0.0780    0.1824    0.3390    0.4492    0.5656
0.0264    0.0534    0.0939    0.1224    0.1356


These values may be put on a grid, in the same manner as in Figure 2 for Example 3 in "Random Vectors and Joint Distributions".

5. As in the case of canonic for a single random variable, it is often useful to have a function version of the procedure jcalc to provide the freedom to name the outputs conveniently. function[x,y,t,u,px,py,p] = jcalcf(X,Y,P) The quantities x,y,t,u,px,py, and p may be given any desired names.

Joint absolutely continuous random variables

In the single-variable case, the condition that there are no point mass concentrations on the line ensures the existence of a probability density function, useful in probability calculations. A similar situation exists for a joint distribution for two (or more) variables. For any joint mapping to the plane which assigns zero probability to each set with zero area (discrete points, line or curve segments, and countable unions of these) there is a density function.

Definition

If the joint probability distribution for the pair {X,Y} assigns zero probability to every set of points with zero area, then there exists a joint density function fXY with the property

(8.18)

We have three properties analogous to those for the single-variable case:

(8.19)

At every continuity point for fXY, the density is the second partial

(8.20)

Now

(8.21)

A similar expression holds for FY(u). Use of the fundamental theorem of calculus to obtain the derivatives gives the result

(8.22)

Marginal densities. Thus, to obtain the marginal density for the first variable, integrate out the second variable in the joint density, and similarly for the marginal for the second variable.

Example 8.10Marginal density functions

Let . This region is the triangle bounded by u=0, u=t, and t=1 (see Figure 8.8)

(8.23)
(8.24)

where Q is the common part of the triangle with the strip between t=0.5 and t=0.75 and above the line u=0.5. This is the small triangle bounded by u=0.5, u=t, and t=0.75. Thus

(8.25)
Example 8.11Marginal distribution with compound expression

The pair has joint density on the region bounded by t=0, t=2, u=0, and (see Figure 8.9). Determine the marginal density fX.

SOLUTION

Examination of the figure shows that we have different limits for the integral with respect to u for 0≤t≤1 and for 1<t≤2.

• For 0≤t≤1

(8.26)
• For 1<t≤2

(8.27)

We may combine these into a single expression in a manner used extensively in subsequent treatments. Suppose and . Then IM(t)=1 for tM (i.e., 0≤t≤1) and zero elsewhere. Likewise, IN(t)=1 for tN and zero elsewhere. We can, therefore express fX by

(8.28)

Discrete approximation in the continuous case

For a pair with joint density fXY, we approximate the distribution in a manner similar to that for a single random variable. We then utilize the techniques developed for a pair of simple random variables. If we have n approximating values ti for X and m approximating values uj for Y, we then have n·m pairs , corresponding to points on the plane. If we subdivide the horizontal axis for values of X, with constant increments dx, as in the single-variable case, and the vertical axis for values of Y, with constant increments dy, we have a grid structure consisting of rectangles of size dx·dy. We select ti and uj at the midpoint of its increment, so that the point is at the midpoint of the rectangle. If we let the approximating pair be , we assign

(8.29)

As in the one-variable case, if the increments are small enough,

(8.30)

The m-procedure tuappr calls for endpoints of intervals which include the ranges of X and Y and for the numbers of subintervals on each. It then prompts for an expression for , from which it determines the joint probability distribution. It calculates the marginal approximate distributions and sets up the calculating matrices t and u as does the m-process jcalc for simple random variables. Calculations are then carried out as for any joint simple pair.

Example 8.12Approximation to a joint continuous distribution
(8.31)

Determine .

>> tuappr
Enter matrix [a b] of X-range endpoints  [0 1]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  3*(u <= t.^2)
Use array operations on X, Y, PX, PY, t, u, and P
>> M = (t <= 0.8)&(u > 0.1);
>> p = total(M.*P)          % Evaluation of the integral with
p =   0.3355                % Maple gives 0.3352455531


The discrete approximation may be used to obtain approximate plots of marginal distribution and density functions.

Example 8.13Approximate plots of marginal density and distribution functions

on the triangle bounded by u=0, u≤1+t, and u≤1–t.

>> tuappr
Enter matrix [a b] of X-range endpoints  [-1 1]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  400
Enter number of Y approximation points  200
Enter expression for joint density  3*u.*(u<=min(1+t,1-t))
Use array operations on X, Y, PX, PY, t, u, and P
>> fx = PX/dx;                % Density for X  (see Figure 8.10)
% Theoretical (3/2)(1 - |t|)^2
>> fy = PY/dy;                % Density for Y
>> FX = cumsum(PX);           % Distribution function for X (Figure 8.10)
>> FY = cumsum(PY);           % Distribution function for Y
>> plot(X,fx,X,FX)            % Plotting details omitted


These approximation techniques useful in dealing with functions of random variables, expectations, and conditional expectation and regression.

8.3. Problems On Random Vectors and Joint Distributions*

Two cards are selected at random, without replacement, from a standard deck. Let X be the number of aces and Y be the number of spades. Under the usual assumptions, determine the joint distribution and the marginals.

Let X be the number of aces and Y be the number of spades. Define the events ASi, Ai, Si, and Ni, i=1,2, of drawing ace of spades, other ace, spade (other than the ace), and neither on the i selection. Let P(i,k)=P(X=i,Y=k).

P(2,2)=P()=0

% type npr08_01
% file npr08_01.m
% Solution for Exercise 1.
X = 0:2;
Y = 0:2;
Pn = [132  24   0; 864 144  6; 1260 216 6];
P = Pn/(52*51);
disp('Data in Pn, P, X, Y')

npr08_01         % Call for mfile
Data in Pn, P, X, Y    % Result
PX = sum(P)
PX =  0.8507    0.1448    0.0045
PY = fliplr(sum(P'))
PY =  0.5588    0.3824    0.0588


Two positions for campus jobs are open. Two sophomores, three juniors, and three seniors apply. It is decided to select two at random (each possible pair equally likely). Let X be the number of sophomores and Y be the number of juniors who are selected. Determine the joint distribution for the pair {X,Y} and from this determine the marginals for each.

Let Ai,Bi,Ci be the events of selecting a sophomore, junior, or senior, respectively, on the ith trial. Let X be the number of sophomores and Y be the number of juniors selected.

Set

% file npr08_02.m
% Solution for Exercise 2.
X = 0:2;
Y = 0:2;
Pn = [6 0 0; 18 12 0; 6 12 2];
P = Pn/56;
disp('Data are in X, Y,Pn, P')
npr08_02
Data are in X, Y,Pn, P
PX = sum(P)
PX =  0.5357    0.4286    0.0357
PY = fliplr(sum(P'))
PY =  0.3571    0.5357    0.1071


A die is rolled. Let X be the number that turns up. A coin is flipped X times. Let Y be the number of heads that turn up. Determine the joint distribution for the pair {X,Y}. Assume P(X=k)=1/6 for 1≤k≤6 and for each k, P(Y=j|X=k) has the binomial (k,1/2) distribution. Arrange the joint matrix as on the plane, with values of Y increasing upward. Determine the marginal distribution for Y. (For a MATLAB based way to determine the joint distribution see Example 7 from "Conditional Expectation, Regression")

P(X=i,Y=k)=P(X=i)P(Y=k|X=i)=(1/6)P(Y=k|X=i).

% file npr08_03.m
% Solution for Exercise 3.
X = 1:6;
Y = 0:6;
P0 = zeros(6,7);       % Initialize
for i = 1:6            % Calculate rows of Y probabilities
P0(i,1:i+1) = (1/6)*ibinom(i,1/2,0:i);
end
P = rot90(P0);         % Rotate to orient as on the plane
PY = fliplr(sum(P'));  % Reverse to put in normal order
disp('Answers are in X, Y, P, PY')
npr08_03            % Call for solution m-file
Answers are in X, Y, P, PY
disp(P)
0         0         0         0         0    0.0026
0         0         0         0    0.0052    0.0156
0         0         0    0.0104    0.0260    0.0391
0         0    0.0208    0.0417    0.0521    0.0521
0    0.0417    0.0625    0.0625    0.0521    0.0391
0.0833    0.0833    0.0625    0.0417    0.0260    0.0156
0.0833    0.0417    0.0208    0.0104    0.0052    0.0026
disp(PY)
0.1641  0.3125  0.2578  0.1667  0.0755  0.0208  0.0026


As a variation of Exercise 3., Suppose a pair of dice is rolled instead of a single die. Determine the joint distribution for the pair {X,Y} and from this determine the marginal distribution for Y.

% file npr08_04.m
% Solution for Exercise 4.
X = 2:12;
Y = 0:12;
PX = (1/36)*[1 2 3 4 5 6 5 4 3 2 1];
P0 = zeros(11,13);
for i = 1:11
P0(i,1:i+2) = PX(i)*ibinom(i+1,1/2,0:i+1);
end
P = rot90(P0);
PY = fliplr(sum(P'));
disp('Answers are in X, Y, PY, P')
npr08_04
Answers are in X, Y, PY, P
disp(P)
Columns 1 through 7
0         0         0         0         0         0         0
0         0         0         0         0         0         0
0         0         0         0         0         0         0
0         0         0         0         0         0         0
0         0         0         0         0         0    0.0005
0         0         0         0         0    0.0013    0.0043
0         0         0         0    0.0022    0.0091    0.0152
0         0         0    0.0035    0.0130    0.0273    0.0304
0         0    0.0052    0.0174    0.0326    0.0456    0.0380
0    0.0069    0.0208    0.0347    0.0434    0.0456    0.0304
0.0069    0.0208    0.0312    0.0347    0.0326    0.0273    0.0152
0.0139    0.0208    0.0208    0.0174    0.0130    0.0091    0.0043
0.0069    0.0069    0.0052    0.0035    0.0022    0.0013    0.0005
Columns 8 through 11
0         0         0    0.0000
0         0    0.0000    0.0001
0    0.0001    0.0003    0.0004
0.0002    0.0008    0.0015    0.0015
0.0020    0.0037    0.0045    0.0034
0.0078    0.0098    0.0090    0.0054
0.0182    0.0171    0.0125    0.0063
0.0273    0.0205    0.0125    0.0054
0.0273    0.0171    0.0090    0.0034
0.0182    0.0098    0.0045    0.0015
0.0078    0.0037    0.0015    0.0004
0.0020    0.0008    0.0003    0.0001
0.0002    0.0001    0.0000    0.0000
disp(PY)
Columns 1 through 7
0.0269    0.1025    0.1823    0.2158    0.1954    0.1400    0.0806
Columns 8 through 13
0.0375    0.0140    0.0040    0.0008    0.0001    0.0000


Suppose a pair of dice is rolled. Let X be the total number of spots which turn up. Roll the pair an additional X times. Let Y be the number of sevens that are thrown on the X rolls. Determine the joint distribution for the pair {X,Y} and from this determine the marginal distribution for Y. What is the probability of three or more sevens?

% file npr08_05.m
% Data and basic calculations for Exercise 5.
PX = (1/36)*[1 2 3 4 5 6 5 4 3 2 1];
X = 2:12;
Y = 0:12;
P0 = zeros(11,13);
for i = 1:11
P0(i,1:i+2) = PX(i)*ibinom(i+1,1/6,0:i+1);
end
P = rot90(P0);
PY = fliplr(sum(P'));
disp('Answers are in X, Y, P, PY')
npr08_05
Answers are in X, Y, P, PY
disp(PY)
Columns 1 through 7
0.3072    0.3660    0.2152    0.0828    0.0230    0.0048    0.0008
Columns 8 through 13
0.0001    0.0000    0.0000    0.0000    0.0000    0.0000


The pair has the joint distribution (in m-file npr08_06.m):

(8.32)
(8.33)

Determine the marginal distributions and the corner values for FXY. Determine P(X+Y>2) and P(XY).

npr08_06
Data are in X, Y, P
jcalc
Enter JOINT PROBABILITIES (as on the plane)  P
Enter row matrix of VALUES of X  X
Enter row matrix of VALUES of Y  Y
Use array operations on matrices X, Y, PX, PY, t, u, and P
disp([X;PX]')
-2.3000    0.2300
-0.7000    0.1700
1.1000    0.2000
3.9000    0.2020
5.1000    0.1980

disp([Y;PY]')
1.3000    0.2980
2.5000    0.3020
4.1000    0.1900
5.3000    0.2100
jddbn
Enter joint probability matrix (as on the plane)  P
To view joint distribution function, call for FXY
disp(FXY)
0.2300    0.4000    0.6000    0.8020    1.0000
0.1817    0.3160    0.4740    0.6361    0.7900
0.1380    0.2400    0.3600    0.4860    0.6000
0.0667    0.1160    0.1740    0.2391    0.2980
P1 = total((t+u>2).*P)
P1 =  0.7163
P2 = total((t>=u).*P)
P2 =  0.2799


The pair has the joint distribution (in m-file npr08_07.m):

(8.34)
 t = -3.1 -0.5 1.2 2.4 3.7 4.9 u = 7.5 0.009 0.0396 0.0594 0.0216 0.044 0.0203 4.1 0.0495 0 0.1089 0.0528 0.0363 0.0231 -2.0 0.0405 0.132 0.0891 0.0324 0.0297 0.0189 -3.8 0.051 0.0484 0.0726 0.0132 0 0.0077

Determine the marginal distributions and the corner values for FXY. Determine P(1≤X≤4,Y>4) and P(|XY|≤2).

npr08_07
Data are in X, Y, P
jcalc
Enter JOINT PROBABILITIES (as on the plane)  P
Enter row matrix of VALUES of X  X
Enter row matrix of VALUES of Y  Y
Use array operations on matrices X, Y, PX, PY, t, u, and P
disp([X;PX]')
-3.1000    0.1500
-0.5000    0.2200
1.2000    0.3300
2.4000    0.1200
3.7000    0.1100
4.9000    0.0700
disp([Y;PY]')
-3.8000    0.1929
-2.0000    0.3426
4.1000    0.2706
7.5000    0.1939
jddbn
Enter joint probability matrix (as on the plane)  P
To view joint distribution function, call for FXY
disp(FXY)
0.1500    0.3700    0.7000    0.8200    0.9300    1.0000
0.1410    0.3214    0.5920    0.6904    0.7564    0.8061
0.0915    0.2719    0.4336    0.4792    0.5089    0.5355
0.0510    0.0994    0.1720    0.1852    0.1852    0.1929
M = (1<=t)&(t<=4)&(u>4);
P1 = total(M.*P)
P1 =  0.3230
P2 = total((abs(t-u)<=2).*P)
P2 =  0.3357


The pair has the joint distribution (in m-file npr08_08.m):

(8.35)
 t = 1 3 5 7 9 11 13 15 17 19 u = 12 0.0156 0.0191 0.0081 0.0035 0.0091 0.007 0.0098 0.0056 0.0091 0.0049 10 0.0064 0.0204 0.0108 0.004 0.0054 0.008 0.0112 0.0064 0.0104 0.0056 9 0.0196 0.0256 0.0126 0.006 0.0156 0.012 0.0168 0.0096 0.0056 0.0084 5 0.0112 0.0182 0.0108 0.007 0.0182 0.014 0.0196 0.0012 0.0182 0.0038 3 0.006 0.026 0.0162 0.005 0.016 0.02 0.028 0.006 0.016 0.004 -1 0.0096 0.0056 0.0072 0.006 0.0256 0.012 0.0268 0.0096 0.0256 0.0084 -3 0.0044 0.0134 0.018 0.014 0.0234 0.018 0.0252 0.0244 0.0234 0.0126 -5 0.0072 0.0017 0.0063 0.0045 0.0167 0.009 0.0026 0.0172 0.0217 0.0223

Determine the marginal distributions. Determine FXY(10,6) and P(X>Y).

npr08_08
Data are in X, Y, P
jcalc
- - - - - - - - -
Use array operations on matrices X, Y, PX, PY, t, u, and P
disp([X;PX]')
1.0000    0.0800
3.0000    0.1300
5.0000    0.0900
7.0000    0.0500
9.0000    0.1300
11.0000    0.1000
13.0000    0.1400
15.0000    0.0800
17.0000    0.1300
19.0000    0.0700
disp([Y;PY]')
-5.0000    0.1092
-3.0000    0.1768
-1.0000    0.1364
3.0000    0.1432
5.0000    0.1222
9.0000    0.1318
10.0000    0.0886
12.0000    0.0918
F = total(((t<=10)&(u<=6)).*P)
F =   0.2982
P = total((t>u).*P)
P =   0.7390


Data were kept on the effect of training time on the time to perform a job on a production line. X is the amount of training, in hours, and Y is the time to perform the task, in minutes. The data are as follows (in m-file npr08_09.m):

(8.36)
 t = 1 1.5 2 2.5 3 u = 5 0.039 0.011 0.005 0.001 0.001 4 0.065 0.07 0.05 0.015 0.01 3 0.031 0.061 0.137 0.051 0.033 2 0.012 0.049 0.163 0.058 0.039 1 0.003 0.009 0.045 0.025 0.017

Determine the marginal distributions. Determine FXY(2,3) and P(Y/X≥1.25).

npr08_09
Data are in X, Y, P
jcalc
- - - - - - - - - - - -
Use array operations on matrices X, Y, PX, PY, t, u, and P
disp([X;PX]')
1.0000    0.1500
1.5000    0.2000
2.0000    0.4000
2.5000    0.1500
3.0000    0.1000
disp([Y;PY]')
1.0000    0.0990
2.0000    0.3210
3.0000    0.3130
4.0000    0.2100
5.0000    0.0570
F = total(((t<=2)&(u<=3)).*P)
F =   0.5100
P = total((u./t>=1.25).*P)
P =   0.5570


For the joint densities in Exercises 10-22 below

1. Sketch the region of definition and determine analytically the marginal density functions fX and fY.

2. Use a discrete approximation to plot the marginal density fX and the marginal distribution function FX.

3. Calculate analytically the indicated probabilities.

4. Determine by discrete approximation the indicated probabilities.

fXY(t,u)=1 for 0≤t≤1, 0≤u≤2(1–t).

(8.37)

Region is triangle with vertices (0,0), (1,0), (0,2).

(8.38)
(8.39)
(8.40)
(8.41)
(8.42)
tuappr
Enter matrix [a b] of X-range endpoints  [0 1]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  200
Enter number of Y approximation points  400
Enter expression for joint density  (t<=1)&(u<=2*(1-t))
Use array operations on X, Y, PX, PY, t, u, and P
fx = PX/dx;
FX = cumsum(PX);
plot(X,fx,X,FX)          % Figure not reproduced
M1 = (t>0.5)&(u>1);
P1 = total(M1.*P)
P1 =  0                  % Theoretical = 0
M2 = (t<=0.5)&(u>0.5);
P2 = total(M2.*P)
P2 =  0.5000             % Theoretical = 1/2
P3 = total((u<=t).*P)
P3 =  0.3350             % Theoretical = 1/3


fXY(t,u)=1/2 on the square with vertices at .

(8.43)

The region is bounded by the lines u=1+t, u=1–t, u=3–t, and u=t–1

(8.44)
(8.45)
(8.46)
(8.47)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  0.5*(u<=min(1+t,3-t))& ...
(u>=max(1-t,t-1))
Use array operations on X, Y, PX, PY, t, u, and P
fx = PX/dx;
FX = cumsum(PX);
plot(X,fx,X,FX)          % Plot not shown
M1 = (t>1)&(u>1);
PM1 = total(M1.*P)
PM1 =  0.2501            % Theoretical = 1/4
M2 = (t<=1/2)&(u>1);
PM2 = total(M2.*P)
PM2 =  0.0631            % Theoretical = 1/16 = 0.0625
M3 = u<=t;
PM3 = total(M3.*P)
PM3 =  0.5023            % Theoretical = 1/2


fXY(t,u)=4t(1–u) for 0≤t≤1, 0≤u≤1.

(8.48)

Region is the unit square.

(8.49)
(8.50)
(8.51)
(8.52) P 3 = ∫010t 4 t ( 1 – u ) d u d t = 5 / 6
tuappr
Enter matrix [a b] of X-range endpoints  [0 1]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  4*t.*(1 - u)
Use array operations on X, Y, PX, PY, t, u, and P
fx = PX/dx;
FX = cumsum(PX);
plot(X,fx,X,FX)           % Plot not shown
M1 = (1/2<t)&(t<3/4)&(u>1/2);
P1 = total(M1.*P)
P1 =  0.0781              % Theoretical = 5/64 = 0.0781
M2 = (t<=1/2)&(u>1/2);
P2 = total(M2.*P)
P2 =  0.0625              % Theoretical = 1/16 = 0.0625
M3 = (u<=t);
P3 = total(M3.*P)
P3 =  0.8350              % Theoretical = 5/6 = 0.8333


for 0≤t≤2, 0≤u≤2.

(8.53)

Region is the square .

(8.54)
(8.55)
(8.56) P 3 = ∫020t ( t + u ) d u d t = 1 / 2
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  (1/8)*(t+u)
Use array operations on X, Y, PX, PY, t, u, and P
fx = PX/dx;
FX = cumsum(PX);
plot(X,fx,X,FX)
M1 = (t>1/2)&(u>1/2);
P1 = total(M1.*P)
P1 =  0.7031              % Theoretical = 45/64 = 0.7031
M2 = (t<=1)&(u>1);
P2 = total(M2.*P)
P2 =  0.2500              % Theoretical = 1/4
M3 = u<=t;
P3 = total(M3.*P)
P3 =  0.5025              % Theoretical = 1/2


fXY(t,u)=4ue–2t for

(8.57)

Region is strip bounded by

(8.58)
(8.59)
(8.60)
tuappr
Enter matrix [a b] of X-range endpoints  [0 3]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  400
Enter number of Y approximation points  200
Enter expression for joint density  4*u.*exp(-2*t)
Use array operations on X, Y, PX, PY, t, u, and P
M2 = (t > 0.5)&(u > 0.5)&(u<3/4);
p2 = total(M2.*P)
p2 =  0.1139            % Theoretical = (5/16)exp(-1) = 0.1150
p3 = total((t<u).*P)
p3 =  0.7047            % Theoretical = 0.7030


for 0≤t≤2, 0≤u≤1+t.

(8.61)

Region bounded by

(8.62)
(8.63)
(8.64)
(8.65) FXY ( 1 , 1 ) = ∫0101 fXY ( t , u ) d u d t = 3 / 44
(8.66)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 3]
Enter number of X approximation points  200
Enter number of Y approximation points  300
Enter expression for joint density  (3/88)*(2*t+3*u.^2).*(u<=1+t)
Use array operations on X, Y, PX, PY, t, u, and P
fx = PX/dx;
FX = cumsum(PX);
plot(X,fx,X,FX)
MF = (t<=1)&(u<=1);
F = total(MF.*P)
F =   0.0681            % Theoretical = 3/44 = 0.0682
M1 = (t<=1)&(u>1);
P1 = total(M1.*P)
P1 =  0.1172            % Theoretical = 41/352 = 0.1165
M2 = abs(t-u)<1;
P2 = total(M2.*P)
P2 =  0.9297           % Theoretical = 329/352 = 0.9347


fXY(t,u)=12t2u on the parallelogram with vertices

(8.67)

Region bounded by

(8.68)
(8.69)
(8.70)
(8.71) P 3 = 1 – P 2 = 13 / 16
tuappr
Enter matrix [a b] of X-range endpoints  [-1 1]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  400
Enter number of Y approximation points  200
Enter expression for joint density  12*u.*t.^2.*((u<=t+1)&(u>=t))
Use array operations on X, Y, PX, PY, t, u, and P
p1 = total((t<=1/2).*P)
p1 =  0.4098                % Theoretical = 33/80 = 0.4125
M2 = (t<1/2)&(u<=1/2);
p2 = total(M2.*P)
p2 =  0.1856                % Theoretical = 3/16  = 0.1875
P3 = total((u>=1/2).*P)
P3 =  0.8144                % Theoretical = 13/16 = 0.8125


for 0≤t≤2, 0≤u≤min{1,2–t}

(8.72)

Region is bounded by

(8.73)
(8.74)
(8.75)
(8.76)
(8.77)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  400
Enter number of Y approximation points  200
Enter expression for joint density  (24/11)*t.*u.*(u<=2-t)
Use array operations on X, Y, PX, PY, t, u, and P
M1 = (t<=1)&(u<=1);
P1 = total(M1.*P)
P1 = 0.5447             % Theoretical = 6/11 = 0.5455
P2 = total((t>1).*P)
P2 =  0.4553            % Theoretical = 5/11 = 0.4545
P3 = total((t<u).*P)
P3 =  0.2705            % Theoretical = 3/11 = 0.2727


for 0≤t≤2, 0≤u≤max{2–t,t}

(8.78)

Region is bounded by

(8.79)
(8.80)
(8.81)
(8.82)
(8.83)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  (3/23)*(t+2*u).*(u<=max(2-t,t))
Use array operations on X, Y, PX, PY, t, u, and P
M1 = (t>=1)&(u>=1);
P1 = total(M1.*P)
P1 =  0.2841
13/46                 % Theoretical = 13/46 = 0.2826
P2 = total((u<=1).*P)
P2 =  0.5190             % Theoretical = 12/23 = 0.5217
P3 = total((u<=t).*P)
P3 =  0.6959             % Theoretical = 16/23 = 0.6957


, for 0≤t≤2, 0≤u≤min{2,3–t}

(8.84) P ( X ≥ 1 , Y ≥ 1 ) , P ( X ≤ 1 , Y ≤ 1 ) , P ( Y < X )

Region has two parts: (1) 0≤t≤1,0≤u≤2 (2) 1<t≤2,0≤u≤3–t

(8.85)
(8.86)
(8.87)
(8.88)
(8.89)
(8.90)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  (12/179)*(3*t.^2+u).* ...
(u<=min(2,3-t))
Use array operations on X, Y, PX, PY, t, u, and P
fx = PX/dx;
FX = cumsum(PX);
plot(X,fx,X,FX)
M1 = (t>=1)&(u>=1);
P1 = total(M1.*P)
P1 =  2312            % Theoretical = 41/179 = 0.2291
M2 = (t<=1)&(u<=1);
P2 = total(M2.*P)
P2 =  0.1003           % Theoretical = 18/179 = 0.1006
M3 = u<=min(t,3-t);
P3 = total(M3.*P)
P3 =  0.7003            % Theoretical = 1001/1432 = 0.6990


for 0≤t≤2, 0≤u≤min{1+t,2}

(8.91)

Region is in two parts:

1. (2)

(8.92) fX ( t ) = I [ 0 , 1 ] ( t ) ∫0 1 + t fXY ( t , u ) d u + I ( 1 , 2 ] ( t ) ∫02 fXY ( t , u ) d u =
(8.93)
(8.94) fY ( u ) = I [ 0 , 1 ] ( u ) ∫02 fXY ( t , u ) d t + I ( 1 , 2 ] ( u ) ∫2 u – 1 fXY ( t , u ) d t =
(8.95)
(8.96)
(8.97)
(8.98)
(8.99)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  200
Enter number of Y approximation points  200
Enter expression for joint density  (12/227)*(3*t+2*t.*u).* ...
(u<=min(1+t,2))
Use array operations on X, Y, PX, PY, t, u, and P
M1 = (t<=1/2)&(u<=3/2);
P1 = total(M1.*P)
P1 =  0.0384             % Theoretical = 139/3632 = 0.0383
M2 = (t<=3/2)&(u>1);
P2 = total(M2.*P)
P2 =  0.3001             % Theoretical = 68/227 = 0.2996
M3 = u<t;
P3 = total(M3.*P)
P3 =  0.6308             % Theoretical = 144/227 = 0.6344


for 0≤t≤2, 0≤u≤min{2t,3–t}

(8.100)

Region bounded by

(8.101)
(8.102)
(8.103)
(8.104)
(8.105) P 3 = ∫020 t / 2 ( t + 2 u ) d u d t = 4 / 13
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 2]
Enter number of X approximation points  400
Enter number of Y approximation points  400
Enter expression for joint density  (2/13)*(t+2*u).*(u<=min(2*t,3-t))
Use array operations on X, Y, PX, PY, t, u, and P
P1 = total((t<1).*P)
P1 = 0.3076             % Theoretical = 4/13 = 0.3077
M2 = (t>=1)&(u<=1);
P2 = total(M2.*P)
P2 =  0.3844            % Theoretical = 5/13 = 0.3846
P3 = total((u<=t/2).*P)
P3 =  0.3076             % Theoretical = 4/13 = 0.3077


for 0≤u≤1.

(8.106) P ( 1 / 2 ≤ X ≤ 3 / 2 , Y ≤ 1 / 2 )

Region is rectangle bounded by

(8.107)
(8.108)
(8.109)
(8.110)
tuappr
Enter matrix [a b] of X-range endpoints  [0 2]
Enter matrix [c d] of Y-range endpoints  [0 1]
Enter number of X approximation points  400
Enter number of Y approximation points  200
Enter expression for joint density  (3/8)*(t.^2+2*u).*(t<=1) ...
+ (9/14)*(t.^2.*u.^2).*(t > 1)
Use array operations on X, Y, PX, PY, t, u, and P
M = (1/2<=t)&(t<=3/2)&(u<=1/2);
P = total(M.*P)
P =  0.1228          % Theoretical = 55/448 = 0.1228

Solutions