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(a) The sample points (x, y) for which probabilities are different from zero are indicated in Fig. 2-8. The probabilities associated with these points, given by c(2x y), are shown in Table 2-6. Since the grand total, 42c, must equal 1, we have c 1 > 42.
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Table 2-6 X 0 1 2 Totals S Y 0 0 2c 4c 6c 1 c 3c 5c 9c 2 2c 4c 6c 12c 3 3c 5c 7c 15c Totals T 6c 14c 22c 42c
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Fig. 2-8
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(b) From Table 2-6 we see that
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CHAPTER 2 Random Variables and Probability Distributions
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(c) From Table 2-6 we see that
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a a f (x, y)
x 1y 2
(2c 24c
as indicated by the entries shown shaded in the table.
3c 24 42
4c)(4c 4 7
2.9. Find the marginal probability functions (a) of X and (b) of Y for the random variables of Problem 2.8.
(a) The marginal probability function for X is given by P(X x) f1(x) and can be obtained from the margin totals in the right-hand column of Table 2-6. From these we see that
P(X 1 7 1 3 11 21
f1 (x)
6c 14c 22c
1>7 x 1>3 x 11>21 x
0 1 2
Check:
f2(y) and can be obtained from the margin
(b) The marginal probability function for Y is given by P(Y y) totals in the last row of Table 2-6. From these we see that
f2(y)
6c 9c 12c 15c
1>7 3>14 2>7 5>14
y y y y
0 1 2 3
Check:
3 14
5 14
2.10. Show that the random variables X and Y of Problem 2.8 are dependent.
If the random variables X and Y are independent, then we must have, for all x and y, P(X x, Y y) P(X x)P(Y y)
But, as seen from Problems 2.8(b) and 2.9, P(X so that 2, Y 1) 5 42 P(X P(X 2, Y 2) 1) 2 P(X 11 21 P(Y 2)P(Y 1) y) > 42 cannot be expressed as a 1) 3 14
The result also follows from the fact that the joint probability function (2x function of x alone times a function of y alone.
2.11. The joint density function of two continuous random variables X and Y is f (x, y) (a) Find the value of the constant c. (b) Find P(1 X 2, 2 Y 3).
cxy 0
0 x 4, 1 otherwise 3, Y
(c) Find P(X
(a) We must have the total probability equal to 1, i.e.,
f(x, y) dx dy
CHAPTER 2 Random Variables and Probability Distributions
Using the definition of f (x, y), the integral has the value
4 5 4
3 0 y
cxy dx dy
c3 c3
Then 96c 1 and c 1 > 96.
xy2 5 2 dx 0 2 y 1 12x dx
xydyR dx
c(6x2) 2
25x 2 96c
x dx 2
(b) Using the value of c found in (a), we have
2, 2
3 1 y
1 2 96 3x 1 2 96 3x
5x dx 1 2
xy dx dy 96
xy dyR dx
1 2 96 3x
xy2 3 2 dx 1 2 y 2 5 128
5 x2 2 2 a b 192 2 1
3, Y
3 3 y
xy dx dy 96 1 B3
1 4 96 3x 1 4 96 3x
xydyR dx
1 4 96 3x
xy2 2 2 dx 3 2 y 1
3x dx 2
7 128
2.12. Find the marginal distribution functions (a) of X and (b) of Y for Problem 2.11.
(a) The marginal distribution function for X if 0 x
4 is
F1(x)
x) uv B3
1 96 5 v
3 ` v
f (u, v) dudv
3 0 v
dudv x2 16
1 x 96 3u
For x 4, F1(x) 1; for x 0, F1(x)
uvdvR du
0. Thus
F1(x)
0 x2>16 1
x x x
0 4 4
As F1 (x) is continuous at x
0 and x
4, we could replace
in the above expression.
CHAPTER 2 Random Variables and Probability Distributions
(b) The marginal distribution function for Y if 1
5 is
F2( y)
3 ` v
f(u, v) dudv y2 24 1
For y 5, F2( y) 1. For y 1, F2( y) 0. Thus
3 0 v
uv dudv 96
F2( y)
As F2( y) is continuous at y 1 and y
0 (y2 1
1)>24
y y y
1 5 5
in the above expression.
5, we could replace
2.13. Find the joint distribution function for the random variables X, Y of Problem 2.11.
From Problem 2.11 it is seen that the joint density function for X and Y can be written as the product of a function of x alone and a function of y alone. In fact, f (x, y) f1(x)f2( y), where
f1 (x)
c1x 0
0 x 4 otherwise
f2(y)
c2 y 0
1 y 5 otherwise
and c1c2 c 1 > 96. It follows that X and Y are independent, so that their joint distribution function is given by F(x, y) F1(x)F2( y). The marginal distributions F1(x) and F2( y) were determined in Problem 2.12, and Fig. 2-9 shows the resulting piecewise definition of F(x, y).
2.14. In Problem 2.11 find P(X
Fig. 2-9
In Fig. 2-10 we have indicated the square region 0 x 4, 1 y 5 within which the joint density function of X and Y is different from zero. The required probability is given by
6 f (x, y) dx dy
CHAPTER 2 Random Variables and Probability Distributions
where 5 is the part of the square over which x over 5, this probability is given by
2 3 x 1
3, shown shaded in Fig. 2-10. Since f(x, y)
xy > 96
3 0 y
xy dx dy 96 1 2 96 3x 1 2 96 3x
xy2 3 x 2 dx 0 2 y 1
xy dyR dx
1 2 [x(3 192 3x 0
1 48
Fig. 2-10
Change of variables 2.15. Prove Theorem 2-1, page 42.
The probability function for U is given by
g(u)
P[f(X)
c(u)]
f [c(u)]
In a similar manner Theorem 2-2, page 42, can be proved.
2.16. Prove Theorem 2-3, page 42.
Consider first the case where u (x) or x (u) is an increasing function, i.e., u increases as x increases (Fig. 2-11). There, as is clear from the figure, we have (1) or
u2 x2
P(u1
P(x1
3u g(u) du
3x f (x) dx
Fig. 2-11
CHAPTER 2 Random Variables and Probability Distributions
Letting x
(u) in the integral on the right, (2) can be written
u2 u2
3u g(u) du
3u f [c (u)] cr(u) du
This can hold for all u1 and u2 only if the integrands are identical, i.e.,
g(u)
f [c(u)]cr(u)
This is a special case of (34), page 42, where cr(u) 0 (i.e., the slope is positive). For the case where cr(u) 0, i.e., u is a decreasing function of x, we can also show that (34) holds (see Problem 2.67). The theorem can also be proved if cr(u) 0 or cr(u) 0.
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