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2.17. Prove Theorem 2-4, page 42.
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We suppose first that as x and y increase, u and v also increase. As in Problem 2.16 we can then show that P(u1 or Letting x
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v), y
2(u,
v) in the integral on the right, we have, by a theorem of advanced calculus,
3v 3v g(u, v) du dv
3u 3 f [c1 (u, v), c2(u, v)]J du dv
u 2 v2 v1
where is the Jacobian. Thus
'(x, y) '(u, v)
g(u, v)
which is (36), page 42, in the case where J
f [c1(u, v), c2(u, v)]J
0. Similarly, we can prove (36) for the case where J 0.
2.18. The probability function of a random variable X is f (x) e 2 0
x 1, 2, 3, c otherwise
X4 1.
Find the probability function for the random variable U
Since U X4 1, the relationship between the values u and x of the random variables U and X is given by 4 2u 1, where u 2, 17, 82, . . . and the real positive root is taken. Then the required u x4 1 or x probability function for U is given by
g(u)
2 e 0
2u 1
u 2, 17, 82, . . . otherwise
using Theorem 2-1, page 42, or Problem 2.15.
2.19. The probability function of a random variable X is given by f (x) e x2 >81 0 3 x 6 otherwise
1 (12 3
Find the probability density for the random variable U
CHAPTER 2 Random Variables and Probability Distributions
We have u 3 (12 x) or x 12 3u. Thus to each value of x there is one and only one value of u and conversely. The values of u corresponding to x 3 and x 6 are u 5 and u 2, respectively. Since cr(u) dx>du 3, it follows by Theorem 2-3, page 42, or Problem 2.16 that the density function for U is
g(u)
(12 0 3u)2
3u)2 >27
2 u 5 otherwise 3u)3 5 2 243 2
Check:
5 (12
2.20. Find the probability density of the random variable U Problem 2.19.
X 2 where X is the random variable of
We have u x2 or x !u. Thus to each value of x there corresponds one and only one value of u, but to each value of u 2 0 there correspond two values of x. The values of x for which 3 x 6 correspond to values of u for which 0 u 36 as shown in Fig. 2-12. As seen in this figure, the interval 3 x 3 corresponds to 0 u 9 while 3 x 6 corresponds to 9 u 36. In this case we cannot use Theorem 2-3 directly but can proceed as follows. The distribution function for U is
G(u)
Now if 0 u 9, we have G(u) P(U
P(X2
P( ! u
! u)
3 1u f (x) dx
Fig. 2-12
But if 9
36, we have G(u) P(U u) P( 3 X !u) 3 3 f (x) dx
CHAPTER 2 Random Variables and Probability Distributions
Since the density function g(u) is the derivative of G(u), we have, using (12), e f ( !u) 2 !u 0 f ( !u) f ( !u) 2 !u 0 9 u u 9 36
g(u)
otherwise
Using the given definition of f (x), this becomes !u>81 !u>162 0 0 u 9 9 u 36 otherwise
g(u) Check:
9 !u 30 81 du
36 !u 39 162 du
2.21. If the random variables X and Y have joint density function f (x, y) e xy>96 0
0 x 4, 1 otherwise X 2Y.
2u 3>2 2 9 243 0
u 3>2 2 36 243 9
(see Problem 2.11), find the density function of U
Method 1 Let u x 2y, v x, the second relation being chosen arbitrarily. Then simultaneous solution yields x v, y 1 (u v). Thus the region 0 x 4, 1 y 5 corresponds to the region 0 v 2 2 u v 10 shown shaded in Fig. 2-13.
Fig. 2-13
The Jacobian is given by 'x 'u 4 'y 'u 0 1 2 1 2 'x 'v 4 'y 'v 1 1 2
CHAPTER 2 Random Variables and Probability Distributions
Then by Theorem 2-4 the joint density function of U and V is g(u, v) e v(u 0 v)>384 2 u v otherwise 10, 0 v 4
The marginal density function of U is given by
g1(u)
g 3v
v(u v) dv 384 0 v(u v) dv 384 0 v(u v) dv 384 u 10
2 6 10
u u u
6 10 14
otherwise
as seen by referring to the shaded regions I, II, III of Fig. 2-13. Carrying out the integrations, we find (u 2)2(u 4)>2304 (3u 8)>144 d (348u u 3 2128)>2304 0 2 u 6 6 u 10 10 u 14 otherwise
g1(u)
A check can be achieved by showing that the integral of g1 (u) is equal to 1. Method 2 The distribution function of the random variable X P(X 2Y u)
2Y is given by 6 f (x, y) dx dy
2y u
x 2y u 0 x 4 1 y 5
xy dx dy 96
For 2
6, we see by referring to Fig. 2-14, that the last integral equals
u 2 (u 0 x)>2 1
xy dx dy 96
The derivative of this with respect to u is found to be (u the result of Method 1 for 6 u 10, etc.
2)2(u
x(u x)2 768
4) >2304. In a similar manner we can obtain
x R dx 192
Fig. 2-14
Fig. 2-15
CHAPTER 2 Random Variables and Probability Distributions
2.22. If the random variables X and Y have joint density function f (x, y) e xy>96 0 0 x 4, 1 otherwise XY 2, V y 5
(see Problem 2.11), find the joint density function of U
X2Y.
Consider u xy2, v x2y. Dividing these equations, we obtain y >x u >v so that y ux >v. This leads to the simultaneous solution x v2>3 u 1>3, y u2>3 v 1>3. The image of 0 x 4, 1 y 5 in the uv-plane is given by 0 which are equivalent to v2 This region is shown shaded in Fig. 2-15. The Jacobian is given by
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