EXAMPLE 6.5 Solution Let U be a uniform random variable on the interval (0, 1). Find a transformation G(U) such that G(U) possesses an exponential distribution with mean B. If U possesses a uniform distribution on the interval (0, 1), then the distribution function of U (see Exercise 4.38) is given by 0, u < 0, 0 ≤u ≤ 1, Fu (u) = U, 1, u > 1. Let Y denote a random variable that has an exponential distribution with mean p. Then (see Section 4.6) Y has distribution function Fy (y) = {i-e-x/P. Notice that Fy (y) is strictly increasing on the interval [0, ∞). Let 0 < u < 1 and observe that there is a unique value y such that Fy (y) = u. Thus, Fy¹ (u), 0 < u < 1, is well defined. In this case, Fy (y) = 1 - e-y/B = u if and only if y = -ß ln(1-u) = Fy¹ (u). Consider the random variable Fy¹(U) = -8 ln(1-U) and observe that, if y > 0, P(Fy¹ (U) ≤ y) = P[-8 ln(1 - U) ≤y] = P[ln(1 - U) ≥ -y/B] = P(U ≤ 1-e-/B) = 1- e-y/B. y < 0, y ≥ 0. Also, P[F'(U) ≤y] = 0 if y ≤ 0. Thus, Fy'(U) = -8 ln(1 - U) possesses an exponential distribution with mean ß, as desired.
EXAMPLE 6.5 Solution Let U be a uniform random variable on the interval (0, 1). Find a transformation G(U) such that G(U) possesses an exponential distribution with mean B. If U possesses a uniform distribution on the interval (0, 1), then the distribution function of U (see Exercise 4.38) is given by 0, u < 0, 0 ≤u ≤ 1, Fu (u) = U, 1, u > 1. Let Y denote a random variable that has an exponential distribution with mean p. Then (see Section 4.6) Y has distribution function Fy (y) = {i-e-x/P. Notice that Fy (y) is strictly increasing on the interval [0, ∞). Let 0 < u < 1 and observe that there is a unique value y such that Fy (y) = u. Thus, Fy¹ (u), 0 < u < 1, is well defined. In this case, Fy (y) = 1 - e-y/B = u if and only if y = -ß ln(1-u) = Fy¹ (u). Consider the random variable Fy¹(U) = -8 ln(1-U) and observe that, if y > 0, P(Fy¹ (U) ≤ y) = P[-8 ln(1 - U) ≤y] = P[ln(1 - U) ≥ -y/B] = P(U ≤ 1-e-/B) = 1- e-y/B. y < 0, y ≥ 0. Also, P[F'(U) ≤y] = 0 if y ≤ 0. Thus, Fy'(U) = -8 ln(1 - U) possesses an exponential distribution with mean ß, as desired.
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Please explain example 6.5
![to a
In some instances, it is possible to find a transformation that, when applied
random variable with a uniform distribution on the interval (0, 1), results in a random
variable with some other specified distribution function, say, F(y). The next example
illustrates a technique for achieving this objective. A brief discussion of one practical
use of this transformation follows the example.
EXAMPLE 6.5 Let U be a uniform random variable on the interval (0, 1). Find a transformation
G(U) such that G(U) possesses an exponential distribution with mean ß.
Solution
If U possesses a uniform distribution on the interval (0, 1), then the distribution
function of U (see Exercise 4.38) is given by
u < 0,
0≤u ≤ 1,
Fu (u)
=
0,
E
u,
1, u > 1.
Let Y denote a random variable that has an exponential distribution with mean p.
Then (see Section 4.6) Y has distribution function
0,
Fr(y) = { 1 - e
1-e-y/B,
y < 0,
y ≥ 0.
Notice that Fy (y) is strictly increasing on the interval [0, ∞). Let 0 < u < 1
and observe that there is a unique value y such that Fy (y) = u. Thus, Fy¹ (u),
0 < u < 1, is well defined. In this case, Fy (y) = 1 - ey/B = u if and only if
y = -ß ln(1-u) = Fy¹ (u). Consider the random variable F¹(U) = -ß ln(1 – U)
and observe that, if y > 0,
-1
P(Fy'(U) ≤ y) = P[-ß ln(1 - U) ≤y]
= P[ln(1 - U) ≥ −y/B]
= P(U ≤ 1-e-y/B)
= 1- e-y/B.
Also, P[F¹(U) ≤ y] = 0 if y ≤ 0. Thus, Fy¹ (U) = −ß ln(1 - U) possesses an
exponential distribution with mean ß, as desired.
Computer simulations are frequently used to evaluate proposed statistical tech-
niques. Typically, these simulations require that we obtain observed values of random
variables with a prescribed distribution. As noted in Section 4.4, most computer
systems contain a subroutine that provides observed values of a random variable
U that has a uniform distribution on the interval (0, 1). How can the result of
Example 6.5 be used to generate a set of observations from an exponential distribution](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6389447-1237-4af0-b5c6-eb1260425b55%2Ffc608517-3408-45c3-9f8e-f3dcf319e10d%2Faq0fg6d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:to a
In some instances, it is possible to find a transformation that, when applied
random variable with a uniform distribution on the interval (0, 1), results in a random
variable with some other specified distribution function, say, F(y). The next example
illustrates a technique for achieving this objective. A brief discussion of one practical
use of this transformation follows the example.
EXAMPLE 6.5 Let U be a uniform random variable on the interval (0, 1). Find a transformation
G(U) such that G(U) possesses an exponential distribution with mean ß.
Solution
If U possesses a uniform distribution on the interval (0, 1), then the distribution
function of U (see Exercise 4.38) is given by
u < 0,
0≤u ≤ 1,
Fu (u)
=
0,
E
u,
1, u > 1.
Let Y denote a random variable that has an exponential distribution with mean p.
Then (see Section 4.6) Y has distribution function
0,
Fr(y) = { 1 - e
1-e-y/B,
y < 0,
y ≥ 0.
Notice that Fy (y) is strictly increasing on the interval [0, ∞). Let 0 < u < 1
and observe that there is a unique value y such that Fy (y) = u. Thus, Fy¹ (u),
0 < u < 1, is well defined. In this case, Fy (y) = 1 - ey/B = u if and only if
y = -ß ln(1-u) = Fy¹ (u). Consider the random variable F¹(U) = -ß ln(1 – U)
and observe that, if y > 0,
-1
P(Fy'(U) ≤ y) = P[-ß ln(1 - U) ≤y]
= P[ln(1 - U) ≥ −y/B]
= P(U ≤ 1-e-y/B)
= 1- e-y/B.
Also, P[F¹(U) ≤ y] = 0 if y ≤ 0. Thus, Fy¹ (U) = −ß ln(1 - U) possesses an
exponential distribution with mean ß, as desired.
Computer simulations are frequently used to evaluate proposed statistical tech-
niques. Typically, these simulations require that we obtain observed values of random
variables with a prescribed distribution. As noted in Section 4.4, most computer
systems contain a subroutine that provides observed values of a random variable
U that has a uniform distribution on the interval (0, 1). How can the result of
Example 6.5 be used to generate a set of observations from an exponential distribution
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