Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(1) and U is given by U = aY + b, the moment-generating function of U is em(at). If Y has mean u and variance o?, use the moment-generating function of U to derive the mean and variance of U. Recall: my(t) = E(etY) and whent = 0 then my(0) = E(e°Y) = E(1) = 0 and d*m(t)] = tk- dik and M) = F(*) = E (*) %3D = E (X et×). Similarly, %3D diz M(t) () = E (X² c²X). = E %3D
Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(1) and U is given by U = aY + b, the moment-generating function of U is em(at). If Y has mean u and variance o?, use the moment-generating function of U to derive the mean and variance of U. Recall: my(t) = E(etY) and whent = 0 then my(0) = E(e°Y) = E(1) = 0 and d*m(t)] = tk- dik and M) = F(*) = E (*) %3D = E (X et×). Similarly, %3D diz M(t) () = E (X² c²X). = E %3D
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 31E
Related questions
Question
![4.137
Show that the result given in Exercise 3.158 also holds for continuous random variables. That
is, show that, if Y is a random variable with moment-generating function m(t) and U is given
by U = aY +b, the moment-generating function of U is em(at). If Y has mean u and
variance o?, use the moment-generating function of U to derive the mean and variance of U.
Recall: my(t) = E(etY) and when t = 0 then my(0) = E(eoY) = E(1) = 0
and
d*m(t)]
dık
and
d
dt
- E ()
= E (X e² *).
Similarly,
M(1) = E (**)
= E
E (X²e+x).
=
1
Hence, in general we get
d"
M(t) = E (e* *)
%3D
dt"
dt"
uP
dt"
= E
- E (X" e² ×).
If we set t = 0 in the nth derivative, we get
d"
= E (X" e²×)\,-o = E (X").
It=0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd74c38e2-bc89-4af1-ad7f-05db36328edb%2F0ed8bbbf-9ed4-4e8d-80fa-65f861459995%2Fobcdx3w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4.137
Show that the result given in Exercise 3.158 also holds for continuous random variables. That
is, show that, if Y is a random variable with moment-generating function m(t) and U is given
by U = aY +b, the moment-generating function of U is em(at). If Y has mean u and
variance o?, use the moment-generating function of U to derive the mean and variance of U.
Recall: my(t) = E(etY) and when t = 0 then my(0) = E(eoY) = E(1) = 0
and
d*m(t)]
dık
and
d
dt
- E ()
= E (X e² *).
Similarly,
M(1) = E (**)
= E
E (X²e+x).
=
1
Hence, in general we get
d"
M(t) = E (e* *)
%3D
dt"
dt"
uP
dt"
= E
- E (X" e² ×).
If we set t = 0 in the nth derivative, we get
d"
= E (X" e²×)\,-o = E (X").
It=0
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 4 images
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage