Solution Found!
The moment-generating function of a normally distributed
Chapter 4, Problem 139E(choose chapter or problem)
The moment-generating function of a normally distributed random variable, π, with mean π and variance \(\sigma^{2}\) was shown in Exercise 4.138 to be \(m(t)=e^{\mu t+(1 / 2) t^{2} \sigma^{2}}\). Use the result in Exercise 4.137 to derive the moment-generating function of \(π =\). What is the distribution of π? Why?
Equation Transcription:
Text Transcription:
2
m(t)=e^mu t+(1/2)t^2sigma^2
X=-3Y+4
Questions & Answers
QUESTION:
The moment-generating function of a normally distributed random variable, π, with mean π and variance \(\sigma^{2}\) was shown in Exercise 4.138 to be \(m(t)=e^{\mu t+(1 / 2) t^{2} \sigma^{2}}\). Use the result in Exercise 4.137 to derive the moment-generating function of \(π =\). What is the distribution of π? Why?
Equation Transcription:
Text Transcription:
2
m(t)=e^mu t+(1/2)t^2sigma^2
X=-3Y+4
ANSWER:
Solution:
Step 1 of 2:
We know that the moment generating function of U is:
Β
The moment generating function of a normally distributed random variable Y, with mean Β and variance
The claim is to derive the moment generating function of X = -3Y + 4. The distribution of X.