Solution Found!
Use the result of Exercise 2.5-5 to find the mean and
Chapter 2, Problem 5E(choose chapter or problem)
Let the moment-generating function M(t) of X exist for \(-h<t<h\). Consider the function \(R(t)=\ln M(t)\). The first two derivatives of R(t) are, respectively,
\(R^{\prime}(t)=\frac{M^{\prime}(t)}{M(t)} \text { and } R^{\prime \prime}(t)=\frac{M(t) M^{\prime \prime}(t)-\left[M^{\prime}(t)\right]^{2}}{[M(t)]^{2}}\)
Setting , show that
(a) .
(b) .
Equation Transcription:
.
Text Transcription:
-h < t < h
R(t) = ln M(t)
R prime prime (t) = M(t)M prime prime (t) - [M prime (t)]^2/[M(t)]^2
Nu = R prime (0)
Sigma^2 = R prime prime (0)
Questions & Answers
QUESTION:
Let the moment-generating function M(t) of X exist for \(-h<t<h\). Consider the function \(R(t)=\ln M(t)\). The first two derivatives of R(t) are, respectively,
\(R^{\prime}(t)=\frac{M^{\prime}(t)}{M(t)} \text { and } R^{\prime \prime}(t)=\frac{M(t) M^{\prime \prime}(t)-\left[M^{\prime}(t)\right]^{2}}{[M(t)]^{2}}\)
Setting , show that
(a) .
(b) .
Equation Transcription:
.
Text Transcription:
-h < t < h
R(t) = ln M(t)
R prime prime (t) = M(t)M prime prime (t) - [M prime (t)]^2/[M(t)]^2
Nu = R prime (0)
Sigma^2 = R prime prime (0)
ANSWER:
Answer:
Step 1 of 6:
From exercise 2.5 - 5.
and
Put t = 0.
Then,
=
=
= .
Therefore, =
=
=
=
Therefore,
=
and
is the moment generating function.