Solution Found!
Let r (t) = ln[m(t)] and r (k) (0) denote the kth
Chapter 3, Problem 162E(choose chapter or problem)
Let \(r(t)=\ln [m(t)] and r^{(k)}(0)\) denote the \(k \text { th }\) derivative of \(r(t)\) evaluated for \(t=0\). Show that \(r^{(1)}(0)=\mu_{1}^{\prime}=\mu\) and \(r^{(2)}(0)=\mu_{2}^{\prime}-\left(\mu_{1}^{\prime}\right)^{2}=\sigma^{2}\)
[Hint: \(m(0)=1\).]
Equation Transcription:
th
Text Transcription:
r(t)=ln[m(t)]
r^(k)(0)
kth
r(t)
t=0
r^(1)(0)=mu'_1=mu
r^(2)(0)=mu'_2-(mu'1)^2=sigma^2
m(0)=1
Questions & Answers
QUESTION:
Let \(r(t)=\ln [m(t)] and r^{(k)}(0)\) denote the \(k \text { th }\) derivative of \(r(t)\) evaluated for \(t=0\). Show that \(r^{(1)}(0)=\mu_{1}^{\prime}=\mu\) and \(r^{(2)}(0)=\mu_{2}^{\prime}-\left(\mu_{1}^{\prime}\right)^{2}=\sigma^{2}\)
[Hint: \(m(0)=1\).]
Equation Transcription:
th
Text Transcription:
r(t)=ln[m(t)]
r^(k)(0)
kth
r(t)
t=0
r^(1)(0)=mu'_1=mu
r^(2)(0)=mu'_2-(mu'1)^2=sigma^2
m(0)=1
ANSWER:
Solution:
Step 1 of 2:
It is given that r(t)=ln[m(t)].
Using this, we need to show that =
and
=
.