My Math Forum  

Go Back   My Math Forum > Science Forums > Economics

Economics Economics Forum - Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance


Reply
 
LinkBack Thread Tools Display Modes
April 27th, 2014, 11:51 PM   #1
Member
 
Joined: Oct 2009

Posts: 50
Thanks: 0

how to derive log approximation?

Hello!

This is probably an easy question. I'm trying to find out how to derive the log approximation of an exchange rate percentage change.
So the actual change is: ( S[t+1] - St )/St
How do you derive from this that an approximation is log(St+1) - log(St)?

Thank you!
questioner1 is offline  
 
May 2nd, 2014, 08:52 PM   #2
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,412
Thanks: 1024

Quote:
Originally Posted by questioner1 View Post
So the actual change is: ( S[t+1] - St )/St
(S[t + 1] - St) / St
= (St + S - St) / St
= S / St
= 1 / t

...what are you asking?
Denis is offline  
May 2nd, 2014, 09:30 PM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,654
Thanks: 2632

Math Focus: Mainly analysis and algebra
I think the OPs equation is not as you worked on, but

$$\frac{s_{t+1} - s_t}{s_t} = \frac{s_{t+1}}{s_t}-1$$

Although,
$$\log{\frac{s_{t+1} - s_t}{s_t}} = \log{\left(s_{t+1} - s_t\right)} - \log{s_t}$$

I don't know what the question is either.

Last edited by v8archie; May 2nd, 2014 at 09:33 PM.
v8archie is online now  
May 3rd, 2014, 11:03 PM   #4
Member
 
Joined: Oct 2009

Posts: 50
Thanks: 0

Quote:
Originally Posted by v8archie View Post
I think the OPs equation is not as you worked on, but

$$\frac{s_{t+1} - s_t}{s_t} = \frac{s_{t+1}}{s_t}-1$$

Although,
$$\log{\frac{s_{t+1} - s_t}{s_t}} = \log{\left(s_{t+1} - s_t\right)} - \log{s_t}$$

I don't know what the question is either.
Sorry, I didn't know how to make mathematical symbols, and I haven't formulated my question perfectly clearly, sorry.

The thing is, my textbook does this "derivation"/aproximation (the factis,I don't know what their doing...):
They assume:
$${r-r*} = \frac{S_{t+1}-S_{t}}{S_t}$$
And conclude from this that (This step I don't get):
$$r-r* = \log{S_{t+1}} - \log{S_t}$$

I understand your derivation, but its different from my textbook, since they don't take the log on both sides.
They're effectively saying:
$$\frac{S_{t+1}-S_{t}}{S_t} = \log{S_{t+1}} - \log{S_t}$$
And I don't understand why that would be true (or approximately true, since I don't know if its an aproximation or not.)

I hope my question is clearer now. Thanks for the help, either way!

Last edited by questioner1; May 3rd, 2014 at 11:10 PM.
questioner1 is offline  
Reply

  My Math Forum > Science Forums > Economics

Tags
approximation, derive, log



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Derive Solarmew Applied Math 16 May 2nd, 2012 09:09 PM
Derive this dudetheman Calculus 2 January 18th, 2012 11:44 AM
how do I derive this dmxnemesis Calculus 4 October 20th, 2010 03:38 PM
How to derive e? NeuroFuzzy Calculus 14 September 18th, 2010 10:35 PM
Derive -x/y instereo911 Calculus 2 October 8th, 2009 02:55 PM





Copyright © 2019 My Math Forum. All rights reserved.