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
November 18th, 2007, 04:47 PM   #1
STV
Member
 
Joined: Aug 2007

Posts: 93
Thanks: 0

A Basic Problem

For each period n an investment will either gain or lose 10% with probability of 50% each way. As n goes to infinity, what is the geometric mean return on the investment?
STV is offline  
 
November 18th, 2007, 04:58 PM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 20,640
Thanks: 2082

The mean return doesn't depend on n; it's zero.
skipjack is offline  
November 18th, 2007, 05:26 PM   #3
STV
Member
 
Joined: Aug 2007

Posts: 93
Thanks: 0

Quote:
Originally Posted by skipjack
The mean return doesn't depend on n; it's zero.
\

The mean return on the investment is not zero, as is proven by if n=2 the mean result is (1+10%)(1-10%) = -1%
STV is offline  
November 18th, 2007, 05:44 PM   #4
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
n = 0: value 1, mean = 1
n = 1: value 0.9 (1/2), 1.1 (1/2), mean = 1
n = 2: value 0.81 (1/4), 0.99 (1/2), 1.21 (1/4), mean = 1

An inductive proof should show that the mean remains at 1. I believe the median tends toward 0, but I'd have to check.
CRGreathouse is offline  
November 18th, 2007, 05:56 PM   #5
STV
Member
 
Joined: Aug 2007

Posts: 93
Thanks: 0

Quote:
Originally Posted by CRGreathouse
n = 0: value 1, mean = 1
n = 1: value 0.9 (1/2), 1.1 (1/2), mean = 1
n = 2: value 0.81 (1/4), 0.99 (1/2), 1.21 (1/4), mean = 1

An inductive proof should show that the mean remains at 1. I believe the median tends toward 0, but I'd have to check.
Yes the expectation remains at 1, but the geometric mean return (think about what $1 would compound at) is negative. The law of large numbers would lead to 1.1^(.5n)*0.9^(.5n). The limit of this for large n is zero
STV is offline  
November 18th, 2007, 09:48 PM   #6
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
Originally Posted by STV
Yes the expectation remains at 1, but the geometric mean return (think about what $1 would compound at) is negative. The law of large numbers would lead to 1.1^(.5n)*0.9^(.5n). The limit of this for large n is zero
I included the compounding in my calculations. Note that the high end is $1.21 and the low end is $0.81 -- 21 cent max gain in 2 rounds vs. 19 cent max loss in 2 rounds. This is made up for with a low median.

Your 'law of large numbers' shows the median for even n, not the mean.
CRGreathouse is offline  
November 19th, 2007, 06:30 AM   #7
STV
Member
 
Joined: Aug 2007

Posts: 93
Thanks: 0

Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by STV
Yes the expectation remains at 1, but the geometric mean return (think about what $1 would compound at) is negative. The law of large numbers would lead to 1.1^(.5n)*0.9^(.5n). The limit of this for large n is zero
I included the compounding in my calculations. Note that the high end is $1.21 and the low end is $0.81 -- 21 cent max gain in 2 rounds vs. 19 cent max loss in 2 rounds. This is made up for with a low median.

Your 'law of large numbers' shows the median for even n, not the mean.
Yeah, I should have phrased the OP better - what I was getting at was the expected return of any investor who plays the game which is that they lose all their money. This is a function of a random variable and at n=10,000 the chances of at least breaking even is 1 in 3.7 million

If this binomial distribution is taken to its limit, can anyone formulate the difference between the expected geometric mean return of the investor (i.e. the compounded rate of return) and the expected arithmetic average return (which is zero)? What continuous distribution describes the wealth of the investor as n goes to infinity?
STV is offline  
November 19th, 2007, 11:11 AM   #8
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
Originally Posted by STV
Yeah, I should have phrased the OP better - what I was getting at was the expected return of any investor who plays the game which is that they lose all their money. This is a function of a random variable and at n=10,000 the chances of at least breaking even is 1 in 3.7 million
I didn't check that, but that matches my expectation.

Quote:
Originally Posted by STV
If this binomial distribution is taken to its limit, can anyone formulate the difference between the expected geometric mean return of the investor (i.e. the compounded rate of return) and the expected arithmetic average return (which is zero)? What continuous distribution describes the wealth of the investor as n goes to infinity?
As n increases without bound:
The arithmetic mean is 1. The geometric mean is 0. The wealth distribution is uniformly 0.
CRGreathouse is offline  
November 19th, 2007, 05:57 PM   #9
STV
Member
 
Joined: Aug 2007

Posts: 93
Thanks: 0

Actually the geometric mean return diverges from the arithmetic mean by half the variance the resulting wealth distribution is lognormal


If you think about it this is a potent argument for minimizing your volatility by diversifying your investments - it actually will lead to a slightly higher compound return.
STV is offline  
November 20th, 2007, 05:53 AM   #10
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
Originally Posted by STV
Actually the geometric mean return diverges from the arithmetic mean by half the variance the resulting wealth distribution is lognormal
It's log-normal for finite n. The limit of the cdf as n tend s to infinity is uniformly zero.

Quote:
Originally Posted by STV
If you think about it this is a potent argument for minimizing your volatility by diversifying your investments - it actually will lead to a slightly higher compound return.
That assumes that return is not correlated with volatility, which seems like a very bad assumption to me.
CRGreathouse is offline  
Reply

  My Math Forum > Science Forums > Economics

Tags
basic, problem



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Basic word problem Chandreyee Elementary Math 8 September 11th, 2013 11:40 PM
Basic PHP Math problem simy202 Computer Science 1 January 17th, 2011 09:01 PM
(Hopefully) Basic integration problem shynthriir Calculus 3 May 27th, 2009 10:01 AM
basic problem balibalo Advanced Statistics 2 February 11th, 2008 04:37 AM
Basic Calculus Problem VDSL_2 Calculus 2 September 9th, 2007 01:54 AM





Copyright © 2019 My Math Forum. All rights reserved.