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 November 18th, 2007, 04:47 PM #1 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?
 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.
November 18th, 2007, 05:26 PM   #3
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Quote:
 Originally Posted by skipjack The mean return doesn't depend on n; it's zero.
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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%

 November 18th, 2007, 05:44 PM #4 Global Moderator     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.
November 18th, 2007, 05:56 PM   #5
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 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 November 18th, 2007, 09:48 PM #6 Global Moderator 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.

November 19th, 2007, 06:30 AM   #7
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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?

November 19th, 2007, 11:11 AM   #8
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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.

 November 19th, 2007, 05:57 PM #9 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.
November 20th, 2007, 05:53 AM   #10
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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.

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