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Old 01-12-2007, 08:13 PM
pzhon pzhon is offline
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Default IBM\'s January challenge

This month's "Ponder This" challenge from IBM is related to Kelly's work, and may be of interest to many here.

<ul type="square">Suppose some future event (such as a horse race) has n possible outcomes each with associated contract a(i) (which can be bought and sold) which will pay off 1 unit if the ith possibility occurs. The other n-1 contracts will become worthless. Suppose contract a(i) has market price p(i). Then p(i) can be interpreted as the market's estimated probability that the ith event will occur. In an efficient market (which we assume) the sum of the p(i) will be 1.

Suppose you believe the true probability of the ith possibility occurring is q(i) (for i=1,...,n). Clearly the q(i) should also sum to 1 and we assume this. Suppose you have X units to invest and you wish to do this in such a way that the expected value (ie the weighted average over the possible outcomes) of the logarithm of your payoff after the event is maximized. How many of each of the n contracts should you buy? [/list] <font color="white">By the way, I disagree with the statement that the prices summing to 1 is a consequence of efficiency. </font>
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Old 01-12-2007, 11:00 PM
AaronBrown AaronBrown is offline
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Default Re: IBM\'s January challenge

You buy q(i)*X/p(i) of contract i. Your expected log wealth is ln(X) + SUM q(i)*[ln(q(i)) - ln(p(i))].
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Old 01-13-2007, 03:27 AM
PairTheBoard PairTheBoard is offline
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Default Re: IBM\'s January challenge

[ QUOTE ]
You buy q(i)*X/p(i) of contract i. Your expected log wealth is ln(X) + SUM q(i)*[ln(q(i)) - ln(p(i))].

[/ QUOTE ]

X is your Bankroll?

So if say for some i, q(i)=1/2 and p(i)=1/8 you buy

[(1/2)X]/(1/8) = 4X

or 4 times your Bankroll worth of contract i?

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Old 01-13-2007, 04:52 AM
PairTheBoard PairTheBoard is offline
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Default Re: IBM\'s January challenge

[ QUOTE ]
You buy q(i)*X/p(i) of contract i. Your expected log wealth is ln(X) + SUM q(i)*[ln(q(i)) - ln(p(i))].

[/ QUOTE ]

Assuming X is something that makes sense, it's still not clear what you're doing here. Do you pick the i for which q(i)/p(i) is largest and only buy that contract? Or do you buy some of each contract i for which q(i)&gt;p(i). If the former how do you know that's optimal? If the latter the computation looks a lot more messy.

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Old 01-13-2007, 04:56 AM
PairTheBoard PairTheBoard is offline
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Default Re: IBM\'s January challenge

Can you Sell Contracts for which q(i)&lt;p(i)?

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