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I suspect that there is a mistake in the Wikipedia article on the St Petersburg paradox, and I would like to see if I am right before modifying the article.

In the section "Solving the paradox", the formula for computing of the expected utility of the lottery for a log utility function is given to be

$$E(U)=\sum_{k=1}^\infty \frac{(\ln(w+2^{k-1}-c)-\ln(w))}{2^k} <\infty$$

I do not see why the term $\ln(w)$ should be inside the summation (neither why it should be divided by $2^k$ by the way). Do you see anything I overlooked which would justify this?

Related question : Maximum amount willing to gamble given utility function $U(W)=\ln(W)$ and $W=1000000$ in the game referred to in St. Petersberg's Paradox?

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Martin Van der Linden
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    Hint: $\sum_{k=1}^\infty \frac{\ln(w)}{2^k} = \ln(w)$, so inside or outside without the fraction would be fine.That's assuming you think the correct formula should be: $$E(U)=\left(\sum_{k=1}^{\infty} \frac{\ln(w+2^{k-1}-c)}{2^k}\right)-\ln(w)$$ These two formulas are equal – Thomas Andrews Nov 17 '13 at 03:27
  • Many thanks Thomas. This solves my problem. Could you make it an answer so that I can accept it and close the question? – Martin Van der Linden Nov 17 '13 at 03:47

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Presumably you mean that the formula should be:

$$E(U)=\left(\sum_{k=1}^{\infty} \frac{\ln(w+2^{k-1}-c)}{2^k}\right)-\ln(w)$$

But since $\sum_{k=1}^\infty \frac{1}{2^k} = 1$, this is identical to the formula from the Wikipedia page.

Thomas Andrews
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