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Consider the following 3-period-market-model:

The discounted price of the risky asset $S$:

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How can I find an arbitrage opportunity in this model?

I know that there would be no arbitrage if we replace the first $8$ by something in $(8,12)$ or if we replace the second $8$ by something in $(5,8)$ but I don't know how I can explicitly state the arbitrage opportunity in the given market. So I'm looking for a portfolio which is an arbitrage opportunity.

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The arbitrage strategy is: if the stock is at 8 at $t=1$ buy it else do nothing. Then sell it at $t=2$.

Either the stock has increased to 12 and you made a profit or it is still worth $8$ and your PnL is 0.

So you are guaranteed not to lose any money but you have a non zero probability of making money (equal to the probability of the stock price increasing to 12 conditional on it being 8 at $t=1$).

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Isn't it as simple as: 1) wait one period 2) if the stock is at 8, buy it. If the stock is at 2 , do nothing. This is an arbitrage strategy, since there is a positive probability of a risk free profit.

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  • $\begingroup$ In addition, although not stated in the OP, if there is a risk-free rate $r$, you can short the asset $S$ at the first period if $S=2$ and buy it back for the same price at the next period, and the same if $S=12$ at the second period (and earn $r$ by lending the shorted amount). $\endgroup$ – Daneel Olivaw Jun 22 at 17:54
  • $\begingroup$ seems to me that the first answer is not an arbitrage of first or second type. For that you should recall the first fundamental theorem of asset pricing. The comment from Daniel is correct to me, while the first answer resembles a risky-arbitrage, which is something else. $\endgroup$ – Vitomir Jun 22 at 19:24
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    $\begingroup$ @vitomir, I think an opportunity to buy a stock at 8, with a non zero chance of being either 8 or 12 in the next period, is an arbitrage. Do you disagree? $\endgroup$ – dm63 Jun 22 at 21:06
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    $\begingroup$ @daniel olivaw, the OP says that the prices are discounted prices. I think it affects your argument doesn't it? $\endgroup$ – dm63 Jun 22 at 21:08

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