So, if I knew ahead of time what your price was going to be tomorrow, I would have a clear arbitrage strategy. But that's not a fair example. The question is: what do I know at t=0 about what your price could be at t=1?
A better example is this: suppose you're creating a probabilistic forecast of the chance of a coin coming up Heads twice. Suppose this represents a win. You say, "according to my model, P[H1 and H2] = 0.3." And suppose you also say P[H1] = 0.4
Then I ask you: imagining the first flip comes up Heads, what will your price for the second Heads be then? The only arbitrage-free price you can quote me is your conditional probability: P[H2 | H1] = P[H1 and H2]/P[H1] = 0.75. Anything else allows me to buy and sell bets and make a riskless profit.
Now, note that if H1 occurs, P[H2 | H1] is now the updated chance of getting H1 and H2. You observed some information and updated your probability accordingly. Also note that according to your model, the chance of H1 happening was 0.4, so the possible prices of "H1 and H2" after one coin flip were 0.75 with probability 0.4 and 0 with probability 0.6 (if the first flip were tails, "H1 and H2" is impossible). So the expected price of "H1 and H2" after one flip is (.75)(.4) + (0)(.6) = 0.3, which is also your price at t=0. That means the probability/price is a martingale according to your probability assignments, which, according to the Fundamental Theorem of Asset Pricing, again means no arbitrage is possible, unless we disagree on what events have probability zero.
So, the election forecasts are like this except we don't get to see all the inner workings. Nate doesn't quote things like the probability of the polls moving by a certain amount each day. We just get to see things like P[H1 and H2] before and after the first coin flip. But as long as it's possible that those come from a consistent set of conditional probabilities and Bayesian updating, there's no chance for arbitrage.
A better example is this: suppose you're creating a probabilistic forecast of the chance of a coin coming up Heads twice. Suppose this represents a win. You say, "according to my model, P[H1 and H2] = 0.3." And suppose you also say P[H1] = 0.4
Then I ask you: imagining the first flip comes up Heads, what will your price for the second Heads be then? The only arbitrage-free price you can quote me is your conditional probability: P[H2 | H1] = P[H1 and H2]/P[H1] = 0.75. Anything else allows me to buy and sell bets and make a riskless profit.
Now, note that if H1 occurs, P[H2 | H1] is now the updated chance of getting H1 and H2. You observed some information and updated your probability accordingly. Also note that according to your model, the chance of H1 happening was 0.4, so the possible prices of "H1 and H2" after one coin flip were 0.75 with probability 0.4 and 0 with probability 0.6 (if the first flip were tails, "H1 and H2" is impossible). So the expected price of "H1 and H2" after one flip is (.75)(.4) + (0)(.6) = 0.3, which is also your price at t=0. That means the probability/price is a martingale according to your probability assignments, which, according to the Fundamental Theorem of Asset Pricing, again means no arbitrage is possible, unless we disagree on what events have probability zero.
So, the election forecasts are like this except we don't get to see all the inner workings. Nate doesn't quote things like the probability of the polls moving by a certain amount each day. We just get to see things like P[H1 and H2] before and after the first coin flip. But as long as it's possible that those come from a consistent set of conditional probabilities and Bayesian updating, there's no chance for arbitrage.