Replication Pricing and Risk-Neutral Valuation in a One-Period Binomial Model

# easy one step option replication


# easy one step option replication












I have the following question. Consider that you have a stock currently trading at 100\$. In one month it can jump to 120\$ with probability 99% and go to 80\$ with probability 1%. How much is call option with strike at 110$ worth now?

So when you do replication portfolio consisting of 0.25 stock and -20\$ gives exactly the same payoff as the option, hence the option should be worth 5\$. But again consider someone who wants to sell you security that in 99% cases gives you 10\$ and in remaining 1% gives you 0\$. How much would you be willing to pay for it? My whole intuition says it should be close to 10\$, but again replication tells 5\$.

Can someone comment on this? I feel really annoyed by this, as I tend to believe the replication model (because it replicates option with certainty), but it contradicts my intuition.

## Answer by Matt Wolf (score 6, accepted)

https://quant.stackexchange.com/a/4833

While another user touched on the hedging argument in order to reconcile your intuition with the correct value of the option he went off track (imho). I like to focus entirely on the hedging issue because it is key in understanding the differences in intuition and the fair price of such option. Unfortunately I have hardly ever found a simple 1-2 paragraph explanation in any of the PhD level papers. (should grad school students not start with the most basic understanding and be able to communicate such basics?). Anyway, here it goes:

2 important concepts apply that make all the difference:

1) In order to apply the fundamentals of derivatives pricing in a discrete space several conditions have to be met. One such condition is the condition of complete markets (or in the binomial case, completeness of the binomial model). The theorem of completeness states that "every derivative security can be replicated by trading in the underlying stock and money market. In a complete market, every derivative security has a unique price that precludes arbitrage" (Shreve, Stochastic Calculus for Finance I, p. 14, ed. 2004).

So, as long as those conditions are met, especially the one that requires the ability to trade in the underlying asset then you can price a derivative security as the discounted expected value applying risk-neutral probabilities. Thus, in your example the price of the option should be `$5`.

2) Now, why should the price be `$5` regardless of the real-world probabilities of attaining each future stock node? I think this is exactly where the intuition leads most astray. The answer is that if you use real-world probabilities to price an option then you also must discount the future expected payoffs at the appropriate real-world discount rate. An option of a stock that has a probability of 99% of going from 100-> 120 and 1% of going from 100-> 80 requires a much higher return and thus the option on such asset must be discounted at a much higher rate. Obviously it is hard to estimate the required returns investors demand on specific investment opportunities, and even hard to estimate the required return, used to discount future expected values, for the option of such underlying assets. That is the whole point of trying to price derivatives with the construct of risk-neutral probabilities because the discounting will be done through the risk-free interest rate one can receive as yield by investing in the money market/bond market (much can even be discussed about what risk-free nowadays means...). The finding that has been rewarded with a Nobel Price has exactly been that: The proof that using risk-neutral probabilities and constructing risk-less portfolios that make one indifferent about future outcomes will lead to the same solution than going the hard way by using real-life probabilities and discounting future expected cash flows at investors' required returns.

## Answer by Andrew (score 1)

https://quant.stackexchange.com/a/4824

the reason the derivative is worth that much is because it is replicating it with stock. in your example, you have a security that in 99% cases gives you \$10 and else \$0. However, what is the underlying that you use to hedge? Nothing. Therefore it must be worth the expected value or 9.90

## Answer by Andrew (score 0)

https://quant.stackexchange.com/a/4830

The reason is because we can hedge out uncertainty by using the stock.

In our case, let's say a stock is trading at 100, with 75% chance it will be 110, and with 25% chance it will be 90. How much would you sell a 100 strike call for?

In 75% of cases, the payoff is 10 else 0. You would think the value of this should be 7.50. Wrong.

However, now view it from the following perspective:

The second I buy the call, I can short the stock. We can derive what the risk-neutral probabilities are and in our case, it will be 50% up and 50% down. Now we can evaluate the value of the call as:

$$(S_{up} -K)^+ * p_{up} + (S_{down} -K)^+ * p_{down}$$

In our case, it will be $(110-100)*.5 = 5$. After I sell this asset for 5 dollars, I do the following: borrow half a share of stock.

In the case that it goes up to 110, I will lose exactly 5 dollars because I now have to buy back half a share of stock at 110 when I sold it at 100. $(110-100)*.5 = 5$, but I sold it to you for 5. I break even. Now in the case that the stock goes down to 90, I make exactly 5 dollars from the stock, but I have to pay you 10 dollars. However, I sold it to you originally for 5 and I also made 5. Again I break even. Therefore, ALL uncertainty is lost and the value of this option MUST be 5 or else, if you are willing to buy it for 7.50, I will sell it to you all day (quite literally :) ).

## Answer by experquisite (score 0)

https://quant.stackexchange.com/a/4841

You simply have to think about this enough to become convinced, it is unintuitive but correct. Next you convince yourself that this does not mean you cannot believe options to be cheap or expensive, if you have a different opinion about the future return distribution of the underlying.