# Why don't real-world probabilities affect the price of a call in a 1-step binomial model?

I was a bit hesitant to post this question because it seems so basic...but I wasn't able to figure it out on my own.

Say we setup a one-step binomial tree with $S_0=100$, $S_u=110$ and $S_d=90$, where $S_u$ and $S_d$ are the up and down possibilities for the stock price at time $T=1$. Let $K=100$ be the strike price of a call, and $r=10%$ be the continuously compounded risk-free interest rate.

Using a replicating portfolio (with some quantity $\Delta$ of the stock and borrowed money), I find the price of the call to be c_0 = 9.28\$$(\Delta=0.5). Now I understand that I don't need to know what the real-world probabilities are (of S_u and S_d), since the replicating portfolio...replicates the option payoff no matter the outcome. But just for fun, let's say Pr(S_1=S_u)=1\% and Pr(S_1=S_d)=99\%, in which case, on average, the call at time 1 would be worth 0.01*10 = 0.1. How would anyone be willing to pay 9.28 for that ? I'm pretty sure I'm missing something very basic, I hope someone can explain what it is. • Are you certain of the value of your call option? I may be wrong but it should come to 9.09. In this case, the RN probabilities themselves are 1 and 0! The option should be a certain payoff discounted by 10%. Dec 30, 2015 at 5:02 • Oh my bad ! Sd was supposed to be 90, thx for your comment! Dec 30, 2015 at 5:34 • You defined r=10. Does that mean if I invest 100 on the risk-free asset at T=0 I will get 110 back at T=1? If so, isn't it an arbitrage opportunity? Oct 12, 2016 at 13:08 • This is thoroughly answered in here, in Part 1: Discrete single-period model. Mar 14 at 18:28 ## 7 Answers It's a pitty that you don't show in your question how you get to your value for c_0 but the idea is that you build a portfolio X_0 = \Delta S_0 - \lambda and you infer the values for \Delta and \lambda so that X_1 = c_1 both in the up and down scenario. Then, because of the law of one price, X_0 = c_0. So for us X_1 = \Delta S_1 + (1+r) \lambda and we want it to be equal to \max(S_1 - K, 0 ) in both up and down case :$$ \Delta S_u + (1+r)\lambda = \max(S_u - K, 0 ) \Delta S_d + (1+r)\lambda = \max(S_d - K, 0 )$$This yields (using you values for S_d, S_0 = K and S_u):$$ \Delta S_u - \Delta S_d = \max(S_u - K, 0 ) - \max(S_d - K, 0 ) = S_u - K \Delta = \frac{ S_u - K }{S_u - S_d} = \frac{ 10}{20}=50\% $$Plugging this back into the second equation:$$ 50\% \cdot S_d + (1+r)\lambda = \max(S_d - K, 0 ) = 0\lambda = 50\% \cdot -\frac{S_d}{1+r} = -40.91$$This means that, by buying 50% of the share at time zero and borrowing 40.91 in the money market at time, you will replicate exactly the payoff of a call option at time 1. You did not mention probabilities at all here, they are completely irrelevant. The value of the replication portfolio at time 0 is:$$X_0 = \Delta S_0 - \lambda = 50\% \cdot S_0 - 40.91 = 50-40.91 = 9.09 =c_0$$and hence because holding X is exactly the same as holding the call itself, both should have the same value. If some option trader agrees to sell that call for less that \9.09, say \2.00, buy the call and sell X, you know that whatever you make with the call you will lose on X, but the time 0 profit of 9.09 - 2.00 = 7.09 is locked and will not change whatever the market does. In terms of probability, the only thing you need is to agree that \mathbb{P}(S_1=S_u) > 0, \mathbb{P}(S_1=S_u) > 0 and \mathbb{P}(S_1 \in \{S_u, S_d\} ) = 1. • Thank you SRKX, that was exactly the way I did it (except I used continuously compounded interest rate). I have to learn how to write mathematical formulas here.. Dec 31, 2015 at 7:37 "But just for fun, let's say Pr(S1=Su)=1% and Pr(S1=Sd)=99%, in which case, on average, the call at time 1 would be worth 0.01*10 = 0.1. How would anyone be willing to pay 9.28 for that ? I'm pretty sure I'm missing something very basic, I hope someone can explain what it is." How would anyone pay 100 for the stock given these probabilities? You don't seem to question that. And this is the 'basic' link that you are missing. Real world probabilities have been already incorporated in the pricing of the spot (loosely speaking). Derivatives are priced in relation to the stock: if the stock is unintuitive, as per your example, then derivatives will also appear stupid. The answer to this question has to do with two things: (i) risk aversion; and (ii) the current stock price, S_0. In short, the current stock price reflects the risk aversion of the average investor. For example, if you compute the weighted average of the two future stock prices (using the probabilities), you get the risk-neutral stock price, which is typically not the same as the current stock price, S_0, that you see in the binomial tree. The reason is that the current stock price reflects the average investor's risk aversion. Put differently, suppose that we change the probabilities from, say, 50/50 to 60/40 (60% probability to the upper node). This will then immediately be reflected in the current stock price, as an increase in S_0. That, in turn, will increase the price of the replicating portfolio, and hence also the price of the call option. (Note that you cannot change the probabilities without allowing the current stock price to change. If you think about that for a second, it will become obvious to you.) Conclusion: The current stock price already contains information about the probabilities, and therefore, we don't need them to price options. To use this method of valuing options, we assume risk neutral probability for calculating expectation of payoff and discount it with the risk free rate to arrive at the final value. If real world probability is used for expectation, we need to discount using real world (unknown) discount rate. In either case, we can't mix real world and risk neutral measures. So, you can't take real probabilities to calculate a 0.1 payoff expectation and discount with a risk free rate. Since we will be using risk free rate almost always for discounting purposes, we don't care about the real world probability (unless it is 0) • Thx for the answer Romeet, but it still leaves me unsatisfied. If I wanted to sell you that option (knowing what the real probabilities are), would you honestly buy it for 9 ? I think the kind of answer I'm looking for is: your scenario is impossible because...[insert reason here] Dec 30, 2015 at 18:29 Real probability has its use in deriving the stochastic discount factor. The fundamental equation in asset pricing is p = E[mX]. In the two-period discrete time world, we have p = \sum_{s \in S} Pr(s) m(s) X(s) for all possible states. Now consider the binomial tree world, we have a stock, a bond and a call option. We want to find a m that prices all three assets. Let's first consider pricing stock and bonds. For the stock, 100=S_0 = p_um(u)S_1(u) + p_dm(d)S_1(d) = 0.99m(u)110+0.01m(d)90, and for the bond 1/(1+0.1)=B_0 = p_um(u)1 + p_dm(d)1 = 0.99m(u)+0.01m(d). With two equations and two unknowns, you can solve for m. Now you can use this m to price the call option payoff as C_0 = p_um(u)(110-100)+p_dm(d)(100-100) = p_um(u) \cdot 10. More explicitly, the link between risk-neutral and the SDF approach is to consider the following:$$\begin{align*} p &= \sum_{s \in S} Pr(s) m(s) X(s)\\ &= \sum_{s \in S}Pr(s) m(s) \cdot \sum_{s \in S} \frac{Pr(s)m(s)}{\sum_{s \in S} m(s)} X(s)\\ &= \frac{1}{r} \cdot \sum_{s \in S} \frac{Pr(s)m(s)}{\sum_{s \in S} Pr(s)m(s)} X(s)\\ &= \frac{1}{r} \cdot \sum_{s \in S} q(s) X(s)\\ &= \frac{1}{r} E^*[X]\\ \end{align*}$Where$\sum_{s \in S}Pr(s) m(s) = 1/r$and$q(s) =\frac{Pr(s)m(s)}{\sum_{s \in S} Pr(s) m(s)}$. You can check for yourself that$q(s)$is indeed the risk-neutral probability. In a general equilibrium pricing context,$m$is a measure of risk aversion. Referring back to the calculation of$m$, the price of the stock and bond reflects how high/low$m$, i.e. people's risk aversions are. So a set of real probability reflects the set of risk aversion, or$m$in mind. • This answer had me mesmerized, thank you so much ! <br/> In my example I had Pu = 0.01%, which gives m(u) = 90.9. Dec 31, 2015 at 1:05 • It's as if the existence of the bond forces the discount factor to be ridiculously huge. So is the conclusion that the only fair price in this example is a price no one would ever pay ? Dec 31, 2015 at 1:13 • @KarimL It is the setup that is weird. If these sets of prices are offered in real world, the option should still be priced as you have calculated, due to the arbitrage opportunity that would otherwise arise. The question of whether anyone would buy the option is not addressed in the equilibrium. In equilibrium, the markets for all three assets will clear.$m(u)$is high means people discount payoffs in good state a lot, i.e. they don't care much about the up state. They care a lot more about down state, and assigned a very low$m(d)$to the low state. Dec 31, 2015 at 1:30 Your scenario is impossible because you are missing the no-arbitrage hypothesis (no free lunch). Replication Portfolio: -Buy 0.5 Shares and short 1 call : Payoff Up :$ 0.5*110-10 = 45$Payoff Down :$ 0.5*90=45 $(call is not exercised) Present value Portfolio = 45 / (1.1) = 40.91 Then the call price is 9.09 (see @SRKX answer). Initial cost of portfolio :$ 50 - 9.09 = 40.91 $How would anyone be willing to pay 9.09$ for that?

Because, no matters the real probabilities, at a different price there are free lunch possibilities.

• If the price of the call is superior to 9.09 $, let's say 10 dollars the investor can create the replication portfolio for less than 40.91 at 40 (50-10=40) by borrowing 40. At maturity the portfolio worth 45 and he pays back only 44 dollars(40*1.1). He will receive a free lunch of 1 dollar. • If the price of the call is inferior to 9.09$, let's say 9 dollars the investor short sell the portfolio at its initial cost 41 dollars (50-9) and can invest 40.91 at the risk free rate. At maturity he will receive 45 dollars (40.91*1.1), this will cover the price of the portfolio and he will make an initial free benefit (41-40.91=0.09).

In both cases there is no risk and no initial investment so you are violating the no-arbitrage hypothesis. The real probabilities play no role.

I am also a beginner, and puzzled by this question for a while. Now my understanding is that the non-arbitrage pricing theory is not giving you only one price, but a function that maps the strike price and other conditions to a price that a market maker could offer. In other words, in the world of derivative market, you can think of this as a supply curve. What you are describing, stemming from the perspective of a potential buyer of the product, is on the demand side of this market. You could be right that in some extreme case no one wants the market maker's price, so the demand and supply curve do not intersect there.