MATH 425: BINOMIAL TREES

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1 MATH 425: BINOMIAL TREES G. BERKOLAIKO Summary. These notes will discuss: 1-level binomial tree for a call, fair price and the hedging procedure 1-level binomial tree for a general derivative, fair price and the hedging procedure many-level binomial tree for a European option; hedging many-level binomial tree for an American option; hedging short-cut formula for the tree 1. 1-Level Trees Example 1.1. Suppose the current price of an asset is S 0 = 100 and we assume that in one month its price will either be S u = 120 or S d = 90 with unknown probabilities. What is the fair price of a E = 110 call? According to our formulae, = S u E S u S d = = 1 3, therefore the price of the call is C = (S 0 S d = Suppose we sold the call for Let us go through the hedging procedure. We hedge by buying shares. Our initial poisiton is 1 call, 1/3 shares and / cash. At time t = T we sell the shares and make the payoff, if any. if S T = S u if S T = S d carry over sell shares = = make payoff 10 0 Balance Either way we made a small profit of This is because we charged for the call 7 cents above the fair price. Remark 1.1. Reality check: Since is a fraction and it is not possible to buy a fractional number of shares, in order to facilitate hedging the options are actually traded in the batches of 100, sometimes called option contracts. The prices, however, are quoted per single option (1 share worth of option contract. Example 1.2. Suppose, in the setting of the previous example we have a chance to purchase the calls for We would like to benefit from discrepancy between this price and the fair price. We buy 300, 000 calls (or 3, 000 call option contracts and hedge reversing the prescription we followed above. Namely, we short-sell share per option, i.e. we sell 300, 000 = 100, 000 (borrowed shares of stock. Our innitial position is 300, 000 call options, 100, 000 shares and 9, 040, 000 in the bank (proceeds of share sale minus the options premium. At time t = T we close our postion: 1

2 2 G. BERKOLAIKO if S T = S u if S T = S d carry over 9, 040, 000 9, 040, 000 buy back shares , 000 = 12, 000, , 000 = 9, 000, receive payoff , Balance 40, , 000 We made 40, 000 profit. This is simply the difference between the fair price and the price we paid times the number of options we so exploited: 300, = 40, Multi-level trees The 1-level trees rely on the rather unrealistic assumption that there are only two possible price outcomes at time t = T. However, we can chain individual 1-level calculations together to estimate the price of an option when the price of the underlying evolves in a more varied way. u 3 S 0 u 2 S 0 us 0 du 2 S 0 S 0 dus 0 ds 0 d 2 us 0 d 2 S 0 d 3 S0 Figure 1. A recombining tree with L = 3 levels. Suppose the time to expiration T is broken down into L equal 1 intervals of length t and over each time period the price either moves by few percent up or be few percent down. Let d < 1 < u be those price movement multiples, i.e. from S 0 at time t = 0 the price moves to us 0 or to ds 0 at time t = t and then to u(us 0 or d(us 0 and to u(ds 0 or d(ds 0 correspondingly. Note that uds 0 = dus 0 and we can combine two nodes together. Such a tree is called recombining. A tree with L = 3 levels is shown in Fig. Note that at time t = T we now have a whole range of prices and for each we can determine the payoff of the option in question. Consider an arbitrary subtree with one parent node and two children, such as shown in Fig. Suppose we already know the option values in the children nodes, Vu and Vd. Then we can use our 1-level tree formula to determine V prev. This formula involves the quasi-probability q, q = er t S prev Sd. Su Sd Taking into account that we know the value of S in terms of S prev, i.e. the formula for q simplifies to S u = us prev, S d = ds prev, (2.1 q = er t d u d, 1 The intervals do not need to be equal; this allows one to overcome the fact that there is no stock-price movement over a weekend, but the interest on deposits still accrues. Here, for simplicity, we only deal with equal time intervals

3 MATH 425: BINOMIAL TREES 3 Su, Vu... S prev, V prev Sd, Vd Figure 2. A subtree to which level-1 formulae can be applied to derive V prev from Vu, Vd and the prices S and the option value is calculated as (2.2 V prev = e r t ( Vu q + Vd (1 q. We note that in the tree with equal time intervals between nodes, the value of q is the same in every generation. Our results for 1-level trees can be formulated as the following statement. Corollary 2.1. At the node with option price S prev it costs V prev (given by equations (2.1 and (2.2 to set up a replicating portfolio which will have value Vu and Vd at the children nodes Su and Sd correspondingly. The replicating portfolio will contain shares of stock, (2.3 = V u S u Vd Sd and an amount of debt necessary to finance their purchase. We can now recursively apply equation (2.2 to iterate the option value backwards through a given tree, from the (known payoff values to the value of the option at present. This works for any European option (or their combination and, with minor modifications, for any American option. Once the value of the option is obtained, and the option is sold or bought at this value, the - hedging allows one to remove the risk of making a payoff resulting in a loss. Removing the risk of a loss also removes the chance of making a profit (due to no arbitrage principle. A hedging trader (a market-maker usually charges a small commission on the computed fair price and, if the price evolves according to the model, gets to keep it. We will now consider several examples. In all of the examples the values u and d will be given. The other values, such as initial stock price S 0, the interest rate r, expiration T and others will also be given, but in any realistic situation those are known to all interested parties. The values u and d are the important parameters of the valuation model and need to be estimated (or guessed in a somewhat scientific manner; this estimation is an important future topic in this course Example 1: Binomial tree for a European Call. Let us evaluate a European Call with S 0 = 100, E = 105, T = 3 weeks, r = 0.02, u = 1.1, d = 0.9, L = 3 and thus t = 3 weeks = 7. L 365 First we fill in the prices in the entire tree and also the final payoff, see Fig. 3.,

4 4 G. BERKOLAIKO S = 100 V 0 0 = S = = S = 90 0 = S = = S = 99 1 = S = 81 0 = S = = S = = 3.90 S = = 0 S = = 0 Figure 3. Binomial tree for a European Call, Example 1. Initial setup. The payoff is filled in according to the definition of the option being valued, in this case max(s E, 0. For example, 2 = max( , 0 = Next we compute q, q = er t d u d = e0.02 7/ and calculate the payoff in the L = 2 column according to (2.2, , 2 = e r t (28.10q (1 q 16.04, 1 = e r t (3.90q + 0(1 q 1.96, 0 = e r t (0q + 0(1 q = We enter these values into the tree, and carry on recursively, 1 = e r t (16.04q (1 q 9.02, 0 = e r t (1.96q + 0(1 q 0.98, and finally, see Fig. 4, V0 0 = e r t (9.02q (1 q This final value V0 0 = 5.01 is the fair value of the option. Applying Corollary 2.1 recursively, we see that given 5.01 it should be possible to set up a replicating portfolio that produces the correct payoff at T = 3. Note that one would have to rebalance the portfolio at every node. Let us demonstrate this with an example. We start off by selling a call at its fair price 5.01 and calculating according to (2.3 = We buy 0.40 shares of stock (at the price S = 100, borrowing the necessary cash, i.e Our position at the end of week-0 is 1 call option, 0.40 shares of stock and in cash.

5 MATH 425: BINOMIAL TREES 5 S = 100 V 0 0 = 5.01 S = = 9.02 S = 90 0 = 0.98 S = = S = 99 1 = 1.96 S = 81 0 = 0.00 S = = S = = 3.90 S = = 0.00 S = = 0.00 Figure 4. Binomial tree for a European Call, Example 1. Full iteration. Suppose the price went up and we find ourselves at the S = 110 node. The bank added some interest to our debt; we now have in cash (rounded to the nearest cent. Note that the value of our hedge (debt plus stock is = 9.00 is almost exactly equal to the value of the short option, Any discrepancy is due to rounding; for example, using a more precise = would result in 110 ( e r t We now need to adust our hedge. The new = ( /( We already hold 0.40 shares of stock so we need to buy 0.24 more, which adds to our debt. At the end of week-1 our position is 1 call option, 0.64 shares of stock and in cash. Suppose the price goes down and we are at the S = 99 node. Our bank balance is 61.42, is = ( /( To adust our hedge we sell 0.44 shares at S = 99, resulting in overall position of 1 call option, 0.20 shares of stock and in cash. Suppose the final move of the price is up, bringing us to S = node. Our cash held is 17.87, we close our hedging position by selling 0.20 shares at S = , and we make the payoff of The final balance is = 0.01 which is approximately 0, as predicted (the 1 cent discrepancy is due to rounding. The entire process is summarized in Table 1. Some remarks are in order: Calculating all numbers with more precision will result in the final balance closer to 0. For example, to 4 decimal places, the option price is and the deltas are 0 = , 1 = , 2 = , giving the final balance of In the process of hedging, the market-maker is buying high and selling low, which loses money. The option premium is compensating for the loss. The higher the oscillations of the market (volatility, the more premium should be charged to compensate. The option price is the price of volatility. One can also sell an option and hedge, in which case all numbers would be multiplied by 1. This way one can make profit from an option sold below fair price (provided the market obliges and behaves according to model predictions; basically, if the volatility is high enough.

6 6 G. BERKOLAIKO Time Actions Proceeds Position t = 0, S = 100 Sell call for = 0.40 Buy 0.40 shares call, 0.40 shares, cash t = 1, S = 110 Bring forward 1 call, 0.40 shares, cash = 0.64 Buy 0.24 shares call, 0.64 shares, cash t = 2, S = 99 Bring forward 1 call, 0.64 shares, cash = 0.20 Sell 0.44 shares call, 0.20 shares, cash t = 3, S = Bring forward 1 call, 0.20 shares, cash Sell 0.20 shares Make payoff cash Table 1. Hedging process for Example Example 2: Binomial Tree for an American Put. Consider an American put with the following parameters: S 0 = 20, E = 24, r = 0.10, T = 3 weeks. We will use a tree with L = 3 levels, t = 7/365, u = 1.1 and d = 0.9. The risk-neutral probability is q = ( e r t d /(u d = We construct the tree and fill in the payoff max(e S, 0. Using (2.2 we can calculate the values of the option at t = 2 if held to expiration, see Fig. 5. S = 20 V 0 0 = S = 22 1 = S = 18 0 = S = = 1.09 S = = 4.15 S = = 7.75 S = = 0.00 S = = 2.22 S = = 6.18 S = = 9.42 Figure 5. Binomial tree for an American Put, Example 2. First steps However, the holder of the option may choose to exercise early. At the node S = the payoff from an immediate exercise is 0.00, so it makes sense to hold the option. However, at S = 19.80, the immediate exercise gives 4.20, which is more than the value of the option if held to expiration. A prudent holder would choose to exercise. Our valuation needs to take this into account, therefore

7 MATH 425: BINOMIAL TREES 7 S = 20 V0 0 = 4.26, 4.00 S = 22 V1 1 = 2.61, 2.00 S = 18 V0 1 = 5.95, 6.00 S = V2 2 = 1.09, 0.00 S = V1 2 = 4.15, 4.20 S = V0 2 = 7.75, 7.80 S = = 0.00 S = = 2.22 S = = 6.18 S = = 9.42 Figure 6. Binomial tree for an American Put, Example 2. Final result. Time Actions Proceeds Position t = 0, S = 20 Sell put for = 0.85 (Shortsell 0.85 shares put, 0.85 shares, cash t = 1, S = 22 Bring forward 1 put, 0.85 shares, cash = 0.71 Buy 0.14 shares put, 0.71 shares, cash t = 2, S = Bring forward 1 put, 0.71 shares, cash Early Exercise! Make payoff 4.20 Buy 0.71 shares cash Table 2. Hedging process for Example 2 the value of the option at every node is the maximal of the early exercise value and the recursively computed value. At S = the early exercise is again optimal, it gives These adusted values are now used in the round of recursive calculation, for example 1 = e r t (1.09q (1 q 2.61, 0 = e r t (4.20q (1 q The result is shown in Figure 6. Let us now simulate hedging for a short put and the price path Up, Down, Down. When hedging for a written (short put, our calculated delta values are negative, indicating that one needs to short stock in order to hedge. Note that at the node S = the holder of the option should exercise early. The hedging process shown in Table 2 takes this into account and terminates at the step t = 2. Should the holder of the option decide to hold the option to expiration, the market-maker will make 0.05 e r t

8 8 G. BERKOLAIKO extra (modulo rounding errors since at the early exercise node the holder essentially sacrifices 5 cents by choosing the suboptimal course of action. Check it yourself! 2.3. European tree shortcut formula. To recap, we calculate the value of a given Europeanstyle option in a multi-level tree by using the payoff formula for the last level of the tree (the future time t = T and then propagating the values back to the present time using the formula ( V n 1 1 = e r t V n q + V 1(1 n q, in the notation of Figure 7. Here q is some value between 0 and 1. If our tree has many levels it would pre preferably to transition from the last level straight to the sought value V0 0. It turns out to be possible. 2 3 V1 1 V2 3 V0 0 V1 2 V0 1 V Figure 7. Binomial tree with notation for option values at different points. Let us describe the heuristics. For V0 0 we have ( V0 0 = e r t V1 1 q + V0 1 (1 q. Substituting the values for V1 1 and V0 1, namely ( V1 1 = e r t V2 2 q + V1 2 (1 q, ( V0 1 = e r t V1 2 q + V0 2 (1 q, we get (after some simplification After another level of substitution, we will get V 0 0 = e r2 t ( 2 q q(1 q + 0 (1 q 2. V 0 0 = e r3 t ( 3 q q 2 (1 q q(1 q (1 q 3, after which we should be able to recognize to Newton s binomial style of the formula. After all L levels we ought to have ( L (2.4 V0 0 = e (V rl t L L q L + LVL 1q L L 1 (1 q + V L 2 L 2q L 2 (1 q Using L t = T and the summation notation this can be shortened as the following formula.

9 MATH 425: BINOMIAL TREES 9 Lemma 2.1. In a European-style binomial tree evaluation with L levels, the option value V0 0 is given by (2.5 V 0 0 = e rt Proof. We will prove the formula (2.6 V 0 0 = e rn t L n V L ( L V n ( n q (1 q L. q (1 q n by induction and then substitute n = L. We have checked the formula for n = 1, 2, 3 already. Assuming it is valid for n = k 1 we would like to derive it for n = k. We have k 1 ( k 1 V0 0 = e r(k 1 t q (1 q k 1 V k 1 k 1 ( ( = e r(k 1 t e r t V+1q k + V k k 1 (1 q q (1 q k 1 ( k 1 ( k 1 k 1 ( k 1 = e rk t V+1 k q +1 (1 q k 1 + V q k (1 q k ( k ( k 1 k ( k 1 = e rk t Vs k q s (1 q k s + V q k (1 q k, s 1 s=1 where in the first sum we made the substitution s = + 1 and in the second we extended the sum to include = k since the binomial ( k 1 k is zero anyway. We are now going to similarly extend the first sum to include s = 0 and to rename s as again. ( k ( k 1 k ( k 1 V0 0 = e rk t V k q (1 q k + V q k (1 q k 1 k (( ( k 1 k 1 = e rk t V k + q (1 q k 1 = e rk t k V k ( k q (1 q k, where we used the property ( ( ( k 1 k 1 k + =. 1 We have now established the validity of the statement for n = k and thus finished the proof. Some remarks are in order. This idea will not work for American-style options as at every node we have an extra operation (comparison with the immediate payoff that will break the recursive nature of the process. The final formula still is not fully explicit since it involves iteration over L values. It is thus said to have order L complexity. Propagating through the whole tree takes L 2 operations (which is a lot more than L.

10 10 G. BERKOLAIKO The form of the final formula is the (present value of the expectation of the final payoff with respect to the binomial distribution induced by q. We will use this similarity and the central limit theorem for the binomial to simplify the formula further in the limit L.