Unit 4, lesson 32

Transaction costs and no arbitrage price intervals

How proportional bid-ask costs turn one replication price into an interval, how superreplication with covering hedges gives seller and buyer price bounds, and how shadow prices certify no arbitrage inside them.

  • Video 32
  • Incompleteness, early exercise and transaction costs

A dealer is asked for a two-way quote on a cash-settled call while the stock trades at 99 bid and 101 ask. Without trading costs one replicating portfolio would fix a single price, but now buying the hedge, selling it later and covering a short position all happen at different prices. The dealer needs the lowest price at which selling the call can be hedged without risk, and the highest price at which buying it can be hedged the same way. This lesson computes both bounds in a one-period model, shows that they bracket the frictionless price, and checks with shadow prices that no price between them creates an arbitrage.

Key terms

Bid-ask spread

Definition At each date tt the stock can be bought at the ask price StaS^{a}_t and sold at the bid price StbS^{b}_t, where 0<Stb≤Sta0 < S^{b}_t \le S^{a}_t. The bid-ask spread is Sta−StbS^{a}_t - S^{b}_t, and the mid-price is (Sta+Stb)/2(S^{a}_t + S^{b}_t)/2. On this page the cash account BtB_t has no friction. It starts at B0=1B_0 = 1 and borrows and lends at the same growth factor R>0R > 0 per period, so B1=RB_1 = R. A position of θ>0\theta > 0 shares bought and sold again at once loses θ(Sta−Stb)\theta(S^{a}_t - S^{b}_t). With a spread the cash flow of a trade depends on its direction, so a portfolio no longer has one value that is linear in its holdings, and the self-financing condition must use the cash actually paid or received.

In plain words Buying costs the higher price and selling brings in the lower one. The mid-price says neither what a purchase costs nor what a sale brings in.

Example In the constructed example of this lesson the mid-price is 100 today, with bid 99 and ask 101, a spread of 2. Buying one share and selling it straight back loses 2 even though the mid-price has not moved. At the end of the period the mid-price is 80 or 120, with bid 79.2 and ask 80.8 in the first state and bid 118.8 and ask 121.2 in the second. The cash account grows by the factor R=1.05R = 1.05 over the period.

Proportional transaction costs

Definition In a model with proportional transaction costs there is a mid-price process St>0S_t > 0 and a constant λ\lambda with 0≤λ<10 \le \lambda < 1, and

Stb=(1−λ) St,Sta=(1+λ) St.\begin{gathered} S^{b}_t = (1 - \lambda)\, S_t, \\[0.4em] S^{a}_t = (1 + \lambda)\, S_t. \end{gathered}

The value λ=0\lambda = 0 is the frictionless market. Write x+=max⁡(x,0)x^{+} = \max(x, 0) and x−=max⁡(−x,0)x^{-} = \max(-x, 0). Changing a stock holding by Δθ\Delta\theta shares at date tt costs the cash

(Δθ)+Sta−(Δθ)−Stb=St Δθ+λSt∣Δθ∣,\begin{aligned} &(\Delta\theta)^{+} S^{a}_t - (\Delta\theta)^{-} S^{b}_t \\[0.4em] &\quad = S_t\, \Delta\theta + \lambda S_t \lvert \Delta\theta \rvert, \end{aligned}

and liquidating a holding θ\theta at date tt, by selling a long holding at the bid or buying back a short holding at the ask, brings in

Lt(θ)=θ+Stb−θ−Sta=θSt−λSt∣θ∣.\begin{aligned} L_t(\theta) &= \theta^{+} S^{b}_t - \theta^{-} S^{a}_t \\[0.4em] &= \theta S_t - \lambda S_t \lvert \theta \rvert. \end{aligned}

The trading cost is convex in Δθ\Delta\theta, the liquidation value is concave in θ\theta, and both are multiplied by kk when the position is multiplied by a number k>0k > 0. In the one-period model of this page a portfolio (Δ,b)(\Delta, b) buys Δ\Delta shares at date 0, a short sale when Δ<0\Delta < 0, and holds bb in the cash account, a debt when b<0b < 0. Its initial cost is

c(Δ,b)=S0Δ+λS0∣Δ∣+b,c(\Delta, b) = S_0 \Delta + \lambda S_0 \lvert \Delta \rvert + b,

and at date 1 it brings in L1,j(Δ)+bRL_{1,j}(\Delta) + bR in terminal state jj, where L1,jL_{1,j} is the liquidation value at the prices of that state.

In plain words Each share traded costs an extra fraction λ\lambda of the mid-price, whichever way it is traded. A purchase and a sale of the same size cancel in shares but not in cash, and closing a long position and closing a short position use opposite sides of the market.

Example With λ=0.01\lambda = 0.01, buying 50/99≈0.50505150/99 \approx 0.505051 shares today costs 101⋅50/99=5050/99≈51.010101101 \cdot 50/99 = 5050/99 \approx 51.010101, which the formula writes as 100⋅50/99+0.01⋅100⋅50/99100 \cdot 50/99 + 0.01 \cdot 100 \cdot 50/99. Selling them at the end of the period brings in 79.2⋅50/99=4079.2 \cdot 50/99 = 40 or 118.8⋅50/99=60118.8 \cdot 50/99 = 60. A short sale of 50/101≈0.49505050/101 \approx 0.495050 shares brings in 99⋅50/101=4950/101≈49.00990199 \cdot 50/101 = 4950/101 \approx 49.009901 today, and buying the shares back costs 80.8⋅50/101=4080.8 \cdot 50/101 = 40 or 121.2⋅50/101=60121.2 \cdot 50/101 = 60.

Covering hedge

Definition Let a claim pay the cash amount XjX_j in terminal state jj. A covering hedge of XX is a portfolio (Δ,b)(\Delta, b) whose liquidation cash at maturity covers the claim, so that

L1,j(Δ)+bR≥XjL_{1,j}(\Delta) + bR \ge X_j

in every terminal state jj. A seller of XX needs a covering hedge of XX. A buyer of XX who wants to lock in the purchase needs a covering hedge of the negative claim −X-X, which pays −Xj-X_j, because the claim together with that hedge leaves a nonnegative amount in every state. A covering hedge that holds with equality in every state replicates the claim after costs. Long holdings are valued at the terminal bid and short holdings at the terminal ask, so a hedge computed from mid-prices is in general not a covering hedge.

In plain words A covering hedge is a position in stock and cash that, once closed at the prices actually available, leaves enough cash to pay what is owed whatever happens.

Example The claim is a call on the mid-price with strike 100, settled in cash, so X=(0,20)X = (0, 20) in the states with mid-price 80 and 120. To cover it with a long position, match the claim at the terminal bids. The holding is 20/(118.8−79.2)=50/9920/(118.8 - 79.2) = 50/99 shares, whose sale brings in 40 or 60, so a debt of 40 at maturity, which is a loan of 40/1.05=800/21≈38.09523840/1.05 = 800/21 \approx 38.095238 today, leaves 0 and 20. To cover −X=(0,−20)-X = (0, -20), sell short 20/(121.2−80.8)=50/10120/(121.2 - 80.8) = 50/101 shares and put 800/21800/21 in the cash account. Buying the shares back costs 40 or 60 and the cash pays 40, which leaves 0 and −20-20. Both hedges cover their claims exactly. With the frictionless hedge ratio 20/(120−80)=0.520/(120 - 80) = 0.5 and the same cash amounts, neither position covers its claim, as the second question below shows for the seller.

Superreplication

Definition Superreplication means covering a claim at the lowest possible initial cost. The superreplication price of XX is

V‾(X)=inf⁡{c(Δ,b):L1(Δ)+bR≥X},\begin{aligned} \overline{V}(X) = \inf \bigl\{ &c(\Delta, b) : \\ &L_1(\Delta) + bR \ge X \bigr\}, \end{aligned}

where the inequality must hold in every terminal state. For a fixed holding Δ\Delta the cheapest cash position is b=max⁡j(Xj−L1,j(Δ))/Rb = \max_j (X_j - L_{1,j}(\Delta))/R, so V‾(X)\overline{V}(X) is the infimum over Δ\Delta of

f(Δ)=S0Δ+λS0∣Δ∣+max⁡jXj−L1,j(Δ)R.\begin{aligned} f(\Delta) &= S_0 \Delta + \lambda S_0 \lvert \Delta \rvert \\[0.4em] &\quad + \max_j \frac{X_j - L_{1,j}(\Delta)}{R}. \end{aligned}

In this one-period model ff is convex and piecewise linear, so a point where its slope changes from negative to positive is a minimum. Matching the claim in every state gives a covering hedge, and this slope check shows whether it is the cheapest one. A consistent price system that values the claim at the hedge’s cost, as in the last card, is another certificate of optimality. With λ=0\lambda = 0 and a claim that can be replicated, the superreplication price in a market without arbitrage is the replication price.

In plain words Among all stock and cash positions that keep the seller safe, find the cheapest. Its cost is the least amount of money from which the claim can be paid in every state.

Example Take the call with λ=0.01\lambda = 0.01. For Δ≥0\Delta \ge 0 the shares are sold at the bids, so the cash at maturity must be at least −79.2Δ-79.2\Delta in the down state and 20−118.8Δ20 - 118.8\Delta in the up state. For 0≤Δ≤50/990 \le \Delta \le 50/99 the up state binds, the cost is 101Δ+(20−118.8Δ)/1.05101\Delta + (20 - 118.8\Delta)/1.05, and its slope is 101−118.8/1.05=−85/7101 - 118.8/1.05 = -85/7. For Δ≥50/99\Delta \ge 50/99 the down state binds and the slope is 101−79.2/1.05=179/7101 - 79.2/1.05 = 179/7. For Δ<0\Delta < 0 the shares are bought back at the asks, the up state binds, and the slope is 99−121.2/1.05=−115/799 - 121.2/1.05 = -115/7, so going short only raises the cost. The minimum is at Δ=50/99\Delta = 50/99,

V‾=101⋅5099−80021=8950693≈12.914863,\begin{aligned} \overline{V} &= 101 \cdot \tfrac{50}{99} - \tfrac{800}{21} \\[0.4em] &= \tfrac{8950}{693} \approx 12.914863, \end{aligned}

which is above the frictionless replication price 250/21≈11.904762250/21 \approx 11.904762, obtained from the hedge ratio 0.5 and the same loan 800/21800/21.

Seller and buyer price bounds

Definition The seller price bound of a claim XX is V‾(X)\overline{V}(X), and the buyer price bound is

V‾(X)=−V‾(−X),\underline{V}(X) = -\overline{V}(-X),

the largest price a buyer can pay for XX and still cover the position. Three facts follow from the definitions. If the claim can be sold at a price p>V‾(X)p > \overline{V}(X), selling it and buying a covering hedge that costs less than pp leaves positive cash today and a nonnegative amount in every state at maturity, an arbitrage. If the claim can be bought at p<V‾(X)p < \underline{V}(X), buying it and setting up a covering hedge of −X-X is an arbitrage in the same way. If the stock and cash market has no arbitrage, then V‾(X)≤V‾(X)\underline{V}(X) \le \overline{V}(X). Otherwise hedges of XX and of −X-X could be found whose total cost is negative, and combining positions never raises the trading cost and never lowers the liquidation cash, so the combined position would cost less than zero and bring in at least X−X=0X - X = 0 in every state, which is an arbitrage. The bounds come from covering arguments only, and prices between them cannot be excluded by those arguments. When λ=0\lambda = 0, the market has no arbitrage and XX can be replicated, both bounds equal the replication price.

In plain words The seller bound is the cheapest safe way to deliver the claim, and the buyer bound is the most a buyer can pay and still be safe. Outside the interval someone can lock in a riskless profit. The width of the interval comes from the trading sides of the hedges, not from an error margin attached to a mid-price value.

Example For the call with λ=0.01\lambda = 0.01, the buyer bound is the short-sale proceeds 4950/1014950/101 minus the cash 800/21800/21 set aside, which is 23150/2121≈10.91466323150/2121 \approx 10.914663. The interval runs from 10.91466310.914663 to 12.91486312.914863, a width of about 2.0002002.000200. It contains the frictionless price 11.90476211.904762, which is not its midpoint 11.91476311.914763. For 0≤λ≤1/150 \le \lambda \le 1/15 hedges of the same form stay optimal, and

V‾=50 1+λ1−λ−80021,V‾=50 1−λ1+λ−80021.\begin{aligned} \overline{V} &= 50\,\frac{1 + \lambda}{1 - \lambda} - \frac{800}{21}, \\[0.4em] \underline{V} &= 50\,\frac{1 - \lambda}{1 + \lambda} - \frac{800}{21}. \end{aligned}

The interval widens as the spread grows, to about 9.9439789.943978 and 13.94557813.945578 at λ=0.02\lambda = 0.02, and both bounds tend to 250/21250/21 as λ→0\lambda \to 0. For λ>1/15\lambda > 1/15 the stock no longer helps the seller, whose bound becomes 20/1.05=400/21≈19.04761920/1.05 = 400/21 \approx 19.047619, the cost of holding 20 in cash.

Shadow price

Definition A shadow price is a process S~t\tilde S_t with

Stb≤S~t≤StaS^{b}_t \le \tilde S_t \le S^{a}_t

at every date and in every state, together with a probability measure Q\mathbb{Q} that gives every state positive probability and makes S~t/Bt\tilde S_t/B_t a Q\mathbb{Q}-martingale. The pair is called a consistent price system. In one period with growth factor RR the martingale condition reads S~0=EQ[S~1]/R\tilde S_0 = \mathbb{E}^{\mathbb{Q}}[\tilde S_1]/R. If a consistent price system exists, the bid-ask market has no arbitrage. To see this, compare each trade with a frictionless market at the shadow price. Purchases cost at least the shadow price and sales bring in at most the shadow price, so c(Δ,b)≥ΔS~0+bc(\Delta, b) \ge \Delta \tilde S_0 + b and L1(Δ)≤ΔS~1L_1(\Delta) \le \Delta \tilde S_1 in every state, hence

EQ[L1(Δ)+bR]≤R (ΔS~0+b)≤R c(Δ,b).\begin{aligned} &\mathbb{E}^{\mathbb{Q}}\bigl[L_1(\Delta) + bR\bigr] \\[0.4em] &\quad \le R\,(\Delta \tilde S_0 + b) \\[0.4em] &\quad \le R\, c(\Delta, b). \end{aligned}

A portfolio with c(Δ,b)≤0c(\Delta, b) \le 0 whose liquidation cash is nonnegative in every state therefore has liquidation cash zero in every state, because Q\mathbb{Q} gives every state positive probability. The same chain shows that every covering hedge of XX costs at least EQ[X]/R\mathbb{E}^{\mathbb{Q}}[X]/R. So every claim value EQ[X]/R\mathbb{E}^{\mathbb{Q}}[X]/R from a consistent price system lies between V‾(X)\underline{V}(X) and V‾(X)\overline{V}(X), and trading the claim at such a value adds no arbitrage. The converse, that no arbitrage produces a consistent price system, is not proved here, and the stronger notion that asks for shadow prices strictly inside the spread belongs to the dynamic theory of transaction costs, which is outside this lesson.

In plain words A shadow price is a frictionless price path that stays inside the spread at every node and has a pricing measure. Real trades are never better than trades at the shadow price, so an arbitrage in the real market would also be an arbitrage in the shadow market, which has none.

Example The seller bound of the call comes from the shadow price equal to the ask 101 today and to the bids 79.2 and 118.8 at maturity, with up probability

qU=1.05⋅101−79.2118.8−79.2=179264≈0.678030,\begin{aligned} q^{U} &= \frac{1.05 \cdot 101 - 79.2}{118.8 - 79.2} \\[0.4em] &= \frac{179}{264} \approx 0.678030, \end{aligned}

and claim value qU⋅20/1.05=8950/693q^{U} \cdot 20/1.05 = 8950/693. The buyer bound comes from the bid 99 today and the asks 80.8 and 121.2 at maturity, with up probability qL=(1.05⋅99−80.8)/40.4=463/808≈0.573020q^{L} = (1.05 \cdot 99 - 80.8)/40.4 = 463/808 \approx 0.573020 and claim value 23150/212123150/2121. Both probabilities lie strictly between 0 and 1, so both are consistent price systems even though their shadow prices sit on the edges of the spread. Unlike the open interval of the frictionless incomplete market in Market completeness and nonunique prices, the endpoints here can be quoted without creating an arbitrage. For a weight θ\theta with 0≤θ≤10 \le \theta \le 1, mix the two systems state by state,

qjθ=(1−θ) qjL+θ qjU,qjθS~1,jθ=(1−θ) qjLS~1,jL+θ qjUS~1,jU,\begin{aligned} q^{\theta}_j &= (1 - \theta)\, q^{L}_j + \theta\, q^{U}_j, \\[0.4em] q^{\theta}_j \tilde S^{\theta}_{1,j} &= (1 - \theta)\, q^{L}_j \tilde S^{L}_{1,j} \\[0.4em] &\quad + \theta\, q^{U}_j \tilde S^{U}_{1,j}, \end{aligned}

where qjLq^{L}_j and qjUq^{U}_j are the probabilities of terminal state jj in the two systems, and set S~0θ=(1−θ)⋅99+θ⋅101\tilde S^{\theta}_0 = (1 - \theta) \cdot 99 + \theta \cdot 101. Each S~1,jθ\tilde S^{\theta}_{1,j} is a weighted average of two prices in the same spread, the martingale condition holds because it holds for both systems, and the claim value is (1−θ)V‾+θV‾(1 - \theta)\underline{V} + \theta\overline{V}. So every price in the closed interval comes from a consistent price system. Mixing the probabilities and the shadow prices with unrelated weights would in general break the martingale condition. At θ=1/2\theta = 1/2 the shadow price today is 100, the up probability is about 0.6255250.625525, and the claim value is about 11.91476311.914763.

Check yourself

  1. In the same model with λ=0.01\lambda = 0.01, a put on the mid-price with strike 100, settled in cash, pays 20 in the state with mid-price 80 and 0 in the state with mid-price 120. Find its seller and buyer price bounds and the hedges that attain them.
Show the answer

The seller covers X=(20,0)X = (20, 0), which falls as the stock rises, with a short position of ss shares bought back at the asks and cash bb. Matching both states gives 1.05 b−80.8 s=201.05\,b - 80.8\,s = 20 and 1.05 b−121.2 s=01.05\,b - 121.2\,s = 0, so s=20/40.4=50/101s = 20/40.4 = 50/101 and b=60/1.05=400/7≈57.142857b = 60/1.05 = 400/7 \approx 57.142857. The short sale brings in 99⋅50/101=4950/10199 \cdot 50/101 = 4950/101 today, so

V‾=4007−4950101=5750707≈8.132956.\begin{aligned} \overline{V} &= \tfrac{400}{7} - \tfrac{4950}{101} \\[0.4em] &= \tfrac{5750}{707} \approx 8.132956. \end{aligned}

The buyer covers −X=(−20,0)-X = (-20, 0) with a long position of 50/9950/99 shares sold at the bids for 40 or 60 and a loan of 400/7400/7, which needs 60 at maturity. That hedge costs 101⋅50/99−400/7=−4250/693101 \cdot 50/99 - 400/7 = -4250/693, so the buyer bound is 4250/693≈6.1327564250/693 \approx 6.132756. The same slope check as for the call shows that both hedges are the cheapest. The interval contains the frictionless put price 38⋅20/1.05=50/7≈7.142857\tfrac38 \cdot 20/1.05 = 50/7 \approx 7.142857.

  1. A classmate hedges the sold call with the frictionless hedge, 0.50.5 shares bought at the ask and a loan of 800/21800/21. Show that this portfolio does not cover the call, find the cheapest covering hedge that keeps 0.50.5 shares, and compare its cost with the seller bound.
Show the answer

At maturity the 0.50.5 shares are sold at the bids for 39.639.6 or 59.459.4, and the loan requires 40, which leaves −0.4-0.4 in the down state and 19.419.4 in the up state, short of the claim by 0.40.4 and 0.60.6. With 0.50.5 shares the cash at maturity must be at least max⁡(0−39.6, 20−59.4)=−39.4\max(0 - 39.6,\ 20 - 59.4) = -39.4, so the loan can be at most 39.4/1.05≈37.52381039.4/1.05 \approx 37.523810. The cost is

0.5⋅101−39.41.05≈50.5−37.523810=12.976190,\begin{aligned} &0.5 \cdot 101 - \tfrac{39.4}{1.05} \\[0.4em] &\quad \approx 50.5 - 37.523810 = 12.976190, \end{aligned}

which exceeds the seller bound 12.91486312.914863 by about 0.0613280.061328. That difference is the size 85/785/7 of the slope of the cost function below 50/9950/99 times the shortfall 50/99−1/2=1/19850/99 - 1/2 = 1/198 in the holding, which gives 85/1386≈0.06132885/1386 \approx 0.061328. The frictionless hedge ignores that shares are sold at the bid, so it either fails to cover or costs more than needed.

  1. A dealer offers to buy the call from you at 13.20. Construct an arbitrage. Then explain why a bid of 12.50 cannot be exploited this way, and give a consistent price system that shows no strategy exploits it.
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Sell the call to the dealer for 13.20, buy 50/9950/99 shares at 101 and borrow 800/21800/21. This costs 8950/693≈12.9148638950/693 \approx 12.914863, so about 0.2851370.285137 is left today. At maturity the shares are sold at the bids for 40 or 60, the loan takes 40, and the call takes 0 or 20, which leaves exactly 0 in both states. Putting the 0.2851370.285137 in the cash account turns this into a sure gain of about 0.2993940.299394 at maturity from a position that cost nothing.

A bid of 12.50 is below the seller bound, so selling the call does not pay for any covering hedge. To show that no strategy works, take the up probability q=12.5⋅1.05/20=21/32q = 12.5 \cdot 1.05/20 = 21/32, which values the call at 12.50, with shadow prices equal to the bids 79.2 and 118.8 at maturity. The martingale condition gives the shadow price today

S~0=2132⋅118.8+1132⋅79.21.05=280528≈100.178571,\begin{aligned} \tilde S_0 &= \frac{\tfrac{21}{32} \cdot 118.8 + \tfrac{11}{32} \cdot 79.2}{1.05} \\[0.4em] &= \frac{2805}{28} \approx 100.178571, \end{aligned}

which lies between the bid 99 and the ask 101. Both state probabilities are positive, so this is a consistent price system that also prices the call at the dealer’s bid, and by the shadow price argument no combination of selling calls at 12.50 and trading the stock and cash is an arbitrage.

How this lesson connects