Fin Case Work3.1 Please check the instruction attachment below. Lecture Note 3.1: Robust Bounds on Options Prices We now turn our attention to options mar

Click here to Order a Custom answer to this Question from our writers. It’s fast and plagiarism-free.

Please check the instruction attachment below.

Lecture Note 3.1: Robust Bounds on Options Prices

We now turn our attention to options markets.

Over the next weeks we will examine in detail the real-world details of

pricing options by no-arbitrage. In doing so, we will review some famil-

iar modelling approaches and examine their robustness. To start, we
will consider the most fundamental properties of options prices that

are extremely robust in that they do not depend on any stochastic

modelling assumption.

We have seen that forwards and swaps can be replicated by portfolios

involving the underlying asset or assets, and risk-free borrowing and

lending. The initial price of such a replicating portfolio then had to be

the price of the derivative.

This is the basis for the options relationships we will examine today.

We will relate prices of options to prices of portfolios of the underlying

asset, bonds, and…other options.

Outline:

I. Background and Assumptions

II. Basic Arbitrage Relationships

III. Put-Call Parity

IV. Relationships with Dividends or Carry Costs

V. Options as a Function of Strike Price

VI. Summary

I. Background.

� I will assume everybody is familiar with how options contracts

work. Here’s the usual notation:

Current date t

Maturity or expiration date T

Price of the underlying asset St
Price of riskless bond Bt,T = e

−r(T−t)

Exercise (strike) price K

Value of a European call c(S, K, B, t, T)

Value of an American call C(S, K, B, t, T)

Value of a European put p(S, K, B, t, T)

Value of an American put P(S, K, B, t, T)

� For today’s purposes we will be thinking about an option on

a traded underlying financial asset (for example a stock or a

currency) that can be easily sold short and has a non-negative

price.

� Let us assume we are considering exchange traded options,

where the terms of the contract are standardized, and there

is a (riskless) central counterparty that clears all trades.

� Also today we are going to assume there are no transactions costs.

� We also –usually – need to assume that all carry costs are

known in advance. Besides interest rates, this includes payouts,

borrowing fees, storage costs, etc.

II. Basic Arbitrage Relations:

� To start, consider an option on a share of stock. Assume there

are no fees for borrowing it.

� All of the following relations were proved by Robert Merton in

1973.

� Of these, (A)-(G) hold regardless of whether or not the stock

pays dividends.

(A) Options never have a negative value:

c(S, K, t, T) ≥ 0 C(S, K, t, T) ≥ 0
p(S, K, t, T) ≥ 0 P(S, K, t, T) ≥ 0

(B) An American option is at least as valuable as the corresponding
European one:

C(S, K, t, T) ≥ c(S, K, t, T)
P(S, K, t, T) ≥ p(S, K, t, T)

(C) A call is never worth more than the stock and a put is never
worth more than the exercise price

C(S, K, t, T) ≤St c(S, K, t, T) ≤ St
P(S, K, t, T) ≤K p(S, K, t, T) ≤ K

Why? Take the bound for p. Suppose p > K. Sell the put,

invest K, and still have cash left today. At T , the worst that

can happen is the stock is worthless. You are obligated to pay

K for something worth zero. But your investment is worth at

least K. So you have left over cash at T too.

� Wait….I made an assumption there: that interest rates are

non-negative!

� Let’s be a little more accurate:

(D) European puts are never worth more than the present value of
the exercise price:

p(S, K, t, T) ≤ K · Bt,T
(and – possibly – Bt,T > 1).

(E) An American option is worth at least its exercised (or “intrin-
sic”) value, i.e., how much you would make if you exercised

today.

C(S, K, t, T) ≥ max[0, St − K]
P(S, K, t, T) ≥ max[0, K − St]

(F) American options are more valuable with more time to maturity
i.e., for T2 > T1,

C(S, K, t, T2) ≥C(S, K, t, T1)
P(S, K, t, T2) ≥P(S, K, t, T1)

Obvious? Then why doesn’t it hold for European options as
well? Or does it?

Consider carefully the steps in the American proof.

� Suppose you can take in cash today selling one-month op-

tions and buying otherwise identical two-month options.

� In one month our short option is worth its intrinsic value.

� If that is negative, and the option is exercised….

� Now suppose you can take in cash today selling one-month

European options and buying two-month ones.

� To have an arbitrage, we have to be CERTAIN we can sell

our long option in one month for at least its intrinsic value.

� What is our remaining option worth? We don’t know, but….

It must be at least as much as its lower no-arbitrage bound.

� Can European options (calls or puts) ever be worth less than

their intrinsic value?

A partial answer is the following:

(G) For a stock that does not pay dividends:

c(S, K, t, T) ≥ max[0, S − K · Bt,T ]
C(S, K, t, T) ≥ max[0, S − K · Bt,T ]

Proof: To prove this, we only need to show (why?)

c(S, K, t, T) ≥ S − K · Bt,T.

If c < S − K · B, we have an arbitrage:
Transaction Cash flow (at t) Payoff (at T)

−c max[0, ST − K]
St −ST

−K · B K
Combined S − K · B − c max[0, ST − K]

−[ST − K]

(H) For European puts on non-dividend-paying stocks, a similar ar-
gument shows that p(S, K, t, T) ≥ max [0, K · B − S] .

� Combining rules (C) and (H) implies that the value of a

European call on a non-dividend-paying stock must lie in

the shaded region:

max
[
0, St − K · Bt,T

]
≤ c(S, K, t, T) ≤ St.

� Combining (D) and (I), we see that the value of a Euro-

pean put on a non-dividend-paying stock must lie within the

shaded region:

max
[
0, K · Bt,T − St

]
≤ p(S, K, t, T) ≤ K · Bt,T.

� Of course if B < 1 then B · K is to the left of K in the
plots.

(I) If that’s true, we can now sometimes conclude that European
calls are worth more than their intrinsic value.

� So we can refine point (G) above:

(G’) For T2 > T1, c(S, K, t, T2) ≥ c(S, K, t, T1) if interest
rates are non-negative and the underlying does not pay a

dividend.

� There’s another important consequnce of the call lower bound:

(J) For calls on a non-dividend-paying stock, if interest rates are
positive, we must have

C(S, K, t, T) = c(S, K, t, T).

� Proof. The options only differ by the right to exercise the

American option before T .Early exercise is clearly subopti-

mal unless we have C(t) ≤ max[0, St − K]. But we just
showed C(t) ≥ max[0, St − Bt,T · K] > max[0, St − K]
if B < 1. Thus, early exercise is never optimal for
an American call on a non-dividend paying stock.
Hence its cash-flows are always equal to those of a Euro-

pean call. Hence they have the same value, by no arbitrage

(again).

� We just learned one optimal exercise policy rule. To solve

the general problem of when to optimally exercise American

options on dividend paying stocks requires a full stochastic

model.

� For European puts, with positive interest rates the lower

boundary is worse than the “intrinsic value”, max[K −S, 0]
but it is better if interest rates are negative.

� What does this tell us about exercising puts early?

III. Put-Call Parity (no payouts)

(A) European case.

� For European options on non-dividend paying stocks:

St = c(S, K, t, T) − p(S, K, t, T) + K · Bt,T.

� Intuition: a certain portfolio of bonds and options has the

same payoff at maturity as a share of stock, so it must have

the same initial price as a share of stock.

� If this relation does not hold there is an arbitrage opportunity.

� From the picture it is also clear that buying a call and selling

a put at strike K is the same as going long a forward on S

with forward price K.

· So (c − p) = Bt,T(Ft,T − K) which says the same thing,

� The Put-Call Parity relation tells us that we can make a

synthetic stock, call, put, and bond for European options on

a non-dividend-paying stock. It even tells us how to make

it:

Synthethic stock: S = c − p + PV (K)
Synthethic call : c = S + p − PV (K)
Synthethic put: p = c − S + PV (K)
Synthethic bond: PV (K) = S − c + p

� Question: Which one has a higher price: an at-the-money
(ATM) European call or an ATM European put on the same

non-dividend-paying stock?

� Question: Suppose you lived in a country with usury laws.
Do you see how a bank could lend money to a company

without ever charging interest?

· This is an example of financial engineering that applies in
Islamic countries.

(B) P/C/P – The American case (still no payouts to S).

� Although we can’t use the same replication argument for

American options, we can use our previous results (assuming

non-negative interest rates) to get a pair of inequalities that

is pretty close to the equality above.

� Step 1: P ≥ p and C = c and European put-call parity
together give

P ≥ C + B · K − S.

� Step 2: By no-arbitrage, also have

c + K ≥ P + S.

Proof. Suppose not. Sell the put and stock, buy the call,

invest K in the bond and have cash left over. Now there are

two possibilities:

1. The put we sold is not exercised early. So our payoff at
T is:

Put plus share: − max [K − ST, 0] − ST = − max [ST, K]
Call : max [ST − K, 0] = max [ST, K] − K
Bond: K/B

Total: K/B − K > 0

2. The put we sold is exercised early. So we buy back our
stock for K from the put holder. But our bond is worth

at least K. The put is gone, and we still have the call.

So all together our position still has positive value.

So in neither case can you lose money. Hence you have an

arbitrage.

� Step 3: Rewrite the inequalities from Steps 1 and 2, use

C = c again, and conclude

S − K ≤ C − P ≤ S − B · K, or

K ≥ S − C + P ≥ B · K.

� Notice that this is almost as good as the European put-

call parity when the options don’t have long to maturity, or

interest rates are not too high.

� You might have already guessed what happens when B > 1.

B · K ≥ S − C + P ≥ K.

I will leave the arguments for these two inequalities as exer-

cises.

� A caution: The real assumption we are making about divi-
dends and borrowing costs here is not just that they are zero,

but also that we know they will stay zero between now and

T .

· When we come across apparent violations of p/c/p in real
data, it is almost always because the markets know some-

thing about future dividend changes that we do not know.

IV. Effects of Dividends and Carry Costs on Arbitrage Restrictions:

� The put-call-parity and lower-bound relations that we derived

above assumed no payouts to the underlying.

� We can relax this if we maintain the assumption that the pay-

out. are known in advance with certainty

� We will treat the cases of discrete payouts and continuous yields

separately.

(A) Stock dividends (discrete payouts).

� As in Lecture 1, to avoid messy formulas involving the div-

idend dates, I will just write PV (D) for the amount you

would have to put in risk-free bonds today to replicate the

payouts that will occur between t and T .

· We are assuming both the payout amounts and dates are
certain.

· And, to remind you, the “payout date” means the ex-
dividend date.

� For any of the arbitrage strategies in the proofs that involved

holding the stock until T , we just lower the cost by the

amount we know we will take in from dividends.

� In other words, replace St everywhere with St − PV (D).

� The lower bounds for options on dividend-paying stocks then

become:

C(K, T) ≥ c(K, T) ≥ max[0, St − PV (D) − K · B]

P(K, T) ≥ p(K, T) ≥ max[0, K · B − St + PV (D)]

· If either of the lower bounds is violated, one constructs
an arbitrage by buying the option and shorting the things

inside the parentheses. (Or going long the things with

minus signs.)

· “Shorting” PV (D) just means borrowing that amount
today and repaying the Ds as they occur.

· Note that we are still assuming it is costless to borrow the
stock in order to short it.

� Put-Call Parity For European options on stocks with known

dividends becomes

c(S, K, t, T) − p(S, K, t, T) = St − PV (D) − K · Bt,T
via the same argument of replacing S by S − PV (D).

� Put-Call Parity for American options on dividend paying stocks

becomes a little trickier. When interest rates are non-negative,

they become:

S − PV (D) − K ≤ C − P ≤ S − K · B.

· The same substiution trick works on the left-hand inequal-
ity (the lower bound on C − P), but not on the upper
bound.

· If the lower bound is violated, our arbitrage involves selling
the stock and selling the put.

* The worst case outcome is that the put is not exercised

until expiration and so we do have to hold the short

position and pay all the dividends.

· If the upper bound is violated, then our enforcement leads
us to buy the stock and sell the call.

* Then the worst case is the call is exercised immediately

and we never get any of the dividends.

· (As in the case without payouts that we saw above, if
B > 1, it multiplies the K on the left side instead of the

right.)

(B) Known percentage yield (continuous flow) to underlying.

� This case encompasses options on currencies, commodities,

and any assets with borrowing fees. It is also not a bad

approximation for baskets of stocks.

� Let’s examine the case of currency options, where the foreign

risk-free rate rf could also be a dividend yield etc.

· Call the domestic rate rd. So B = e
−rd(T−t).

· Now the trick is the following. For each inequality that we
want to prove, re-do the same arguments but instead of

buying (or selling) one unit of S, do the trade for e−rf(T−t)

units.

· Imagine we do this every ∆t = 1 day, adjusting our posi-
tion until T . Two things happen.

1. The size of our position changes by

e−rf(T−t1)−e−rf(T−t0) = e−rf(T−t0) [erf∆t−1] ≈ rf ∆t e−rf(T−t0).

This says the percent change in our position size is rf∆t.

2. But we also get to receive the interest (or pay if we’re
short) equal to

Ste
−rf(T−t0) [erf∆t − 1] ≈ rf ∆t St e−rf(T−t0).

The second part provides exactly the amount of money

needed to do the first part!

· In other words, each period we will re-invest the flow re-
ceived in time interval ∆t back into our holdings of the

underlying.

* The net cash-flow is obviously zero each period.

* The neat thing is, we maintain the same exponential

holding quantity. In particular, our position will equal

exactly one unit at time t.

� I will leave it as an exercise to fill in the details of the indi-

vidual proofs.

· Intuitively, where before we replaced S by S − PV (D),
now we will try to replace St with e

−rf(T−t)·St everywhere.

� The option lower bounds become

C(S, K, t, T) ≥ c(S, K, t, T) ≥ max[0, St·e−rf(T−t)−K·e−rd(T−t)]
P(S, K, t, T) ≥ p(S, K, t, T) ≥ max[0, K·e−rd(T−t)−Ste−rf(T−t)]

� European Put-Call Parity becomes

c − p = St · e−rf(T−t) − K · e−rd(T−t)

Or for a stock with (continuous) dividend yield d and bor-

rowing rate y

c − p = St · e−(d+y)(T−t) − K · e−r(T−t)

� You should try to derive the American put-call parity inequal-

ities yourself.

V. Options Prices as a Function of Strike Price

Consider the set of European options on a single underlying

that all expire at the same T . I will drop all the arguments

to the pricing functions except K, the strike.

(A) Spreads:

Consider the following combinations of options of the

same type but different exercise prices (“vertical spreads”)

Here K2 > K1.

1. Bear Put Spread

� Cost today: p(K2) − p(K1) (≥ 0)

� Since the payoff is positive for all final stock prices, the

position value (or price, or cost) must also be positive at

all points in time. That is:

At t : p(K2) − p(K1) ≥ 0

2. Bull Put Spread

� p(K1) − p(K2) (≤ 0)

3. Bull Put Spread + Bond (with face value K2 − K1)

� p(K1) − p(K2) + (K2 − K1) · B (≥ 0)

� This position always has positive value, which implies that,

for a $ 1 increase in the strike, the price of the put can

increase by no more than the present value of $ 1.

0 ≤
p(K2) − p(K1)

K2 − K1
≤ B => 0 ≤

∂p

∂K
≤ B

4. Bear Call Spread

� c(K2) − c(K1) (≤ 0)

5. Bear Call Spread + Bond (with face value K2 − K1)

� c(K2) − c(K1) + (K2 − K1) · B (≥ 0)

� This position always has positive value, which implies that,

for a $ 1 increase in the strike, the price of the call can

decrease by no more than the present value of $ 1.

−B ≤
c(K2) − c(K1)

K2 − K1
≤ 0 => −B ≤

∂c

∂K
≤ 0

6. Slope restrictions for American options

� For American calls and puts, the arbitrage arguments have

to take into account the possibility that any options you

sold may be exercised against you at any time.

� If this happens you will have to close out the position and

be certain you have a positive payoff.

� This is no problem for the strategy where you buy the

lower strike call or the higher strike put, because if the

other option is exercised early, you can just exercise early

too. So we don’t have to change the conclusions C(K2)−
C(K1) ≤ 0 or 0 ≤ P(K2) − P(K1).

� But for the other strategy where we had to add zero

coupon bonds to the position, we need to know that we

have at least as much cash available whenever the options

are exercised (for example, if they are exercised tomorrow).

� If interest rates are negative so that B (K2 − K1) >
(K2 − K1), our bounds are still correct. However, if B <
1, we need to have (K2 − K1) dollars (not bonds) in our
portfolio.

� Hence, the slope restriction for American options with

positive interest rates become:

−(K2 −K1) ≤ C(K2)−C(K1) ≤ 0 => −1 ≤
∂C

∂K
≤ 0

0 ≤ P(K2) − P(K1) ≤ (K2 − K1) => 0 ≤
∂P

∂K
≤ 1.

(B) Butterfly Spreads and Convexity Restrictions:

Now consider a position involving three strikes, with K1 <

K2 < K3 and K2 = (K1 + K3)/2. That is, the middle

one midway between the other two. For some reason this

is called a “butterfly”.

1. Butterfly Spread (with calls)

� c(K1) − 2c(K2) + c(K3) (≥ 0)

� This spread can also be viewed as a combination of bullish

and bearish call spreads at different exercise prices:

c(K1)−2c(K2)+c(K3) = [c(K3)−c(K2)]−[c(K2)−c(K1)]

2. Butterfly Spread (with puts)

� p(K1) − 2p(K2) + p(K3) (≥ 0)

� And again, this spread can also be viewed as a combination

of bullish and bearish vertical spreads at different exercise

prices:

p(K1)−2p(K2)+p(K3) = [p(K3)−p(K2)]−[p(K2)−p(K1)]

� Also, note that the European put and call butterfly spreads

must have the same price.

Question: How are these payoffs (puts or calls) affected if
there is a yield or dividend on the underlying?

Question: Can butterfly payoffs for American options ever
be negative?

Question: What about if interest rates are negative?

3. Convexity

� Since butterfly spreads always have positive payoffs, the

position value is always positive:

[c(K3) − c(K2)] − [c(K2) − c(K1)] ≥ 0

and since K3 − K2 = K2 − K1 > 0, we have
c(K3) − c(K2)

K3 − K2

c(K2) − c(K1)
K2 − K1

≥ 0

and similarly

p(K3) − p(K2)
K3 − K2


p(K2) − p(K1)

K2 − K1
≥ 0

� This says the change in the slope of c(K) and p(K) is

positive as we move to the right. Or the slopes are in-

creasing as K increases.

� This is the same as saying that c(S, K, t, T) and p(S, K, t, T)

are convex in K.

� For strike prices that are very close together (∆K → 0)
we would write:

∂K

(
∂c(S, K, t, T)

∂K

)
=

∂2c(S, K, t, T)

∂K2
≥ 0

and

∂K

(
∂p(S, K, t, T)

∂K

)
=

∂2p(S, K, t, T)

∂K2
≥ 0

� This says the slope of the slope is positive.

� Graphically, for calls, the plot of price versus exercise price

must look like this:

· What does the plot of price versus exercise price for puts
look like?

� Also, by absence of arbitrage, the price of butterfly spreads

for European puts and calls must be the same:

c(K3) − 2c(K2) + c(K1) = p(K3) − 2p(K2) + p(K1)

=>
∂2c(S, K, t, T)

∂K2
=

∂2p(S, K, t, T)

∂K2

VI. Summary

� Based on static dominating portfolios, we can prove a number

of rules about option prices without making any assump-
tions about the behavior of the underlying security
over time.

� To prove these rules, we show that if they fail, arbitrage oppor-

tunities will be present.

· If we observe failures in real life, it may be because some
subtle assumption we made is violated.

� Most important: Put-Call Parity.

· Tells us how to synthesize either call, put, stock or bond
using the other three.

· Tells us we won’t need separate theories to price (European)
puts and calls.

� Methodological point: whenever analysing a new derivative,

always start by deriving the “common sense” bounds that static

domination requires.

· Sometimes that’s all you need anyway.

· Can guide your intuition in deriving better results based on
dynamic replication.

· Provides sanity check for more “sophisticated” models.

Place your order now for a similar assignment and have exceptional work written by one of our experts, guaranteeing you an A result.

Need an Essay Written?

This sample is available to anyone. If you want a unique paper order it from one of our professional writers.

Get help with your academic paper right away

Quality & Timely Delivery

Free Editing & Plagiarism Check

Security, Privacy & Confidentiality