Advanced Financial Mathematics
Chapter 13
Market marking and delta hedging
These notes are greatly inspired from the book Derivatives Markets
Chapter 13-Advanced Financial Mathematics-UEH-F2023
2
Market makers
• A market maker stands ready to buy from sellers and sell to buyers, ensuring a
liquid market with quick/immediate transactions.
• Proprietary trading operations are conceptually distinct from market making and
are conducted by a financial institution for its own account, following an investment
strategy.
• Clients and proprietary traders of financial institutions profit from market
movements, while market makers profit from transaction fees.
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3
Delta hedging
• Without hedging, a market maker ends up with an arbitrary position
based on customer orders.
• Market makers can control their risk by engaging in delta hedging.
• They calculate the delta of the option they have sold and take a
perfectly opposite position in the underlying asset.
• A delta-hedged position is not typically perfectly costless and requires
capital.
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Option risk in the absence of hedging
• We have already seen the concept that for every options bought/sold (traded),
there are always two parties involved, often a long position versus a short
position, but not always.
• The risk is associated with one of the positions for the option or the strategy in
question.
• Let's suppose a client wants to buy a call option from a market maker, and the
market maker sells the client one with:
• S0 = K = 40$, 𝜎 = 30 %, 𝑟 = 8 %, 𝑇 = 3 months (
91
365
years)
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Detailed calculation
• Call price:
91
𝐶 𝑆 = 40, 𝐾 = 40, 𝜎 = 0.3, 𝑟 = 0.08, 𝑇 − 𝑡 =
,𝛿 = 0
365
= 40 𝑒
91
−0∗365
𝑁 𝑑1 − 40𝑒
91
−0.08∗365
1∗0.58240416=∆𝐶
with 𝑑1 =
𝑙𝑛
𝑆
𝜎2
𝐾
+ 𝑟−𝛿+
2
𝜎 (𝑇−𝑡)
(𝑇−𝑡)
=
𝑁 𝑑2
= 2.7804$
0.5232266
𝑙𝑛
=0
40
40
+
0.32 91
0.08−0+
2 365
0.3
91
365
and 𝑑2 = 𝑑1 − 𝜎 (𝑇 − 𝑡) = 0.20804775 − 0.3
91
365
= 0.20804775
= 0.05825337
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7
• Greeks:
∆𝐶 = 𝑒 −𝛿(𝑇−𝑡) 𝑁 𝑑1 = 𝑒
Γ𝑐 =
𝑒 −𝛿(𝑇−𝑡)
𝜙 𝑑1
𝑆𝜎 (𝑇−𝑡)
=
𝑒
91
−0∗365
91
−0∗365
𝑁 0.20804775 = 0.5824>0
𝜙 0.20804775
40∗0.3
91
365
= 0.06515618>0
𝜃𝑐 = ⋯ ≅ −0.017<0
• For the market maker:
• ∆ < 0: risk is high when the stock price increases.
• 𝜃 > 0: his position is profitable in time.
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8
Discussion
• We assume that the stock price increases by $0.75 to reach $40.75.
• Option price is revaluated:
91
𝐶 𝑆 = 40,75, 𝐾 = 40, 𝜎 = 0.3, 𝑟 = 0.08, (𝑇 − 𝑡) =
,𝛿 = 0
365
= 40.75
91
−0∗365
𝑒
𝑁 𝑑1 − 40𝑒
91
−0.08∗365
1∗0.63007814=∆𝐶
𝜎2
with 𝑑1 =
𝑙𝑛 𝑆 𝐾 + 𝑟−𝛿+ 2 (𝑇−𝑡)
𝜎 (𝑇−𝑡)
=
𝑁 𝑑2
= 3.2352$
0.57231299
=0.01857639
𝑙𝑛 40.75 40 +
0.32 91
365
0.08−0+ 2
0.3
𝑑2 = 𝑑1 − 𝜎 (𝑇 − 𝑡) = 0.33206032 − 0.3
91
365
91
365
= 0.33206032
= 0.18226594
• The market maker has thus lost $2.7804 - $3.2352 = -$0.4548.
• If one day had passed before reaching this new value, then we would have had a lower
value due to the negative theta. However, in this scenario, we are only varying the
underlying asset's value without the effect of time passing.
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10
Delta and Gamma as measures of exposure
• If we had used the delta to approximate the new price of the call option, we would have obtained:
New Call Price = $2.7804 (Call Price with S = $40) + $0.75 (Change in S) * 0.5824 (Delta) = $3.2172.
• However, if we rely solely on the delta, the price change in the option would be estimated as
$0.4368, which underestimates the actual change of $0.4548.
• The reason for this discrepancy is that the delta was measured for an underlying asset with an exact
value of $40. Once it reaches $40.75, the underlying asset has a delta of 0.6301, which is greater
than 0.5824.
• When the value of the underlying asset increases, the delta also increases for a call option.
Therefore, you would need to continually recalculate the new delta (and integrate the delta for
each increment in the underlying asset's value between $40 and $40.75).
• In conclusion, the delta can be useful for small changes in the underlying asset's price, but for larger
changes, it underestimates the change in the option price.
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Question
How to address this issue without getting involved in complex calculations?
• Solution : For larger variations, one will need to consider convexity with gamma,
and thus:
0.0652
2
2.7804 + 0.75 ∗ 0.5824 +
0.75
= 3.2355$
2
∆
option pric
Variation
𝐶
square of variation
2
Γ
of
S
𝑐
of S
with S = 40$
2
• Conclusion : The combination of delta and gamma provides a good
approximation for moderate variations.
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Taylor approximation
𝑓′ 𝑎
𝑓 𝑥 =𝑓 𝑎 +
1!
𝑓 ′′ 𝑎
𝑥−𝑎 +
2!
(3) 𝑎
𝑓
𝑥−𝑎 2+
3!
𝑥−𝑎
3
+⋯
• The delta overestimates the decrease in the option price when the underlying
asset's value decreases.
• In fact, delta will generally provide a slightly lower approximation for the
option's value for any change in the underlying asset's value because it neglects
all the 2nd, 3rd, 4th derivatives, and so on.
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Delta hedging
• We revisit the example from page 385 of the reference book,
section 13.3.
• The market maker sells a call option and hedges the position with
stocks. Based on the previous example, this means buying 0.5824
shares.
• We will now examine the evolution of the market maker's position.
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Day 0 𝒕 = 𝟎/𝟑𝟔𝟓
• Selling a call option on 100 shares:
• Each option (per unit of the underlying) is worth $2.7804. So, 100 * $2.7804 =
$278.04 is received by the market maker.
• The market maker buys 100 * 0.5824 = 58.24 shares for $58.24 * $40 = $2329.60.
• The net initial cost is:
• $2329.60 - $278.04 = $2051.56.
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Day 1 𝒕 = 𝟏/𝟑𝟔𝟓
• 𝑆
= 40.50$
1
365
• 𝐶 𝑆
1
365
= 40.50$, 𝐾 = 40, 𝜎 = 0.3, 𝑟 = 0.08, 𝑇 − 𝑡 =
• For the market maker:
58.24 ∗ 40.50 −
𝑠𝑡𝑜𝑐𝑘 𝑣𝑎𝑙𝑢𝑒
90
,𝛿
365
100 ∗ 3.0621 − 2051.56 ∗ 𝑒
new option price
0.08
365
= 0 = 3.0621$
= 0.5003
• Hence, a profit about 0.50$ in one day
Rebalancing the portfolio is necessary because the delta has changed!
• The new delta is
2
𝑒 −𝛿(𝑇−𝑡) 𝑁
𝜎
𝑙𝑛 𝑆 𝐾 + 𝑟 − 𝛿 +
(𝑇 − 𝑡)
2
𝜎 𝑇
2
=𝑁
0.3
𝑙𝑛 40.50 40 + 0.08 +
2
0.3
Chapter 13-Advanced Financial Mathematics-UEH-F2023
90
365
90
365
= 𝟎. 𝟔𝟏𝟒𝟐
16
Day 1 𝒕 = 𝟏/𝟑𝟔𝟓
• The market maker needs 100 * (0.6142 - 0.5824) = 3.18 new shares, which requires an
additional investment of 3.18 * $40.50 = $128.79.
• When summing up the accumulated initial cost, the day's profit, and the purchase of
new shares, we get:
$2051.56 * 𝑒^(0.08/365) + 0.5003 + $128.79 = $2181.30.
• In other words, if you wanted to enter into an identical position at 𝑡=1/365, you would
have to buy 61.42 shares (with a delta of 0.6142 times 100 shares) at a price of $40.50
each and sell 100 call options at a price of $3.0621 each.
• This would result in a new initial cost on day 1 of:
61.42 * $40.50 - 100 * $3.0621 = $2181.30.
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Day 2 𝒕 = 𝟐/𝟑𝟔𝟓
• At 𝑡 = 2/365:
• The underlying asset's price is $39.25.
• - The option price for the market maker is
C S
2
365
= 39.25$, K = 40, σ = 0.3, r = 0.08, T − t =
• For the market maker: 𝟔𝟏. 𝟒𝟐 ∗ 39.25 − 100 ∗ 2.3282 − 𝟐𝟏𝟖𝟏. 𝟑𝟎 ∗ e
•
•
•
0.08
365
89
, δ = 0 = 2.3282$
365
= −3.8631
Value of their shares at day 2: $61.42 * $39.25.
New value of the option at day 2: 100 * $2.3282.
Interest on the cost of the position at the previous day for the market maker: $2181.30 * 𝑒^(0.08/365).
• This results in a loss of approximately $3.8631 in one day.
• Rebalancing the portfolio is required because the delta has changed! The new delta is:
σ2
0.32 89
ln S K + r − δ +
(T − t)
ln 39.25 40 + 0.08 +
2
2 365
e−δ(T−t) N
=N
σ T
89
0.3
365
= 0.5311
• The market maker needs to sell 8.31 new shares to balance their position since the delta is negative.
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Day 2 𝒕 = 𝟐/𝟑𝟔𝟓
• From the sale of 8.31 shares, the market maker will receive 8.31 * $39.25 = $326.66.
• When summing up the accumulated initial cost, the day's profit, and the repurchase of new
shares, we get:
𝟐𝟏𝟖𝟏. 𝟑𝟎 ∗
0.08
𝑒 365
+ −3.8631 − 𝟑𝟐𝟔. 𝟐𝟔 = 𝟏𝟖𝟓𝟏. 𝟔𝟔$
• This is the value of the position at this moment.
• In other words, if you wanted to enter into an identical position at 𝑡=2/365, you would need
to buy 53.11 shares (with a delta of 0.5311 times 100 shares) at a price of $39.25 each and sell
100 call options at a price of $2.3282 each.
• This would result in a new initial cost on day 2 of:
53.11 * $39.25 - 100 * $2.3282 = $1851.66.
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Day 0
• Selling 100 call options:
• In exchange, buying 58.24 shares with a loan of $2051.56 in addition to the
$278.04 obtained from selling the call options.
• Note: Our borrowing capacity is assumed to be equal to the value of the shares in
our portfolio, which is $2051.56 on day 0.
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Jour 1
The portfolio value rises to $2181.30, and our borrowing capacity increases by $129.74 in one
day.
To remain delta-neutral (delta-hedged), you need to buy 3.18 shares at $40.50 each for a total
of $128.79.
You pay $0.45 in daily interest on the loan.
The difference between $129.74 (borrowing capacity increase) - $128.79 (purchase of shares)
- $0.45 (interest) = $0.50 is the daily profit.
The market valuation profit is equal to the net cash flow of a 100% leveraged position.
Market Valuation Profit = Change in the value of the underlying + Interest expenses for
borrowed funds - Cash outflows for options.
Remarks:
- If the net cash flow is positive, you can pocket the amount.
- If the net cash flow is negative, you need to inject funds.
- A hedged portfolio that never requires fund injections is called self-financing.
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Delta hedging for multiple days
The daily gains and losses result from the combined effect of three elements, always
using the example of the call option.
1. Gamma: For large movements in the value of the underlying asset St, the market
maker loses money.
• For small movements in St, the market maker makes money. The larger the movement, the less
adequate delta hedging becomes.
• When St increases, ∆_𝐶 increases, and the written call option loses value (for the market maker
who has a short position in the option) faster than the proportion ∆_𝐶 of shares that the market
maker holds gains.
• Conversely, when St decreases, ∆_𝐶 decreases, and the written call option by the market maker
gains value more slowly than the proportion ∆_𝐶 of shares that the market maker holds loses.
2. Theta: The option loses value over time in absolute terms, which is advantageous for
the market maker (who has a short position in the call option).
3. Interest: Interest expenses on borrowed funds must be paid.
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Underlying asset price movement and autofinancing portfolio
• For a one-standard deviation (σ) move in the underlying asset's value (up or down), you
will obtain Table 13.3 (p. 389 DM).
• These movements resemble those in the forward tree model (term tree) from Chapter 10.
• For movements of exactly one standard deviation, the portfolio will be self-financed. In
Figure 13.2, this represents the two points where the curve crosses the axis, resulting in a
zero daily profit (moderate underlying asset movement, neither too high nor too low).
Lemma: If 𝑆𝑖 = 𝑆𝑖−1 𝑒 𝑟ℎ±𝜎
ℎ,
the delta-neutral portfolio is auto-financing.
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Mathematics of delta hedging
• The delta (∆), gamma (Γ), and theta (𝜃) influence the profit on a delta-hedged
(delta-neutral) position.
• Note: The other components (𝜎,𝑟,𝛿) do not change, so there's no impact on vega,
rho, and psi.
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Delta-gamma approximation
𝐶(𝑆𝑡 ) +
𝐶𝑎𝑙𝑙 𝑝𝑟𝑖𝑐𝑒
(𝑆𝑡+ℎ − 𝑆𝑡 )
stock price variation
Γ𝐶(𝑆𝑡)
∗ ∆𝐶(𝑆𝑡) + (𝑆𝑡+ℎ − 𝑆𝑡 ) ∗
≅ 𝐶(𝑆𝑡+ℎ )
2
2
• En 13.2.3. on avait pris l’exemple suivant :
0.752
0.0652
2
2.7804 + 0.75 ∗ 0.5824 +
= 3.2355$
∆𝐶
Variation
Prix du call Variation
Γ𝑐
prix de S
avec S = 40$ prix de S
2
au carré
• Et le véritable prix de l’option avec que le prix du sous-jacent soit passé de 40 à 40.75 est de
3.2352$.
• Concrètement, l’approximation delta-gamma indique que (résultat des séries de Taylor) :
• Remarques :
• Ceci ignore l’effet du temps (l’impact de « h ») et fonctionnera donc pour un h petit!
• L’approximation tient également pour une option de vente.
• Le delta et le gamma dépendent du prix de l’action.
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Delta-Gamma-theta approximation
Lemma: For small 𝜖 = 𝑆𝑡+ℎ - 𝑆𝑡 and h, we have
𝐶 𝑆𝑡+ℎ = 𝐶 𝑆𝑡 + 𝜖
≅ 𝐶 𝑆𝑡 , 𝑇 − 𝑡 +
𝜖
Call price
stock price variation
∗ ∆𝐶
𝑆𝑡 ,𝑇−𝑡
Chapter 13-Advanced Financial Mathematics-UEH-F2023
+ 𝜖
2
𝛤𝐶
𝑆𝑡,𝑇−𝑡
2
+ ℎ𝜃(𝑆𝑡,𝑇−𝑡)
29
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30
Market maker profit
• The value of the market maker's position at time t is
∆𝐶
𝑆𝑡 ,𝑇−𝑡
∗ 𝑆𝑡 − 𝐶 𝑆𝑡 , 𝑇 − 𝑡 = ∆𝑡 ∗ 𝑆𝑡 − 𝐶 𝑆𝑡 .
• The change in value from t to t+h is:
≅𝑟ℎ
• ∆𝑡 ∗ 𝑆𝑡+ℎ − 𝑆𝑡 − 𝐶 𝑆𝑡+ℎ − 𝐶 𝑆𝑡
− ∆𝑡 𝑆𝑡 − 𝐶 𝑆𝑡
𝑒 𝑟ℎ − 1
= ∆𝑡 ∗ 𝑆𝑡+ℎ − 𝑆𝑡 − ∆𝑡 ∗ 𝑆𝑡+ℎ − 𝑆𝑡 + 𝑆𝑡+ℎ − 𝑆𝑡
− ∆𝑡 𝑆𝑡 − 𝐶 𝑆𝑡 𝑒 𝑟ℎ − 1
Γ𝐶 𝑆𝑡
=−
∗ 𝑆𝑡+ℎ − 𝑆𝑡
2
2
+ℎ𝜃
𝑆𝑡 ,𝑇−𝑡
2
Γ𝐶 𝑆𝑡
∗
+ ℎ𝜃
2
+ ∆𝑡 𝑆𝑡 − 𝐶 𝑆𝑡
Chapter 13-Advanced Financial Mathematics-UEH-F2023
𝑆𝑡 ,𝑇−𝑡
𝑒 𝑟ℎ − 1
31
Remarques
• This is the profit of the market maker when the price goes from 𝑆𝑡 to 𝑆𝑡+ℎ
within a time interval of length h. We can break down the last equation
into three components:
• Gamma :
Γ𝐶 𝑆𝑡
−
2
• Theta : −ℎ𝜃
∗ 𝑆𝑡+ℎ − 𝑆𝑡
𝑆𝑡 ,𝑇−𝑡
2
∗ 365
• Intérêts : − ∆𝑡 𝑆𝑡 − 𝐶 𝑆𝑡
𝑒 𝑟ℎ − 1
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32
Tableau 13.5 DM
Delta-gamma-theta approximation for different values of 𝑆𝑡+ℎ − 𝑆𝑡
Chapter 13-Advanced Financial Mathematics-UEH-F2023
33
Movement of one standard deviation.
• We revisit the example of a one standard deviation change in the value of the
underlying asset with
𝑆𝑡+ℎ − 𝑆𝑡
2
= 𝜖 2 = 𝜎 2 𝑆𝑡2 ℎ.
• The profit is equal to:
Γ𝐶 𝑆𝑡 2 2
=−
𝜎 𝑆𝑡 + 𝜃 𝑆𝑡,𝑇−𝑡 ℎ − ∆𝑡 𝑆𝑡 − 𝐶 𝑆𝑡 𝑒 𝑟ℎ − 1
2
For exactly one standard deviation of variation, the profit should be approximately
zero
−
Γ𝐶 𝑆𝑡
2
𝜎 2 𝑆𝑡 2 + 𝜃
𝑆𝑡 ,𝑇−𝑡
ℎ − ∆𝑡 𝑆𝑡 − 𝐶 𝑆𝑡 𝑟ℎ = 0.
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34
Black-Scholes PDE
Theorem: Suppose that the stock price S follows the following geometric Brownian
motion with drift r and volatility 𝜎, 𝑖. 𝑒. S solves the following SDE
𝑑𝑆𝑡 = 𝑟𝑆𝑡 𝑑𝑡 + 𝜎𝑆𝑡 𝑑𝑊𝑡
Then, the call option price satisfies the following Black-Scholes partial differential
equation (PDE):
𝛤𝐶 𝑆𝑡 2 2
𝜎 𝑆𝑡 + 𝑟𝑆𝑡 ∆𝑡 + 𝜃 𝑆𝑡,𝑇−𝑡 = 𝑟𝐶 𝑆𝑡 .
2
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35
Frequency of portfolio rebalancing
• In practice, transaction costs introduce frictions, meaning that portfolio rebalancing can be expensive every time the delta changes.
• One can measure the impact of less frequent rebalancing:
1
𝑅ℎ,𝑖 = 𝑆𝑡 2 𝜎 2 Γ𝐶 𝑆𝑡 𝑥𝑖2 − 1 ℎ
2
• With 𝑥𝑖 representing the number of standard deviations of the movement of S and 𝑅ℎ,𝑖 being the return in period i, assuming an
initially delta-neutral position, and h is the time between adjustments in the portfolio.
• Here, we assume that the market maker has sold an option. Hence,
1
𝑉𝑎𝑟[𝑅ℎ,𝑖 ] = 𝑆 2 𝜎 2 Γh 2
2
• For daily rebalancing :
2
1 𝑆2𝜎2Γ
𝑉𝑎𝑟 𝑅 1
=
,1
2 365
365
• For hourly rebalancing :
24
𝑉𝑎𝑟
24
𝑅
𝑖=1
1
,
365∗24
1
=
𝑖=1
1 𝑆2𝜎2Γ
2 24 ∗ 365
𝑉𝑎𝑟 𝑅
2
=
1
,
365
1
24
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36
Delta-neutral hedigng in pratice
• Delta-neutral hedging is not sufficient to completely eliminate risk.
• It is possible to make a position gamma-neutral, but it requires derivative products.
• Static replication of the option is possible, a strategy in which options are used to hedge
options.
• Protection against significant underlying asset movements is possible by buying
insurance, i.e., purchasing deeply out-of-the-money call and put options.
• The problem can be addressed by creating a new derivative product, such as a variance
swap.
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37
Gamma-neutral
• Retake Table 13.6 with 𝑆 = 40, 𝜎 = 0.3, 𝑟 = 0.08, 𝛿 = 0.
3
𝐶𝑎𝑙𝑙 𝐾 = 40, 𝑡 =
12
4
𝐶𝑎𝑙𝑙 𝐾 = 45, 𝑡 =
12
Achat 1,2408 Call(K = 45), Vente
d’un Call(K = 40)
Prix (C)
2.7804
1.3584
-1.0993
Delta Δ
0.5825
0.3285
-0.1749
Gamma Γ
0.0651
0.0524
0
Vega 𝜈
0.0781
0.0831
0.0250
Theta 𝜃
-0.0173
-0.0129
0.0013
Rho 𝜌
0.0513
0.0389
-0.0013
Chapter 13-Advanced Financial Mathematics-UEH-F2023
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Discussion
• If you have a sold option with a strike price (K) of 40 and you want to neutralize the
gamma:
−0,0651 ∗ 1 + 0,0524 ∗ 𝑛𝑐𝑎𝑙𝑙𝑠 = 0
𝑛𝑐𝑎𝑙𝑙 = 1,2408 calls to buy
• Then, you seek to neutralize the delta:
−0,5825 ∗ 1 + 0,3285 ∗ 1,2408 + 𝑛𝑎𝑐𝑡𝑖𝑜𝑛𝑠 ∗ 1 = 0
𝑛𝑎𝑐𝑡𝑖𝑜𝑛𝑠 = 0,1749 shares to buy
• Neutralizing gamma improves profitability, so why not do it?
• Possible responses:
• It requires taking positions in other options, which can become complex and often involves costs.
• Market maker clients buy options that typically have positive gammas (recall, gamma is positive
for both call and put options).
• Therefore, market makers have negative gammas in aggregate. While a market maker could
theoretically seek gamma neutrality, they certainly cannot all be gamma-neutral at the same time.
Chapter 13-Advanced Financial Mathematics-UEH-F2023
39
Chapter 13-Advanced Financial Mathematics-UEH-F2023
40