MATH4514 Financial Economics in Actuarial Science (2023 Spring)
Suggested Solution of Assignment 4
Problem 1 (Required, 30 marks)
We consider a forward start put option on an asset as follows: The options become an 1-year
European put option with strike 0.9𝑆1 at year 1 (so the put options will expire at year 2), where 𝑆1 is
the asset price at year 1.
You are given that
 The current price of the asset is 𝑆0 = 30 and its price process satisfies
𝑑𝑆𝑡 = 𝜇𝑆𝑡 𝑑𝑡 + 𝜎𝑆𝑡 𝑑𝑊𝑡 ,
with 𝜎 = 0.25.
 The annual riskfree interest rate is 𝑟 = 5% convertible continuously.
Find the current price of the forward start put option using risk neutral valuation principle.
Solution
The terminal payoff of the forward put options at time 2 is given by
𝑉2 = max (0.9𝑆
⏟ 1 − 𝑆2 , 0).
𝑋
We first compute the option price at time 1. Under risk neutral probability measure 𝑄, the price
process of the asset is governed by
𝑑𝑆𝑡 = 𝑟𝑆𝑡 𝑑𝑡 + 𝜎𝑆𝑡 𝑑𝑊𝑡𝑄 .
Given the asset price at time 1, the (random) asset price at time 2 (under 𝑄) can be expressed as
𝑆2 = ⏟
𝑆1 𝑒
(0.05−
𝑆1 𝑒
0.252
𝑄
)(1)+0.25𝑊1
2
𝑄
= 𝑆1 𝑒 0.01875+0.25𝑊1 .
𝜎2
𝑄
(𝑟− )(𝑇−𝑡)+𝜎𝑊𝑇−𝑡
2
Note that
 𝑊1𝑄 = 𝑊1𝑄 − 𝑊0𝑄 is normally distributed with mean 0 and variance 1 and
ln 0.9−0.01875
 𝑆2 ≤ 0.9𝑆1 ⇔ 𝑊1𝑄 ≤
≈ −0.49644.
0.25
Using risk neutral valuation principle, the options price at time 1 can be expressed as
𝑉1 = 𝑒 −0.05(2−1) 𝔼𝑄 [𝑉2 (𝑆2)|𝑆1 ]
−0.49644
1 −𝑧 2
−0.05
0.01875+0.25𝑧
(0.9𝑆1 − 𝑆1 𝑒
)(
=𝑒
[∫
𝑒 2 ) 𝑑𝑧]
⏟
√2𝜋
−∞
0.9𝑆1 −𝑆2
−0.49644
= 0.9𝑒 −0.05 𝑆1 ∫
−∞
(
1
√2𝜋
𝑧2
𝑒 − 2 ) 𝑑𝑧 − 𝑆1 𝑒 −0.05 ∫
−∞
−0.496442
= 0.9𝑒
−0.05
𝑆1 (𝑁(−0.496442)
) − 𝑆1 ∫
⏟
=
⏞
−∞
−0.746442
=0.30979128
𝑦=𝑧−0.25
−0.49644
0.9𝑒 −0.05 𝑆1 (0.338543) − 𝑆1 ∫
−∞
1
−
1 −𝑧 2
𝑒 0.01875+0.25𝑧 (
𝑒 2 ) 𝑑𝑧
√2𝜋
(𝑧−0.25)2
2
) 𝑑𝑧
(
𝑒
√2𝜋
1 −𝑦 2
(
𝑒 2 ) 𝑑𝑦
√2𝜋
= 0.9𝑒 −0.05 𝑆1 (0.338543) − 𝑆1 𝑁(−0.746442)
⏟
=02277
= 0.037514𝑆1 .
Given the current price of the asset 𝑆0 , the asset price at time 1 can be expressed as
𝑆1 = 𝑆0 𝑒
(0.05−
0.252
𝑄
)(1)+0.25𝑊1
2
𝑄
= 30𝑒 0.01875+0.25𝑊1 .
By applying risk neutral valuation principle again, the current price of the option at time 0 is found
to be
𝑉0 = 𝑒 −0.05(1−0) 𝔼𝑄 [𝑉1 (𝑆1 )|𝑆0 ]
= 0.037514𝑒 −0.05 𝔼𝑄 [𝑆1 |𝑆0 ]
∞
1 −𝑧 2
−0.05
0.01875+0.25𝑧
= 0.037514𝑒
[∫ 30𝑒
(
𝑒 2 ) 𝑑𝑧]
√2𝜋
−∞
∞
= 30(0.037514) [∫
1
𝑒
−
−∞ √2𝜋
𝑦=𝑧−0.25
=
⏞
(𝑧−0.25)2
2
𝑑𝑧]
∞
1 −𝑦 2
30(0.037514) [∫
𝑒 2 𝑑𝑦] = 1.125423.
⏟−∞ √2𝜋
=1
Problem 2 (Required, 40 marks)
We consider a European put option on an asset. You are given that
 The strike price and maturity date of the put option is 𝑋 and 𝑇 (> 𝑡) respectively.
 The annual riskfree interest rate is 𝑟
 The volatility of the asset is 𝜎.
 The current price of the asset is 𝑆𝑡 . The asset pays continuous dividend at an annual dividend
yield rate 𝑞 ≥ 0 (When 𝑞 = 0, the asset is equivalent to non-dividend paying asset).
(a) Using Black-Scholes model, derive the pricing formula of the put option at time 𝑡 (denoted by
𝑝𝑡 ).
(b) Derive the formula of delta and gamma of the put option.
Solution of (a)
Under risk neutral probability measure, the price process of 𝑆𝑡 satisfies the following equation:
𝑑𝑆𝑡 = (𝑟 − 𝑞)𝑆𝑡 𝑑𝑡 + 𝜎𝑆𝑡 𝑑𝑊𝑡𝑄 ,
where 𝑊𝑡𝑄 is the standard Brownian motion under 𝑄.
Given 𝑆𝑡 , the stock price at maturity date 𝑇 can be expressed as
𝑆𝑇 = 𝑆𝑡 𝑒
(𝑟−𝑞−
𝜎2
𝑄
)(𝑇−𝑡)+𝜎𝑊𝑇−𝑡
2
.
On the other hand, we note that
𝑆𝑇 < 𝑋 ⇔ 𝑆𝑡
𝜎2
𝑄
(𝑟−𝑞− )(𝑇−𝑡)+𝜎𝑊𝑇−𝑡
2
𝑒
𝑄
< 𝑋 ⇔ 𝑊𝑇−𝑡
<
𝑋
𝜎2
ln 𝑆 − (𝑟 − 𝑞 − 2 ) (𝑇 − 𝑡)
𝑡
⏟
𝜎
.
𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑏𝑦 𝑑
𝑄
Combining the fact that 𝑊𝑇−𝑡 has normal distribution with mean 0 and variance 𝑇 − 𝑡, then the
current price of the call option can be computed as
𝑐(𝑡, 𝑆𝑡 ) = 𝑒 −𝑟(𝑇−𝑡) 𝔼𝑄 [max(𝑋 − 𝑆𝑇 , 0) |𝑆𝑡 ]
𝑑
=𝑒
−𝑟(𝑇−𝑡)
∫ (𝑋 − 𝑆𝑡 𝑒
−∞ ⏟
(𝑟−𝑞−
𝜎2
)(𝑇−𝑡)+𝜎𝑧
2
)(
1
√2𝜋(𝑇 − 𝑡)
𝑒
−
𝑧2
2(𝑇−𝑡) ) 𝑑𝑧
𝑋−𝑆𝑇
𝑑
= 𝑋𝑒
−𝑟(𝑇−𝑡)
1
∫
−∞ √2𝜋(𝑇
𝑑
1
= 𝑋𝑒 −𝑟(𝑇−𝑡) ∫
−∞ √2𝜋(𝑇
𝑦=
− 𝑡)
− 𝑡)
𝑒
−
=
⏞
𝑋𝑒
𝑒
−
𝑧2
2(𝑇−𝑡) 𝑑𝑧
− 𝑆𝑡 𝑒
∫
∫
𝑑
√𝑇−𝑡
1
√2𝜋(𝑇 − 𝑡)
∞
− 𝑆𝑡 𝑒 −𝑞(𝑇−𝑡) ∫
𝑑
1
√2𝜋
−∞
= 𝑋𝑒 −𝑟(𝑇−𝑡) 𝑁 (
∞
−𝑞(𝑇−𝑡)
𝑑
𝑧−𝜎(𝑇−𝑡)
𝑧
, 𝑢=
√𝑇−𝑡
√𝑇−𝑡
−𝑟(𝑇−𝑡)
𝑧2
2(𝑇−𝑡) 𝑑𝑧
𝑒
−
𝑢2
2 𝑑𝑢
− 𝑆𝑡 𝑒
1
√2𝜋(𝑇 − 𝑡)
−𝑞(𝑇−𝑡)
∫
𝑒
𝑒
−
𝜎2
𝑧2
(𝑇−𝑡)+𝜎𝑧−
2
2(𝑇−𝑡) 𝑑𝑧
−
(𝑧−𝜎(𝑇−𝑡))
2(𝑇−𝑡)
𝑑−𝜎(𝑇−𝑡)
√𝑇−𝑡
1
√2𝜋
−∞
𝑒
−
2
𝑑𝑧
𝑦2
2 𝑑𝑦
𝑑
𝑑 − 𝜎(𝑇 − 𝑡)
) − 𝑆𝑡 𝑒 −𝑞(𝑇−𝑡) 𝑁 (
)
√𝑇 − 𝑡
√𝑇 − 𝑡
= 𝑋𝑒 −𝑟(𝑇−𝑡) 𝑁(−𝑑2′ ) − 𝑆𝑡 𝑒 −𝑞(𝑇−𝑡) 𝑁(−𝑑1′ )
(Note:
𝑑
√𝑇−𝑡
=
𝑋
𝜎2
ln −(𝑟−𝑞− )(𝑇−𝑡)
𝑆𝑡
2
𝑑−𝜎(𝑇−𝑡)
=
√𝑇−𝑡
𝑑−𝜎(𝑇−𝑡)
𝜎
−𝜎(𝑇−𝑡)
√𝑇−𝑡
=
𝑋
𝑆𝑡
ln −(𝑟−𝑞+
+ 𝜎√𝑇 − 𝑡 = −𝑑1 + 𝜎√𝑇 − 𝑡 =
√𝑇−𝑡
𝜎2
)(𝑇−𝑡)
2
𝜎√𝑇−𝑡
−𝑑2′ )
=−
𝑆
𝜎2
ln 𝑡 +(𝑟−𝑞+ )(𝑇−𝑡)
𝑋
2
𝜎√𝑇−𝑡
= −𝑑1′ and
Solution of (b)
By direct differentiation, we deduce that that
′ 2
′ 2
𝜕
1 −(−𝑑1 ) 𝜕𝑑1′
1 −(𝑑1 )
1
2
𝑁(−𝑑1′ ) = −
𝑒
=−
𝑒 2 (
)
𝜕𝑆𝑡
𝜕𝑆𝑡
𝑆𝑡 𝜎√𝑇 − 𝑡
√2𝜋
√2𝜋
2
𝜕
1 −(−𝑑2′ ) 𝜕𝑑2′
1 −(𝑑1′ )
2
𝑁(−𝑑2′ ) = −
𝑒
=−
𝑒
𝜕𝑆𝑡
𝜕𝑆𝑡
√2𝜋
√2𝜋
2
−2𝑑1′ 𝜎√𝑇−𝑡+𝜎2 (𝑇−𝑡)
2
(
1
)
𝑆𝑡 𝜎√𝑇 − 𝑡
2
2
1 −𝑑1′ 𝑆𝑡 (𝑟−𝑞)(𝑇−𝑡)
1
1 −𝑑1′ 𝑒 (𝑟−𝑞)(𝑇−𝑡)
=−
𝑒 2 ( 𝑒
)(
)=−
𝑒 2 (
)
𝑋
𝑆𝑡 𝜎√𝑇 − 𝑡
𝑋𝜎√𝑇 − 𝑡
√2𝜋
√2𝜋
Then the delta of the put option can be computed as
Δ𝑝 =
𝜕𝑝
𝜕
𝜕
= 𝑋𝑒 −𝑟(𝑇−𝑡)
𝑁(−𝑑2′ ) − 𝑒 −𝑞(𝑇−𝑡) 𝑁(−𝑑1′ ) − 𝑆𝑡 𝑒 −𝑞(𝑇−𝑡)
𝑁(−𝑑1′ )
𝜕𝑆𝑡
𝜕𝑆𝑡
𝜕𝑆𝑡
2
1 −𝑑1′ 𝑒 (𝑟−𝑞)(𝑇−𝑡)
−𝑟(𝑇−𝑡)
= −𝑋𝑒
𝑒 2 (
) − 𝑒 −𝑞(𝑇−𝑡) 𝑁(−𝑑1′ )
𝑋𝜎√𝑇 − 𝑡
√2𝜋
2
′
1 −(𝑑1 )
1
+ 𝑆𝑡 𝑒 −𝑞(𝑇−𝑡)
𝑒 2 (
) = −𝑒 −𝑞(𝑇−𝑡) 𝑁(−𝑑1′ ).
𝑆𝑡 𝜎√𝑇 − 𝑡
√2𝜋
Hence, the gamma of the put option is seen to be
2
𝜕 2𝑝
𝜕
𝜕
1 −(𝑑1′ )
𝑒 −𝑞(𝑇−𝑡)
′)
−𝑞(𝑇−𝑡)
2
Γ𝑝 = 2 =
Δ = −𝑒
𝑁(−𝑑1 =
𝑒
(
).
𝜕𝑆𝑡 𝑝
𝜕𝑆𝑡
𝜕𝑆𝑡
𝑆𝑡 𝜎√𝑇 − 𝑡
√2𝜋
Problem 3 (30 marks + 10 bonus marks))
We consider a risky corporate debt issued by a corporate firm whose the firm asset value 𝐴𝑡 evolves
according to Geometric Brownian process:
𝑑𝐴𝑡 = 𝜇𝐴 𝐴𝑡 𝑑𝑡 + 𝜎𝐴𝑡 𝑑𝑊𝑡 ,
where 𝜇𝐴 is the expected rate of return and 𝜎 is the volatility of the firm asset value process. We
consider a 𝑇-year risky bond issued by this firm. It is given that



The bond has face value 𝐹 and no intermediate coupon payment.
The current level of the firm value is 𝐴𝑡 , where 𝐴𝑡 > 𝐹
If the firm cannot fulfill the debt obligation after 𝑇 year (i.e. 𝐴𝑇 < 𝐹), then the firm will be
liquidized immediately and the remaining firm value will be transferred to the bondholder.
 In the event of default, we assume that the liquidation cost is 𝛼𝐴𝑇 , where 𝛼 ∈ (0,1).
 The annual riskless interest rate is 𝑟 convertible continuously.
(a) (Required) Recall that we derive the pricing formula of similar debt using no arbitrage pricing
principle and pricing formula of European option in the lecture, can we apply the same trick in
this problem? Briefly explain your answer.
(b) (Required) Derive the pricing formula of this risky debt. Express your answer in terms of
𝐴𝑡 , 𝜎, 𝑟, 𝑇 − 𝑡, 𝛼, 𝐹 etc..
(c) (Optional) Show that the bond price decreases with respect to volatility 𝜎. Give a financial
explanation to the result (You may assume that 𝐴𝑡 > 𝐹, i.e. the firm has sufficient amount of
asset which can repay the face value initially).
Solution
(a) Note that the terminal payoff of the risky debt is given by
𝐹
𝑖𝑓 𝐴𝑇 ≥ 𝐹
𝑉𝑇 (𝐴𝑇 ) = {
.
(1 − 𝛼)𝐴𝑇 𝑖𝑓 𝐴𝑇 < 𝐹
Since lim − 𝑉𝑇 (𝐴𝑇 ) = (1 − 𝛼)𝐹 ≠ 𝐹 = lim + 𝑉𝑇 (𝐴𝑇 ) for 𝛼 > 0, so 𝑉𝑇 (𝐴𝑇 ) is not
𝐴𝑇 →𝐹
𝐴𝑇 →𝐹
continuous at 𝐴𝑇 = 𝐹.
On the other hand, the terminal payoff of European options [i.e. max(𝑆𝑇 − 𝑋, 0) (for call
option) or max(𝑋 − 𝑆𝑇 , 0) (for put option)] is a continuous function. Therefore, the
terminal payoff of a portfolio of European options is also continuous (since the terminal
payoff is just the linear combination of the payoffs of European options) and never equals
to the terminal payoff of the risky debt (which is not continuous).
Hence, it is impossible to find such replicating portfolio for risky debt.
(b) Assuming that the risky debt is hedgeable, one can price the risky asset using risk neutral
valuation principle: The current price of the risky debt (denoted by 𝑉𝑡 ) can be expressed as
𝑉𝑡 (𝐴𝑡 ) = 𝑒 −𝑟(𝑇−𝑡) 𝔼𝑄 [𝑉𝑇 (𝐴𝑇 )|𝐴𝑡 ],
where 𝐴𝑡 is governed by the following equation:
𝑑𝐴𝑡 = 𝑟𝐴𝑡 𝑑𝑡 + 𝜎𝐴𝑡 𝑑𝑊𝑡𝑄 .
 Given the current value of 𝐴𝑡 , its future value at maturity date 𝑇 is given by
(𝑟−

𝜎2
𝑄
)(𝑇−𝑡)+𝜎𝑊𝑇−𝑡
2
,
𝐴𝑇 = 𝐴𝑡 𝑒
where
is normally distributed with mean 0 and variance 𝑇 − 𝑡.
On the other hand, we note that
𝑄
𝑊𝑇−𝑡
𝐴𝑇 ≥ 𝐹 ⇔ 𝐴𝑡
𝜎2
𝑄
(𝑟− )(𝑇−𝑡)+𝜎𝑊𝑇−𝑡
2
𝑒
𝑄
≥ 𝐹 ⇔ 𝑊𝑇−𝑡
≥
𝐹
𝜎2
ln 𝐴 − (𝑟 − 2 ) (𝑇 − 𝑡)
𝑡
⏟
𝜎
𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑏𝑦 𝑑
𝐹
𝑖𝑓 𝐴𝑇 ≥ 𝐹
, the current price 𝑉𝑡 is
(1 − 𝛼)𝐴𝑇 𝑖𝑓 𝐴𝑇 < 𝐹
𝑉𝑡 (𝐴𝑡 ) = 𝑒 −𝑟(𝑇−𝑡) 𝔼𝑄 [𝑉𝑇 (𝐴𝑇 )|𝐴𝑡 ]
∞
𝑧2
1
−
−𝑟(𝑇−𝑡)
2(𝑇−𝑡)
=𝑒
[∫ 𝐹 (
𝑒
) 𝑑𝑧
√2𝜋(𝑇 − 𝑡)
𝑑
𝑑
𝑧2
𝜎2
1
−
(𝑟− )(𝑇−𝑡)+𝜎𝑧
2
2(𝑇−𝑡)
+ ∫ (1 − 𝛼)𝐴𝑡 𝑒
(
𝑒
) 𝑑𝑧]
√2𝜋(𝑇 − 𝑡)
−∞
Using the fact that 𝑉𝑇 (𝐴𝑇 ) = {
∞
=𝑒
−𝑟(𝑇−𝑡)
1
[∫ 𝐹 (
𝑒
√2𝜋(𝑇 − 𝑡)
𝑑
−
𝑧2
2(𝑇−𝑡) ) 𝑑𝑧
𝑑
+ (1 − 𝛼)𝐴𝑡 𝑒 𝑟(𝑇−𝑡) ∫ (
−∞
𝑦=
𝑧
𝑧−𝜎(𝑇−𝑡)
, 𝑢=
√𝑇−𝑡
√𝑇−𝑡
=
⏞
∞
𝑒
−𝑟(𝑇−𝑡)
[∫ 𝐹 (
𝑑
√2𝜋(𝑇 − 𝑡)
+ (1 − 𝛼)𝐴𝑡 𝑒 𝑟(𝑇−𝑡) ∫ (
−∞
= 𝑒 −𝑟(𝑇−𝑡) [∫
𝑑
√𝑇−𝑡
𝐹(
1
√2𝜋
= 𝐹𝑒 −𝑟(𝑇−𝑡) (1 − 𝑁 (
𝑦2
𝑒 − 2 ) 𝑑𝑦
𝑑
√𝑇 − 𝑡
√2𝜋(𝑇 − 𝑡)
1
𝑑
∞
1
𝑒
−
𝑒
−
(𝑧−𝜎(𝑇−𝑡))
2(𝑇−𝑡)
) 𝑑𝑧]
𝑧2
2(𝑇−𝑡) ) 𝑑𝑧
1
√2𝜋(𝑇 − 𝑡)
𝑒
−
(𝑧−𝜎(𝑇−𝑡))
2(𝑇−𝑡)
+ (1 − 𝛼)𝐴𝑡 𝑒 𝑟(𝑇−𝑡) ∫
2
) 𝑑𝑧]
𝑑−𝜎(𝑇−𝑡)
√𝑇−𝑡
−∞
)) + (1 − 𝛼)𝐴𝑡 𝑁 (
2
𝑑 − 𝜎(𝑇 − 𝑡)
√𝑇 − 𝑡
(
1
√2𝜋
𝑢2
𝑒 − 2 ) 𝑑𝑢]
)
𝑑
𝑑 − 𝜎(𝑇 − 𝑡)
= 𝐹𝑒 −𝑟(𝑇−𝑡) 𝑁 (−
) + (1 − 𝛼)𝐴𝑡 𝑁 (
).
⏟ √𝑇 − 𝑡
⏟ √𝑇 − 𝑡
𝑑1
𝑑2
Here,
𝐴𝑡
𝜎2
(𝑇
ln
+
(𝑟
−
𝑑
𝐹
2 ) − 𝑡)
𝑑1 = −
=
𝜎√𝑇 − 𝑡
√𝑇 − 𝑡
2
𝐹
𝜎
𝑑 − 𝜎(𝑇 − 𝑡) ln 𝐴𝑡 − (𝑟 + 2 ) (𝑇 − 𝑡)
𝑑2 =
=
= −𝑑1 − 𝜎√𝑇 − 𝑡.
𝜎√𝑇 − 𝑡
√𝑇 − 𝑡
(c) Using the fact that
𝐴𝑡
𝜎2
𝐴𝑡
𝜎2
(𝑇
(𝑇
ln
+
(𝑟
−
)
−
𝑡)
ln
+
(𝑟
+
𝜕𝑑1 −𝜎(𝑇 − 𝑡)
𝐹
2
𝐹
2 ) − 𝑡)
=
−
=−
< 0,
𝜕𝜎
𝜎√𝑇 − 𝑡
𝜎 2 √𝑇 − 𝑡
𝜎 2 √𝑇 − 𝑡
𝜕𝑑2
𝜕𝑑1
=−
− √𝑇 − 𝑡,
𝜕𝜎
𝜕𝜎
when 𝐴𝑡 > 𝐹, we deduce that
.
𝜕𝑉𝑡
1 −𝑑12 𝜕𝑑1
1 −𝑑22 𝜕𝑑2
= 𝐹𝑒 −𝑟(𝑇−𝑡) (
𝑒 2 )(
) + (1 − 𝛼)𝐴𝑡 (
𝑒 2 )(
)
𝜕𝜎
𝜕𝜎
𝜕𝜎
√2𝜋
√2𝜋
1
𝑑2
𝜕𝑑1
− 1
𝑒 2 )(
)
1
(𝑑1 +𝜎√𝑇−𝑡)
2
2
𝜕𝑑1
(
− (1 − 𝛼)𝐴𝑡 (
𝑒
)(
+ √𝑇 − 𝑡)
𝜕𝜎
𝜕𝜎
√2𝜋
√2𝜋
1 −𝑑12 𝜕𝑑1
−𝑟(𝑇−𝑡)
= 𝐹𝑒
(
𝑒 2 )(
)
𝜕𝜎
√2𝜋
1 −𝑑12 +2𝜎√𝑇−𝑡𝑑1 +𝜎2 (𝑇−𝑡) 𝜕𝑑1
2
− (1 − 𝛼)𝐴𝑡 (
𝑒
)(
+ √𝑇 − 𝑡)
𝜕𝜎
√2𝜋
1 −𝑑12 𝜕𝑑1
𝜕𝑑1
= 𝐹𝑒 −𝑟(𝑇−𝑡) (
𝑒 2 )(
− (1 − 𝛼)
− (1 − 𝛼)√𝑇 − 𝑡)
𝜕𝜎
𝜕𝜎
√2𝜋
1 −𝑑12
𝜕𝑑1
−𝑟(𝑇−𝑡)
= 𝐹𝑒
(
𝑒 2 ) (−𝛼
− (1 − 𝛼)√𝑇 − 𝑡) < 0.
𝜕𝜎
√2𝜋
= 𝐹𝑒
−𝑟(𝑇−𝑡)
−
Financially, the probability of default is getting higher when 𝜎 increases (when 𝐴𝑡 > 𝐹). Upon
default, the payoff received by the bondholder is getting lower since 𝐴𝑡 is very likely to take lower
value when 𝜎 is higher. On the other hand, the bondholder payoff remains to be 𝐹 when there is
no default and cannot get any additional benefit when 𝜎 increases. This explains the monotonicity
of 𝑉𝑡 with respect to 𝜎 derived earlier.
Bonus Problem (Optional, 30 marks)
The current price of an asset is 𝑆0 = $30 and its price movement is assumed to follow Black-Scholes
model as follows:
𝑑𝑆𝑡 = 𝜇𝑆𝑡 𝑑𝑡 + 𝜎𝑆𝑡 𝑑𝑊𝑡 ,
where 𝜎 = 0.3. The annual riskfree interest rate is 𝑟 = 4%. The asset pays dividend continuously at
an annual dividend yield rate 𝑞 = 3%.
Questions
(a) We consider a 1-year digital option which pays the holder one unit of the asset at maturity
date if the asset price at the maturity date is above $𝑋 (where 𝑋 > 0). Find the current price
of this digital option, express your answer in terms of 𝑋.
(😊Hint: The formula is in forms of 𝑆0 𝑁(𝑑), where 𝑑 is some number depends on 𝑋 and
𝑁(𝑥) = 𝑃(𝑍 ≤ 𝑥))
(b) We consider 1-year European call option on this asset with strike price $35. However, a
restriction is added to this option and the investor cannot exercise this option when the asset
price is above $40 at maturity date. Using no arbitrage pricing and the pricing formula of call
option, put option and digital option mentioned in (a), find the current price of this option.
(☹Note: You will get 0 mark in (b) if you compute the price using risk neutral valuation
principle).
Solution
(a) Using the risk neutral probability measure 𝑄, the price process of the asset is
𝑑𝑆𝑡 = (𝑟 − 𝑞)𝑆𝑡 𝑑𝑡 + 𝜎𝑆𝑡 𝑑𝑊𝑡𝑄 .
Given 𝑆0 = $30, the asset price after 1 year is seen to be
𝑆1 =
(0.3)2
𝑄
(0.04−0.03−
)(1)+0.3𝑊1
2
30𝑒
⏟
𝑆0 𝑒
(𝑟−𝑞−
𝑄
= 30𝑒 −0.035+0.3𝑊1 ,
𝜎2
𝑄
)𝑇+𝜎𝑊
𝑇
2
𝑊1𝑄 ~𝑁(0,1).
where
Note that the terminal payoff of the digital options is
𝑆 𝑖𝑓 𝑆1 > 𝑋
𝑉1 (𝑆1 ) = { 1
.
0 𝑖𝑓 𝑆1 ≤ 𝑋
and
𝑆1 = 30𝑒
𝑄
−0.035+0.3𝑊1
>𝑋⇔
𝑊1𝑄
>
𝑋
ln 30 + 0.035
⏟
0.3
.
𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑏𝑦 𝑑
Using risk neutral valuation principle, the current price of the digital option can be
expressed as
𝑉0 = 𝑒 −0.04 𝔼𝑄 [𝑉1 (𝑆1 )]
∞
𝑑
1 −𝑧 2
1 −𝑧 2
= 𝑒 −0.04 [∫ ⏟
30𝑒 −0.035+0.3𝑧 (
𝑒 2 ) 𝑑𝑧 + ∫ 0 (
𝑒 2 ) 𝑑𝑧]
√2𝜋
√2𝜋
𝑑
−∞
𝑆1
∞
= 30𝑒
−0.03
∫
1
−
𝑦=𝑧−0.3
∞
(𝑧−0.3)2
−0.03
2
𝑑𝑧 =
⏞ 30𝑒
∫
1
−
𝑦2
2 𝑑𝑦
𝑒
𝑒
√2𝜋
𝑑−0.3 √2𝜋
= 30𝑒 −0.03 (1 − 𝑁(𝑑 − 0.3)) = 30𝑒 −0.03 𝑁(0.3 − 𝑑).
(b) Note that the terminal payoff of this call option can be expressed as
0
𝑖𝑓 𝑆1 > 40
𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑝𝑎𝑦𝑜𝑓𝑓 = {𝑆1 − 35 𝑖𝑓 35 < 𝑆1 ≤ 40
0
𝑖𝑓 𝑆1 ≤ 35
By some algebra, one can rewrite the option’s payoff as
0
𝑖𝑓 𝑆1 > 40
𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑝𝑎𝑦𝑜𝑓𝑓 = {𝑆1 − 35 𝑖𝑓 35 < 𝑆1 ≤ 40
0
𝑖𝑓 𝑆1 ≤ 35
𝑆1 − 35 𝑖𝑓 𝑆1 > 40
𝑆1 − 35 𝑖𝑓 𝑆1 > 40
𝑖𝑓 35 < 𝑆1 ≤ 40
= {𝑆1 − 35 𝑖𝑓 35 < 𝑆1 ≤ 40 − {0
0
𝑖𝑓 𝑆1 ≤ 35
0
𝑖𝑓 𝑆1 ≤ 35
7
1
𝑆1 − 35 𝑖𝑓 𝑆1 > 40
(𝑆1 − 40) + 𝑆1 𝑖𝑓 𝑆1 > 40
8
= {𝑆1 − 35 𝑖𝑓 35 < 𝑆1 ≤ 40 − {8
0
𝑖𝑓 35 < 𝑆1 ≤ 40
0
𝑖𝑓 𝑆1 ≤ 35
0
𝑖𝑓 𝑆1 ≤ 35
𝑆1 𝑖𝑓 𝑆1 > 40
7
1
= max(𝑆1 − 35, 0) − max(𝑆1 − 40 ,0) − × { 0 𝑖𝑓 35 < 𝑆1 ≤ 40 .
8
8
𝑖𝑓 𝑆1 ≤ 35
⏟0
𝑑
𝑑𝑖𝑔𝑖𝑡𝑎𝑙 𝑜𝑝𝑡𝑖𝑜𝑛 𝑋=40
So the options can be replicated by a portfolio consists of
 Holding 1 unit of 1-year European call option with strike price $35
7

Shorting 8 unit of 1-year European call option with strike price $40

Shorting 8 unit of 1-year digital option (in (a)) with 𝑋 = $40.
1
Using no arbitrage pricing principle, the current price of the option is
7
1
𝑉0 = 𝑐(35) − 𝑐(40) − (𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑎𝑙 𝑜𝑝𝑡𝑖𝑜𝑛)
8
8
−0.03(1)
= 30𝑒
𝑁(−0.3305)
− 35𝑒 −0.04(1) 𝑁(−0.6305)
⏟
𝑐(35)
7
− ⏟
30𝑒 −0.03(1) 𝑁(−0.77561) − 35𝑒 −0.04(1) 𝑁(−0.107561)
8
𝑐(35)
1
− 30𝑒
⏟ −0.03 𝑁(−0.77561) ≈ 0.270629.
8
𝑑𝑖𝑔𝑖𝑡𝑎𝑙 𝑜𝑝𝑡𝑖𝑜𝑛 (𝑓𝑟𝑜𝑚 (𝑎))
*Note:
The pricing formula of call option is
𝑐 = 𝑆𝑡 𝑒 −𝑞(𝑇−𝑡) 𝑁(𝑑1 ) − 𝑋𝑒 −𝑟(𝑇−𝑡) 𝑁(𝑑2 ),
where
𝑆
𝜎2
ln 𝑋𝑡 + (𝑟 + 2 ) (𝑇 − 𝑡)
𝑑1 =
,
𝑑2 = 𝑑1 − 𝜎√𝑇 − 𝑡.
𝜎√𝑇 − 𝑡