CHAPTER 4: Binomial trees and risk neutral valuation Flashcards
Derivative
A derivative is an asset whose value depends on the value of another asset, e.g., Call/Put
European/American options.
Consider a 1-year European call option on a stock with strike price £10
Assume that current price of the stock: S_0 = £10 and at the end of one year period price will be Su = £11 or Sd = £9.
Assume 1-year interest rate= 5%.
What should the price c of the option be?
• S_u = £11
/ Option payoff = £1
/
S= £10•<
\
\
• S_d = £9
Option payoff=£0Consider a 1-year European call option on a stock with strike price £10
Assume that current price of the stock: S_0 = £10 and at the end of one year period price will be Su = £11 or Sd = £9.
Assume 1-year interest rate= 5%.
consider a portfolio with δ shares of this stock
If the stock price goes up the portfolio is worth 11δ-1 and if stock price goes down it will be worth 9δ .
SO choosing δ st 11δ-1=9δ ie δ=0.5. Then the value of this portfolio is the same in all possible states of the world.
Portfolio must have a present value one year discounted to the present (9/2)e^-0.05 but the current price of stock in the portfolio is £10/2
so (9/2)e^-0.05 = 5-c so c≈ 0.72
The probability of up or down movements in the stock price plays no role whatsoever
Consider a financial asset which provides no income and a financial derivative on
that asset providing a single payoff t years in the future. The current price of the asset is S in t
years the price of the stock will be either Su (u > 1), resulting in a payoff of P_u from the derivative,
or Sd (0 ≤ d < 1) resulting in a payoff of P_d from the derivative. Let r be the t-year interest rate.
•Asset price =Su
/ Option payoff = P_u
/
S•<
\
\
• Asset price =Sd
Option payoff= P_d
(prices Su and Sd not S_u and S_d)
Consider a financial asset which provides no income and a financial derivative on
that asset providing a single payoff t years in the future….
Construct a portfolio consisting of δ units of the asset and -1 units of the derivative
choose δ st value of the portfolio after 2 years is certain
δ must satisfy:
δSu-P_u = δSd- P_d
⇒ δ = (P_u -P_d)/(S(u-d))
the value of the portfolio in t years
(P_u -P_d)/(S(u-d)) Su - P_u = u [(P_u -P_d)/((u-d))] - P_u
Present value is
e^-rt * (u [(P_u -P_d)/((u-d))] - P_u)
Let x be the price of the derivative, must have equality of present values
x
= (P_u -P_d)/((u-d))- e^-rt * u [(P_u -P_d)/((u-d))] - P_u
= ( (e^-rt)/(u-d))
*
((e^rt -d)P_u + (u -e^rt)P_d)
= e^-rt ( qP_u + (1-q)P_d)
where 0 ≤ q= (e^rt -d)/(u-d) ≤ 1
(q is a probability)
In a world where the probability of the up movement
in the asset price is q, the equation x = e^−rt (qPu + (1 − q)Pd) says that the price of the derivative is
the EXPECTED PRESENT VALUE of its payoff
STOCK PRICE AT TIME t HAS EXPECTED VALUE
E= qSu + (1-q)Sd = qS(u-d) +Sd
= (e^rt -d)/(u-d) +Sd
= (e^rt -d)S +Sd
=e^rt * S
the world where the probability of the up movement in the asset price is q is one in which the
stock price grows on average at the risk-free interest rate
RISK NEUTRAL PROBABILITIES
RISK NEUTRAL VALUATION
We refer to the probabilities q and 1 − q as risk neutral probabilities and to equation
above??
as a
risk neutral valuation
CONSIDER
underlying asset price changes twice each time by a factor of u>1 or d>1.
After two periods basic tree?
After two periods stock price will be Su², Sud =Sdu or Sd².
The derivative expires after the two periods producing payoffs Pᵤᵤ, Pᵤd=Pᵤd and P_dd respectively. assume each period is ∆t years long and interest rates for all periods is r.
• Su², Payoff =Pᵤᵤ
Su /
•<
/ \
S•< >• Sud, Payoff =P ᵤd
\ /
Sd •<
\ • Sd², Payoff =P_ddCONSIDER
underlying asset price changes twice each time by a factor of u>1 or d>1.
After two periods find the value of the derivative?
∆t each period
at node B
• D
B /
•<
/ \
A•< >• E
\ /
C •<
\ • F
To find x, the value of the derivative work away backwards from the end of the tree (2nd PERIOD) to the root (PRESENT)
Value of derivative known at vertices D, E and F (payoffs Pᵤᵤ etc) B and C values of derivative found by considering one period tree:
• Su²
/ Pᵤᵤ
V_B•<
\
• Sud
Pᵤd
The risk neutral probability q of an upward movement is q= (e^r∆t -d)/(u-d).
so value of derivative @ B is V_B = e^-r∆t(qPᵤᵤ + (1-q) P ᵤd)
CONSIDER
underlying asset price changes twice each time by a factor of u>1 or d>1.
After two periods find the value of the derivative?
∆t each period
at node C
• D
B /
•<
/ \
A•< >• E
\ /
C •<
\ • F
the risk neutral probability of an upward movement is given by q= (e^r∆t -d)/(u-d).
the value of the derivative at node c is obtained from
• Sud
/ P ᵤd
V_C•<
\
• Sd²
P_dd
V_C = e^-r∆t(qP ᵤd + (1-q) P_dd)
CONSIDER
underlying asset price changes twice each time by a factor of u>1 or d>1.
After two periods find the value of the derivative?
∆t each period
at node A value of derivative is
• D
B /
•<
/ \
A•< >• E
\ /
C •<
\ • F
• V_B
/
V_A•<
\
• V_C
V_A = e^-r∆t(qV_B + (1-q) V_C)
EXAMPLE: consider a European put option on stock currently traded at £10 with strike price £11 expiring in one year.
interest rates for all periods are 4%
two 6-month period tree with u=5/4 and d=3/4
s=15.62 (PO =0)
/
s=12.2
/ \
s=10 s=9.375 (PO = 1.625)
\ /
s=5.625
\
s= 5.625 (PO = 5.375)
--0.5--0.5–\ years
Risk neutral probability of an upwards movement of stock price
q = (exp(0.5 * 0.04)-(3/4))/(5/4 - 3/4) = 0.5404
VALUE OF DERIVATIVE
@ node B: V_B = e^(-0.05/2)(q0 + (1-q)1.625) = 0.7321
@ node C: V_C = e^(-0.05/2)(q1.625 + (1-q)5.375) = 3.282
@ node A: V_A = e^(-0.05/2)(qV_B + (1-q)V_C) = 1.866
calc values
EXAMPLE:
consider a 18-month European put option with strike £12 on a stock whose current price is £10. Assume interest rates for all periods are 5%. 3 step tree:
u=6/5 , d = 4/5
< .
< < .
<
< < .
< < .
S_A= £10
S_B= £12 , S_C= £8
S_D= £14.4, S_E= £9.6, S_F= £6.4
S_G= £17.28, S_H= £11.58, S_I= £7.68, S_J= £5.12
PO_G = 0 PO_H = 0.48 PO_I = 4.32 PO_J = 6.88
Risk neutral probability of an upwards movement of stock price
q = (exp(0.5 * 0.05)-(4/5))/(6/5 - 4/5) = 0.5633
Values of derivs:
exp(-0.5 * 0.05)((1-q)0.48) =0.2044
exp(-0.5 * 0.05)(q0.48+(1-q)4.32) =2.104
exp(-0.5 * 0.05)(q4.32+(1-q)6.88) =5.304
… V_A = 2.008
V_B= 1.008
V_C = 3.415
EXAMPLE:
consider a 18-month AMERICAN put option with strike £12 on a stock whose current price is £10. Assume interest rates for all periods are 5%. 3 step tree: u =6/5 d=4/5
consider dd in prev: immediate exercise gives payoff = 12-6.4 = 5.6 BIGGER THAN 5.304 (value of option at this node)
/ 7.68 (4.38) (_6.4_) 5.304
\ 5.12 ( 6.88)
SO MODIFY TREE FOR AMERICAN
V_A = (\_\_2.327\_\_) V_B= (\_\_1.1345\_\_) V_C = (\_\_4\_\_) V_D = 2.044 V_E= (\_\_2.4\_\_) V_F= (\_\_5.6\_\_)
consider D: do i exercise? NO, 14.4 is bigger than 12
rather keep derivative worth 0.2044
Consider E: YES, 9.6 less than 12 so change
similarly change F, C, B, A
q = 0.5633
V_E 2.104 less than 12-9.6 = 2.4
V_F 12-6.4 = 5.6 bigger than 5.304 so change
then
V_B= e^(-0.05/2) (qV_D +(1-q)V_E)= 1.1345
so DONT exercise option as 1.1345 not 12-2
V_C = e^(-0.05/2) (qV_E +(1-q)V_F) = 3.7037 LESS than 12-8 =4
exercise
V_A =e^(-0.05/2) (qV_B +(1-q)V_C) = 2.327 bigger than 12-10 =2 so dont exercise
more expensive than european
