log-normal distribution and derivatives

Question 1

stochastic differential equation:

Answer 1

dS=μSdt+σSdW (f(S,t) here)

Question 2

ito’s lemma:

Answer 2

if we have the stochastic differential, df=((∂f/∂t)+μS(∂f/∂S)+(1/2)σ^(2)S^(2)(∂^(2)f/∂S^(2)))dt+σS(∂f/∂S)dW, where df=f(S+dS,t+dt)-f(S,t)
this looks really scary but we are basically just working out each of those derivative bits from a given f=something involving S and substituting them into the complicated bit, leave it as df=whatever but sub the S terms with corresponding f

Question 3

log-normal distribution:

Answer 3

if f=ln(S), the stochastic differential equation is ln(S(t))=ln(S0)+(μ-(σ^(2)/2))t+σW(t) and ln(S(t))~N(ln(S0)+(μ-(σ^(2)/2))t, σ^(2)t)

Question 4

geometric brownian motion:

Answer 4

a stochastic process S(t) follows a gbm if dS=μSdt+σSdW where W is a weiner process and μ and σ are constants, yes this is the same equation as the stochastic differential equation one
means S(t)=S0e^((μ-(σ^(2)/2))t+σW(t))

Question 5

if S(t) follows a geometric brownian motion, the probability density function is:

Answer 5

f↓(St)(s;μ,σ,t)=(1/sσroot(2πt))exp(-(ln(s)-ln(S0)-(μ-(σ^(2)/2))t)^(2)/2σ^(2)t)

Question 6

issues with the geometric brownian motion model:

Answer 6

σ, the variance, isn’t constant irl, and there’s often large Non-normal fluctuations in price from external factors

Question 7

value of a contract with a call option:

Answer 7

C(S,T)=max(S-E,0) where T=t at the expiry date of the contract, so the max value it’ll take, and E=the pre-agreed exercise price, S=share price
call options you are deciding to buy the share or not

Question 8

value of a contract with a put option:

Answer 8

P(S,T)=max(E-S,0), S is share price T is t at expiry date of contract and E is the agreed price
put options you are deciding to sell the share or not

Question 9

position:

Answer 9

the number of financial contracts held, closing out a position means to make that 0, presumably by selling them

Question 10

payoff diagram:

Answer 10

shows the cash flow of all possible scenarios on the expiry date of a contract, for a call option it’s a graph with the line at 0 until E on the x axis where it starts increasing in a straight line (so more shares/assets means more profit now we’re actually selling), for a put option it decreases in a straight line until 0 and remains at 0 - this isn’t the profit gained it’s the value of the option itself

Question 11

profit of a call option (at expiry):

Answer 11

max(S↓(T)-E, 0)-C0e^(rT) where r is the rate of interest if you just invested the money at a bank

Question 12

profit of a put option (at expiry):

Answer 12

max(E-S↓T, 0)-P0e^(rT) where r is the rate of interest if you just invested the money at a bank

Question 13

investor’s expected profit:

Answer 13

using call option as an example -> funky E[C(S,T)]-C0e^(rT)
we are calculating the expected payoff Not the payoff of the expected share price