An Asian option is an option on the average price of the underlying asset. One

particular application of Asian options is hedging exchange rate risk. Asian options

are also popular in the energy OTC market and many commodity markets. Further$-$

more, because an Asian option is based on a price average, an attempt to manipulate

the asset price just before expiration will normally have little effect or no effect on

the option's value. Asian options should therefore be of particular interest in markets

for thinly traded assets.

Consider the case of an Asian call option with discrete arithmetic averaging. An

option with maturity of $T$ years and strike $K$ has the payoff

max$\{\frac{1}{n}{\displaystyle \underset{i=1}{\overset{n}{}}}S(t_i)-K,0\}$

where $t_i=ih$ and $h=\frac{T}{n}.$ Notice that we do not include in the summation the

price of the asset today, i$.$e$.S(t_0).$

Furthermore, assume $S(t_0)=200,r=0\mathrm{.}02,=0\mathrm{.}20,K=220,T=1,n=365.$

a$.$ Use Monte Carlo simulation to obtain one price path starting from $S(t_0)=200$

to $S(t_n).$ To simulate the process, consider its values at each step:

$S(t_{i+1})=S(t_i)exp[(r-\frac{1}{2}{}^2)t+\sqrt[\phantom{2}]{t}{}_t]$

where $t=\frac{T}{n}$ is the time interval, $r$ and $$ are given above, and ${}_t$ is a random

value drawn from a standard normal distribution.

Plot the resulting price path.

b$.$ Build a second plot that has two lines:

$($i$)$ The price path from the previous point

$($ii$)$ The arithmetic average of the price path at each point in time $t_i,$ starting

from $S(t_0).$ That is$,$ your second line on the plot should be:

$S(t_0),\frac{S(t_1)}{1},\frac{S(t_1)+S(t_2)}{2},\frac{S(t_1)+S(t_2)+S(t_3)}{3},...\frac{{\displaystyle \underset{i=1}{\overset{n}{}}}S(t_i)}{n}$

What do you observe? Do you think the price of the Asian option should be

lower, equal, or higher compared with a similar standard option? Why?

c$.$ Price the option by Monte Carlo simulation using $100$ paths. Compute the price

of the option on each path with the formula

$C_0^A=e^{-rT}$max$\{\frac{1}{n}{\displaystyle \underset{i=1}{\overset{n}{}}}S(t_i)-K,0\}$

What is the average of $C_0^A$ over the $100$ simulated prices? State the $95$ percent

confidence interval of this average. $($For this last question, you might need

to revisit statistical notions like "standard error of the mean" and "confidence

interval"$).$

d$.$ Redo the previous point, but now with $10,000$ paths. Compare the confidence

interval obtained in this case with the confidence interval from the previous

point.