2
$\begingroup$

enter image description here

enter image description here

For the solution: Just wanted to ask for $V_{3}(HTH)$ I get $S_{1}(H)$ and $S_{3}(HTH)$ to give me the same maximum value of $16$ so would it also be right if I used $S_{1}(H)$ nstead of $S_{3}(HTH)$? Same applies for $V_{3}(THT)$ for which I get $S_{0}$ and $S_{2}(TH)$ which both give me a maximum value of $8$ so could I have used $S_{3}(TH)$ instead of $S_{0}$?

Any help would be much appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

First, you have to compute the value of the payoff of your lookback option.

For example consider path $THT$. You have:

$S_{0}=8$, $S_{1}=4$, $S_{2}=8$ and $S_{3}=4$

It means that

$$\max\left\{S_{0},S_{1},S_{2},S_{3}\right\}=\max\left\{8,4,8,4\right\}=8$$

So the payoff is of the form

$$V_{3}(THT)=\max_{0\leq n\leq 3}S_{n}-S_{3}=8-4=4$$

It doesn't matter if you choose $S_{0}$ or $S_{2}$ to compute the maximum.


If you compute all $8$ payoffs you have two methods for pricing: replication and risk-neutral pricing method.

In the case of risk-neutral pricing method the martingale measure is of the form $Q(p^{*},1-p^{*})$, where

$$p^{*}=\frac{1+r-d}{u-d}$$

And you have to compute the following steps:

First step:

$C(HH)=\frac{1}{1.25}\left(p^{*}V_{3}(HHH)+(1-p^{*})V_{3}(HHT)\right)$

$C(HT)=\frac{1}{1.25}\left(p^{*}V_{3}(HTH)+(1-p^{*})V_{3}(HTT)\right)$

$C(TH)=\frac{1}{1.25}\left(p^{*}V_{3}(THH)+(1-p^{*})V_{3}(THT)\right)$

$C(TT)=\frac{1}{1.25}\left(p^{*}V_{3}(TTH)+(1-p^{*})V_{3}(TTT)\right)$

Second step:

$C(H)=\frac{1}{1.25}\left(p^{*}C(HH)+(1-p^{*})C(HT)\right)$

$C(T)=\frac{1}{1.25}\left(p^{*}C(TH)+(1-p^{*})C(TT)\right)$

Last step:

$C(0)=\frac{1}{1.25}\left(p^{*}C(H)+(1-p^{*})C(T)\right)$

The price of the lookback option is equal to $C(0)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .