Two scenarios are foreseen for a certain stock after one period, one in which the stock value is $110 E$ and another in which the value is $90E$. Its current value is $S_{0}= 100E$. Furthermore, each operation of selling the stock to the market carries a fee of $2\%$ (there is no fee to buy from the market). A call option is established at a strike price also equal to $100E$. Assuming zero interest rate I want to determine the risk-neutral probability and the fair price option.
I know that the value of the portfolio is equal to:
I know that $r=0$, $S_{0}=100$, $S_{1}(H)=110$, $S_{1}(T)=90$ thus $V_{1}(T)=0$ and $V_{1}(H)=10*0.98=9.8$ because we have a $2\%$ fee when we sell stock.
I also know that: $q'=1-p'$ and $p'=\cfrac{(1+r)S_{0}-S_{1}(T)}{S_{1}(H)-S_{1}(T)}=\cfrac{1}{2}=q'$ thus the risk-neutral probability is $\cfrac{1}{2}$.
To calculate the fair price I used the following formula:
$V_{0}=p'\cfrac{V_{1}(H)}{1+r}+q'\cfrac{V_{1}(T)}{1+r}= 9.8p'+0q'=4.9$
Can someone tell me if what i'm doing here is correct because I wasn't sure where I needed to amply the $2\%$ fee.


The solution for this would be

Risk Neutral Probability $= \frac{(1-d-(1+r)k)}{u-d-(1+r)k}$

Fair Price of the Option $ = \frac{1}{1+r}\left(p\psi{(u)}+(1-p)\psi{(d)}\right)$

where $\psi{(u)} = Max((110-100),0) = 10$

$\psi{(d)} = Max((90-100),0)= 0$

Solution for the said problem is

$p = \frac{(1-.9-(1+0)0.02)}{1.1-0.9-(1+0)0.02} = \frac{0.08}{0.18} = 0.44444$

fair price of the option $= (10*0.4444+0*(1-0.4444)) = 4.44E$

The derivation is lengthy and you can refer to the article below https://www.ekf.vsb.cz/export/sites/ekf/frpfi/cs/archiv/rocnik-2005/prispevky/dokumenty/TT_I_transaction-costs.pdf

They have given the derivation for symmetric but you will have to derive it for assymetric transaction costs.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.