# Calculating the fair price of an option knowing that selling the stock to the market carries a $2\%$ fee

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:
$X_{0}=(1+r)V_{0}+\Delta_{0}(S_{1}-S_{0}(1+r)$,
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.