Pricing options in underlying problem

Question

Let us look at options, which are cash settled, but instead of receiving cash, you receive the proportion from underlying asset with the same value as cash. Moreover, you can pay for these options in underlying asset only.

Example: ETH option expired with strike price \$1000 and the current value of ETH is \$2000. The buyer of the option will receive: (2000 − 1000)/2000 = 1 − 1000/2000 = 0.5ETH.

How can we price this option?

By Black Scholes model

We can calculate the price using BS model, which will be in USD and then divide it by current asset price.

For example:

• S = 20
• K = 20
• r = 0
• volatility = 1.3
• T = 7/365.2425

The resulted value from BS model is: $1.43 = 1.43/S$ ETH $= 1.43/20 = 0.0715$ETH.

Pricing it directly in underlying price

Instead of using BS model, we can price the option directly in the underlying asset.

We can simulate geometric Brownian motion from BS and then calculate the final price as: $$E_{FP} [(FP − K)/FP ]$$ where $FP$ is final price for given path in the process. Applying this to our example, we get:

$0.05758$ ETH $= \\\$0.05758 ∗ S = \\\$1.15$

Problem

This have resulted into to different prices for the same option:

• 0.0715 based on Black Scholes model
• 0.05758 based on second approach

Which one is correct and why the other one is incorrect?

Answer 1

You look at ETHUSD - how many USD per one ETH. If S=K=20 you get (correctly, if vol is 130%) a price of 1.43.

However, if you swap it around to price the option as USDETH (how many ETH per USD), you need to change your strike and notional because Garman Kohlhagen (BS for FX) is is priced in terms of notional in ccy1 (ETH), and premium in ccy2 (USD). If you use USDETH, your notional is now in USD and your premium in ETH.

Using Julia, this looks like this:

function GK(S,K,ccy1,ccy2,σ)
    d1 = (log(S/K) + (ccy2-ccy1+0.5*σ^2)*t)/(σ*sqrt(t))
    d2 = d1 - σ*sqrt(t)
    c = S*exp(-ccy1*t)*N(d1)-K*exp(-ccy2*t)*N(d2)
    p = K*exp(-ccy2*t)*N(-d2) - S*exp(-ccy1*t)*N(-d1)
    return c, p 
end
S = K =  20
ccy1 = ccy2 = 0
σ = 1.3
t = 7/365.2425;
GK(S,K,ccy1,ccy2,σ) # ETHUSD in USD
GK(S,K,ccy1,ccy2,σ) ./S # ETHUSD in ETH
GK(1/S,1/K,ccy1,ccy2,σ).*K # USDETH in ETH

You can read some more details here.

Edit:
You are not pricing the same payoff. In Black Scholes, you have $S_t-K$ (or FP-K) as the original option but you price (FP-K)/FP (you do not divide by current spot), and to get the value of 0.05758 you would actually have a "final price" (FP) of something like 21.222 (for (FP−K)/FP to result in 0.05758). If you now compute 21.222 - 20 (S-K) you get 1.222 ; if you use your 0.05758 * 21.222 you get the same 1.222.

Your MC

def MCSIM(S,t,r,vol,M=1000000, inverted = False):
    #np.random.seed(10)
    dt = t
    nudt = (r - 0.5*vol**2)*dt
    volsdt = vol*np.sqrt(dt)
    lnS = np.log(S)
    Z = np.random.normal(size=M)
    delta_lnSt = nudt + volsdt*Z
    if not inverted:
        lnSt = lnS + delta_lnSt
    else:
        lnSt = lnS - delta_lnSt
# compute Expectation and SE
ST = np.exp(lnSt)
return ST

(as copied from the chat with @Kurt G) gives this value, because you compute np.average([max((FP-K)/FP,0) for FP in MCSIM(S,t,r,σ,M=1000000, inverted = False)]). However, if you think this through, you only compute this value if FP > 20, else it is 0. Since all values smaller than 20 are excluded, this results in an average FP of ~21.43 (and FP-K will be the correct BS price). You can compute this like so np.average([FP if FP >= 20 else 20 for FP in MCSIM(S,t,r,σ,M=1000000, inverted = False)]).

However, the true expected spot np.average([FP for FP in MCSIM(S,t,r,σ,M=1000000, inverted = False)]) is ~20 in your MC run, given you ignore interest rates. I used a convoluted way with comprehensions to make this difference more explicit.

Put differently, you are interested in the ETH price of the option as of today - not as of the expiry date. Hence, you would need to divide by S instead of FP in np.average([max((FP-K)/S,0) for FP in MCSIM(S,t,r,σ,M=1000000, inverted = False)]), which gives 0.0717544967742682 (not using a seed).