Valuing Zero-Strike Calls and the Role of Carry

# Practically, are the prices of 0-strike European calls and stock identical?


# Practically, are the prices of 0-strike European calls and stock identical?












By no-arbitrage, the price of a vanilla European call with $K=0$ should be that of the underlying stock (as selling the call is perfectly hedged by buying the stock). However, is this true in practice?

More concretely, if you ran an options desk and I called you asking me to make me a market on $50M worth of 0-strike calls, would your market be identical to your market on the stock itself? If the answer is no, what factors contribute to your altered quote and in what ways? I'm essentially asking about the ways theory diverges from practice as far as how these things would trade.

The most I can think of is that there's an opportunity cost to hedging a short call because you aren't earning the risk-free rate on the cash you used to buy stock and hold it. Likewise, with the long call, you're paying borrow from the time you sell the stock (to hedge) until the time you cover with the delivered stock at maturity. Is this actually a consideration? Are there any others?

## Answer by user34971 (score 4)

https://quant.stackexchange.com/a/73294

The (call) price is undefined for $K=0$. You can only speak about the price as $K \downarrow 0$. Not the same thing as $\infty$ is not a number.

EDIT:

Following the question of @Kermittfrog, some more clarification.

First of all, assume that the stock price is a positive price process.

By definition a call option's price is $$ C(S_t,K,T) := E_t \left[ (S_T - K)_+ \right] $$ So a zero-strike call option is $$ C(S_t,0,T) = E_t \left[ (S_T)_+ \right] = S_t $$ which is rather trivial. There is no sense / added value in asking the question, in my opinion, whether a zero strike option is equal to the stock price.

Thus a call option, i.e. something that actually has optionality in it, is sensible only for $K>0$ regardless of the model as long as the stock price process is a positive process.

As for the OPs question, if the question is is the price of a call option with strike $K=10^{-9}$ practically equal to the stock price? No I don't think so; it's not equal theoretically, and practical slippages and considerations is another reason the prices will not be equal.

## Answer by AKdemy (score 3)

https://quant.stackexchange.com/a/73277

I do not think anyone actually trades this but in any case, it would follow standard pricing logics.

If you have a strike of truly 0, volatility does not matter anymore. The expected future value of unconditionally receiving the stock equals the forward. However, with options, the stock is received conditionally on the probability N(d2), the (risk neutral) probability of the option expiring in the money. In the case of a zero strike option, this probability is 1, no matter what. You can check out this answer for some related details.

So you know you will use this right for sure (you can buy the underlying for "free"). What matters is the forward at that time (which you discount to get the current options price).

In case of zero dividend and interest rates, the forward will be idenical to spot, and your option is worth the spot value. In case of dividends, you have to look at the forward. Below is a bit of Julia code to showcase this.

```
# load packages
using Pkg,  Distributions, DataFrames, Plots, PlotThemes
# define cdf
N(x) = cdf(Normal(0,1),x)
# generic Black Scholes pricer
function BSM(S,K,t,rf,d,σ)
    d1 = ( log(S/K) + (rf - d + 1/2*σ^2)*t ) / (σ*sqrt(t))
    d2 = d1 - σ*sqrt(t)
    c  = exp(-d*t)S*N(d1) - exp(-rf*t)*K*N(d2) 
  return c
end 

# inputs
s = 4000.00        # spot
k = 4000.00        # strike
σ = 15/100         # IVOL 
t = 90 / 365       # time to maturity in years
r = 0.1 / 100      # interest rates
r = log(1+r*t)/t   # continous rates
d = 1.5/100        # dividends 
d = log(1+d*t)/t   # cont. divs

DataFrame(Call = BSM(s,k,t,r,d,σ)[1])
```

This value should match pricers like Bloomberg's OVME exactly.

Now, if you set set k = 0 you get the discounted fwd.

The following lines compute the option value for different values of IVOL

```
theme(:juno)
k = 0
σ = 0.01:0.01:30  #  0.1 is 10% (meaning 30 is very unrealistically high)
plot(σ, BSM.(s,k,t,r,d,σ), 
                        legend = false, 
                        title = "ATM Option value for different IVOL",
                        xlabel = "IVOL",
                        ylabel = "Option value")
ylims!((0,5000))
```

compared to the value of an ATM option (see here for an explanation of what happens to call and put options if IVOL tends to infinity).