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?

  • $\begingroup$ Buying a call option gives you the right to buy the stock in the future. Assuming European options for simplicity, you will need to think of the price at expiry (the no arbitrage forward), not today's spot. That is also the result you get if you run a simple Black Scholes calculator and set strike to 0 $\endgroup$
    – AKdemy
    Sep 23, 2022 at 7:21
  • $\begingroup$ @AKdemy . I am assuming you have in mind that the stock pays dividends. $\endgroup$
    – Kurt G.
    Sep 23, 2022 at 9:13
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    $\begingroup$ @actinidia Have you seen in practice an options market that quotes prices of far ITM and far OTM options ? If so please post some details . $\endgroup$
    – Kurt G.
    Sep 23, 2022 at 9:14
  • $\begingroup$ @Kurt G., yes, I had dividends in mind. Since you do not yet own the stock with an option, you forgo the dividend income you could get by buying ths stock immediately. $\endgroup$
    – AKdemy
    Sep 23, 2022 at 11:39
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    $\begingroup$ In theory, there is no difference between theory and practice. In practice, there is. Ps: zero-strike equity call options don't really trade in practice. $\endgroup$ Sep 25, 2022 at 8:34

2 Answers 2


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.


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.

  • 1
    $\begingroup$ Hi Frido, just a clarification: You meant that it is undefined under some model (e.g. BS), right? $\endgroup$ Sep 26, 2022 at 7:55
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    $\begingroup$ @Kermittfrog Thanks for the question, pls see my edits. $\endgroup$
    – user34971
    Sep 26, 2022 at 8:34
  • $\begingroup$ "...practical slippages and considerations is another reason the prices will not be equal." You reworded my question in a better way: precisely what I'm asking is what are those considerations? $\endgroup$
    – actinidia
    Sep 27, 2022 at 19:36

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

# 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.

enter image description here

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

enter image description here

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

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")

enter image description here

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). enter image description here

  • $\begingroup$ Does this mean that the market on the call (with nonzero dividends) will differ from the spot market insofar as the market maker has uncertainty about future dividends? Practically, will this impact on the spread be substantial when moving large size? From what I know, unlike the situation when one wants to hedge forward interest rates, there's no liquid market for hedging divs in arbitrary names. $\endgroup$
    – actinidia
    Sep 24, 2022 at 19:33
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    $\begingroup$ To be honest, I thought this is a purely theoretical question but seems there are actually markets like the Australien Stock exchange where LEPOs trade. The ASX explains they are like a forward purchase of shares which is in line with what I showed (ignoring that mathematically speaking K=0 it is not defined in BS model, as @Frido Rolloos pointed out). $\endgroup$
    – AKdemy
    Sep 24, 2022 at 21:26
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    $\begingroup$ If you have questions about an actual market, I will not be able to help but it would help if you add a lot of details in this case. That way someone trading in that specific market will be able to better understand your specific issue. Otherwise, I would say the question lacks clarity and is not focused enough. The explanatory booklet from ASX is already 28 pages for example. $\endgroup$
    – AKdemy
    Sep 24, 2022 at 21:27

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