Why is the price of a call option with $K=0$ equal to the price of the stock $S_0$?

In a case of a call option with strike $K=0$, then payoff at expiration time $T$ is equal to:

$$(S_T-0,0)^{+}=S_T$$

In reality the price of the option on the date of maturity is never equal to the stock price itself regardless of the strike price.

Why?

Having the price of the call option equal to the stock price itself provided that the strike is zero implies that holding the call is equivalent to, i.e. generates the same value as, holding the stock.

However, holding the stock has something that holding the call does not offer, e.g. the right to vote and claim on a share of firm’s property.

Hence, holding the call option is not equivalent to holding the stock. Therefore, the price of the call will always be at least a little lower than the stock price itself.

• I don't think I understand your question, where are you seeing options with $K=0$? – Bob Jansen Jan 9 '15 at 21:16
• It was very difficult to understand what you were asking here, and to determine whether you weren't actually answering you own question in you comment. I tried to make it clearer, feel free to edit if I got the question wrong. – SRKX Jan 12 '15 at 9:19
• @BobJansen: Please don't close this question - there are actually zero-strike options out there - see e.g. here: investment-and-finance.net/derivatives/z/… – vonjd Jan 12 '15 at 10:04
• I didn't know that and I won't. I do agree with SKRX, I'm still a bit confused though so clarification would be good. – Bob Jansen Jan 12 '15 at 10:12
• It's unclear whether you've seen in the market that the price of the option was smaller than the stock's or if it's something you think should be true but isn't in the market.... – SRKX Jan 13 '15 at 8:32

Buying a call at time $t=0$ with strike $K=0$ on a stock whose value is $S_0$ will produce the following cash flows ensure a cash flow at time $t=T$ of $S_T$, because as you mentioned $(S_T - 0, 0 )^{+}=S_T$ because $S_t \geq 0 ~ \forall t$ by definition.

This cash flow is replicable by buying the stock for $S_0$ at $t=0$.

By the law of one price, if there is no arbitrage then the price of the call has to be equal to the price of the replicating portfolio, which yields $c_t = S_t$ indeed.

What you're referring to about voting vote and credit risk is kind of different.

The credit risk part can be adjusted by some kind of CVA, but frankly as a share holder you will come after all debt holders and you probably don't have much to recover in case of bankruptcy.

The voting right part is actually very different. I don't really see why this would add value very much to the price of the stock, but if it was you could "model" it as some kind of dividend yield and you'll miss the opportunity of cashing in these dividends during the call's life. This would make the price of the call indeed lower than $S_0$.

• law of one price assumes absence of arbitrage or something? – BCLC Nov 3 '15 at 12:41
• @BCLC not sure what you mean but I tried to make that part more explicit. – SRKX Nov 4 '15 at 1:50

It is only in models that guarantee positiveness of the stock price, like Black & Scholes, that an option of strike 0 is worth the spot. In other models, like Bachelier's model, where the distribution of the spot is Gaussian, the zero strike option is worth more than the stock.

You need to make a distinction between reality and the model you are considering.

1) In your model, the conclusion is valid: in your model holding the stock is equivalent to holding the zero strike call. This is because you make many implicit assumptions (basically these with zero risk free rate).

2) Yes these assumptions i.e. your model are very simplistic. No it doesn't take into account voting right or counterparty risk so you cannot expect the predictions of this model to match reality on these subjects. In fact it doesn't even take into account two much more important factors - interest rate - dividends (much more important that voting rights)