Since if the option's price is lower than its intrinsic value (eg. strike price - current stock price for puts), then an arbitrage opportunity arises from buying the option at bargain and then exercising it...

Consequently, an option's price will always be greater than or equal to its intrinsic value

Am I right or wrong in this?

  • $\begingroup$ Judging by the list of answers, it is clearly not so trivial for everybody on this forum. Furthermore, it is not so obvious/common to find results for negative rates. $\endgroup$ – jherek Apr 8 '20 at 13:10

You answered your own question with the statement it began with:

"Since if the option's price is lower than its intrinsic value (eg. strike price - current stock price for puts), then an arbitrage opportunity arises from buying the option at bargain and then exercising it..."

An options price cannot be lower than its intrinsic value (for any discernab,le amount of time - assuming markets remain open and transactions are occuring), for the simple reason that it will represent "free money" - it will be arbitraged away immediately, as you quite rightly noted in your question.

  • $\begingroup$ This is assuming this is an American option with early exercise rights. For a European option, this is wrong. $\endgroup$ – jherek Apr 2 '20 at 16:05
  • $\begingroup$ @jherek A trader doesn't need to wait until expiration to realise profits. A reverse collar (or collar) can be used to trivially construct a combination that allows European options selling below their intrinsic value, to be "arbitraged against". $\endgroup$ – Homunculus Reticulli Apr 2 '20 at 16:17
  • $\begingroup$ please take a look at the other answers here. You are clearly mistaken. Olivier gives a concrete example. And it's very easy to find many. $\endgroup$ – jherek Apr 3 '20 at 17:36
  • $\begingroup$ In 20 years+ of trading, I am yet to see this phenomena that Olivier describes. The numbers he uses clearly demonstrate the point he is making - but are those really prices from the market? the point I'm making is that in the real world any such "anomalous" prices will be arbitraged away immediately (if transaction costs allow it). I concede one thing however - my experience is solely from listed (i.e. exchange traded options) - and it is quite possible that away from regulated exchanges, all manner of different beasts roam ;) That said, for all practical purposes, my comment is true. $\endgroup$ – Homunculus Reticulli Apr 3 '20 at 17:51
  • $\begingroup$ The prices from Olivier are not anomalous at all, they are standard. It is not difficult to find examples from the market, consider SPX500 Put options prices quoted on CBOE for example. I am a bit shocked to hear that a trader with 20+ years experience has never seen this, but this is clearly not the subject of the initial question. $\endgroup$ – jherek Apr 4 '20 at 16:36

In kamikaze_pilot's defense, the question is not that naive or simple.

First of all, you need to define what options you are talking about. Consider a digital option for example (which is really fairly vanilla since you can proxy it as a combination of two European calls), which pays 1 of the stock is beyond a certain level at maturity and nothing otherwise. The intrinsic value of the option is just 1 or 0, but would you pay more than 1 to buy such an option ? (what's the point of paying more than whatever you would get at most as a payoff....)

In fact, there is a general point here, a option with a limited upside can have a price lower than the intrinsic value. And this may even apply to European vanilla options, although in very specific circumstances: take a European put option with 0.5y to maturity, 20% implied vol, 100 strike and 5% risk-free rate (ignore all dividends for this example). Now for various spot level, calculate the intrinsic value ( Max(0,strike - spot)) and compare to the option price which you can calculate using Black-Scholes:

spot = 90 --> intrinsic = 10, option price = 11.04 --> intrinsic < option price

spot = 50 --> intrinsic = 50, option price = 47.55 --> intrinsic > option price

spot = 10 --> intrinsic = 90, option price = 87.53 --> intrinsic > option price

Of course, this only happens for deep in-the-money put options. Also it can not happen for American options (there would definitely be an arbitrage here, since you could by the option and exercise it right away).

In terms of arbitrage, consider the following: At time t = 0, your underlying spot is S(0). You buy the stock and a European put option with maturity T and strike K. At time T (maturity), the stock is worth S(T) and your portfolio is either

1) (K - S(T)) + S(T) = K, if S(T) < K

2) or S(T), if S(T) > K

If risk-free rate is r, the accrued cost of the portfolio you set up at time 0 is (P(0) + S(t)) * exp(rT). To avoid arbitrage, you need this cost to be greater than the minimum of 1) and 2) above, which is K. You get (P(0) + S(t)) * exp(rT) > K --> P(0) > K*exp(-rT) - S(0). This is the actual arbitrage condition, which is close to saying the intrinsic value is a lower bound of the put option but not quite.


It is common for the Bid (and sometimes the average of the Bid/Ask) price of deep in the money options to be below the Intrinsic Price.

Download some data and try it.



False. It is not always the case for European options which cannot be exercised early and for Americans when you include transaction costs.

  • $\begingroup$ This is the correct answer. Deep ITM European put values are often less than intrinsic for positive risk-free rates. $\endgroup$ – bcf Jan 11 '17 at 18:54

In reality, european option's premium surely can be less than its intrinsic value which is due to short sell is not free...

  • $\begingroup$ Could you show an example? $\endgroup$ – Bob Jansen Jul 31 '16 at 12:44

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