I'm looking for put-call parity for the call and put digital options, but I don't really know what is digital options and it's difference between binary options.

I found that payoff of the digital call option is: $$ C^b(T) = \begin{cases}0, \; S(T) \leq K \\ 1, \: S(T) > K \end{cases},$$ and payoff for digital put option is: $$ P^b(T) = \begin{cases} 1, \; S(T) \leq K \\ 0, \: S(T) > K \end{cases}.$$ Are they the same for binary options? Does "$^b$" in $C^b$ mean it binary ( = digital) option?

Next I found put-call parity is as follows: $$ C - P = S(t) - Ke^{-r(T-t)},$$ so put-call parity for the call and put digital options would be

$$ \begin{cases} -1 = S - Ke^{-r(T-t)},\; if \; S(T)\leq K \\ 1 = S - Ke^{-r(T-t)},\; if \; S(T)> K \end{cases} \;?$$

  • 1
    $\begingroup$ welcome to QSE! binary options and digital options are the same thing. in theory you have "cash-or-nothing" and "asset-or-nothing" variations. I have only seen the former in practice. $\endgroup$
    – oronimbus
    May 14, 2021 at 16:55
  • $\begingroup$ The standard put-call parity doesn't apply to binary options. But note that here buying both a put and call with the same strike gives you a payoff of $1$ for sure. $\endgroup$
    – fes
    May 14, 2021 at 16:59
  • $\begingroup$ What do you mean @fesman? What kind of put-call parity can I use for binary options then? $\endgroup$
    – Saguro
    May 14, 2021 at 18:40
  • $\begingroup$ $C+P=e^{-r(T-t)}$ because buying a put and call with the same strike gives the same payoff than putting $e^{-r(T-t)}$ in the money market (1). $\endgroup$
    – fes
    May 14, 2021 at 19:27
  • $\begingroup$ Can you be more specific or recommend me article that explains it? Why is there plus sign next to $P$? And does it mean the final put-call parity is always $1 = e^{-r(T-t)}$? $\endgroup$
    – Saguro
    May 14, 2021 at 20:15

1 Answer 1


Generally, I would say it is a bit difficult to look for put-call parity of something you do not know what it is in the first place.

To give more intuition to what fesman wrote, look at your C and P equation. If you have a C+P, you get 1 no matter what. Why +P? That is just because it is the most natural way to look at it.

  • Call: you get 1 if spot ends up above strike
  • Put: you get 1 if spot is below strike.

Combined, call plus put, will give you 1 no matter the outcome. That payoff, will be in the future though, so you discount to today. Furthermore, it can be any monetary value. This is the cash or nothing variant.

The other variant is not important here but intuitively, if you get the asset, further increases or decreases in Spot above or below the strike have an impact on the actual value. Assume the asset has a strike of 100 and you receive either 100 USD or 1 asset. 100 USD will always be 100 USD, no matter the value of the underlying. However, 1 asset will only be worth 100 if spot is equal to strike. If spot ends up 10% above strike, you actually gain 110 USD.

Simplified, it is like flipping a coin (binary means one of two outcomes). Heads you win (1 or any agreed cash payment), tails you lose (get nothing). In a fair coin example, the price will be 0.5. Why? Because your expected outcome is that it falls 50% of the time on heads and 50% on tails. Hence, you will get 1 in 50% of the time, and zero in the other outcome. Assume you agree to play in 1 year today. You can still expect to get 0.5 but the time value of money tells you it needs to be discounted to todays value.

The fair price of a digital options is also the probability of the event happening (S>K for call). At least theoretically. In reality they are usually priced as call spreads.

  • $\begingroup$ @Saguro, if this answered your question please consider also accepting it or comment what's missing. $\endgroup$
    – AKdemy
    Oct 17, 2022 at 0:56

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