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I'm not able to understand why we are working out probability sT is less than K. For example why could we not have done probability sT is more than K? I understand the steps after that but why we were supposed to start with that first step is something that I don't understand yet.

Any help would be much appreciated.

user134785
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1 Answers1

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The payoff of a digital put option is of the form:

$$f(S_{T})=I_{\{K-S_{T}>0\}}$$

It means that the option gives you $1$ iff $K>S_{T}$ and gives you $0$ iff $K\leq S_{T}$.

The price of this option at time $t=0$ in BS model is given by the following formula:

$$C_{0}=\mathbb{E}^{Q}\left[e^{-rT}f(S_{T})\right]=\mathbb{E}^{Q}\left[e^{-rT}I_{\{K-S_{T}>0\}}\right]=e^{-rT}\mathbb{E}^{Q}\left[I_{\{K-S_{T}>0\}}\right]$$

$$=e^{-rT}\mathbb{Q}\left(K-S_{T}>0\right)=e^{-rT}\mathbb{Q}\left(S_{T}<K\right)$$

Leon
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  • Would the payoff of a digital call option be the opposite? So that means the option gives you 1 iff K is less than sT and gives you 0 iff K is more than equal to sT? – user134785 Mar 17 '16 at 21:09
  • Yes. The payoff of digital call option is $f(S_{T})=I_{{S_{T}-K>0}}$. Could you match my above post as a solution to your problem to close this subject? – Leon Mar 18 '16 at 06:47