I want to price Binary Option in Black-Scholes model.

The payoff is of the form $f(S_{T})=I_{\{S_{T}-K>0\}}$.

If we assume that $t=0$ this is easy, because then we have


But how to derive a price at any time $t\in[0,T]$?

For any $t\in[0,T]$ we have something like this:


How to compute $\mathbb{E}^{*}\left[I_{\{S_{T}-K>0\}}|F_{t}\right]$?

My attempt:


I know that $W^{*}_{T}-W^{*}_{t}$ is independent with respect to $F_{t}$ (generated by Brownian Motion).

Does $I_{\{S_{T}-K>0\}}$ is independent with respect to $F_{t}$? Why (if yes)?

If yes $\mathbb{E}^{*}\left[I_{\{S_{T}-K>0\}}|F_{t}\right]=\mathbb{E}^{*}\left[I_{\{S_{T}-K>0\}}\right]=Q(S_{T}-K>0)=\ldots$


3 Answers 3


$I_{\{S_{T}-K>0\}}$ is NOT independent of $\mathcal{F}_{t}$, since \begin{align*} S_T=S_t \, e^{(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^*-W_t^*)}, \end{align*} where $S_t \in \mathcal{F}_t$, though $e^{(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^*-W_t^*)}$ is independent of $\mathcal{F}_t$. However, since $W_T^*-W_t^*$ is independent of $\mathcal{F}_t$, you are able to compute \begin{align*} \mathbb{E}^{*}\left(I_{\{S_{T}-K>0\}}|F_{t}\right) &= Q(S_{T}-K>0 \mid \mathcal{F}_t)\\ &=Q\left(\sigma (W_T^*-W_t^*) \ge \ln\frac{K}{S_t} -(r-\frac{1}{2}\sigma^2)(T-t) \mid \mathcal{F}_t\right)\\ &=Q\left(\frac{W_T^*-W_t^*}{\sqrt{T-t}} \ge \frac{\ln\frac{K}{S_t} -(r-\frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}} \mid \mathcal{F}_t\right)\\ &=1-N\left( \frac{\ln\frac{K}{S_t} -(r-\frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}\right)\\ &=N(d_2), \end{align*} since $\frac{W_T^*-W_t^*}{\sqrt{T-t}}\sim N(0, 1)$ is independent of $\mathcal{F}_t$, while $\frac{\ln\frac{K}{S_t} -(r-\frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}$ is $\mathcal{F}_t$ measurable. Here, $$d_2 = \frac{\ln\frac{S_t}{K} + (r-\frac{1}{2}\sigma^2) (T-t)}{\sigma \sqrt{T-t}}$$

  • $\begingroup$ Could you edit your post and show me how to compute $Q(S_{T}-K>0|F_{t})=N(d_{2})$. The question is how to deal with conditioning with respect to $F_{t}$. $\endgroup$
    – Leon
    Mar 4, 2016 at 7:08
  • $\begingroup$ I think I found a clear justification how to pass from $Q(S_{T}-K>0|F_{t})$ to $N(d_{2})$. Could you check it? $\endgroup$
    – Leon
    Mar 12, 2016 at 18:31

You can also infer the value of your binary option from the value of a European call option with the same strike and time-to-maturity by going long on a call with strike $K$ and time-to-maturity $\tau$, and short on a call with strike $K+\Delta K$ and the same time-to-maturity. If you hold $\frac{1}{\Delta K}$ of that portfolio, then as $\Delta K$ goes to zero, your total payoff becomes closer to the payoff of the binary option.

If you denote by $V_t$ the value of your binary option, and $C_t \left(\tau ,K \right)$ the value of a European call option with strike $K$ and expiring at $t+\tau$, then it is easy to see the following :

$$V_t = \lim_{\Delta K \to 0} \frac{1}{\Delta K} \left( C_t\left(\tau,K\right)-C_t\left(\tau,K+\Delta K\right) \right)$$

Basically the value of the binary option is the opposite of the partial derivative of the price of a European call option of the same strike and time-to-maturity with respect to the strike, ie :

$$V_t = - \frac{\partial C_t}{\partial K} \left( \tau, K \right)$$ $$=-\frac{\partial}{\partial K} \left( e^{-q\tau} S_t \mathcal{N}\left( d_1 \right) - e^{-r \tau} K \mathcal{N}\left( d_2 \right) \right)$$ $$=e^{-r\tau} \mathcal{N} \left( d_2 \right)$$


I think that I found correct answer to my question.

We have the following theorem:

Theorem. If $X$ is independent of $\mathcal{G}$, $Y$ is $\mathcal{G}$ - measurable and $\phi(x,y)$ is bounded function then:


In my problem I want to calculate the following formula:

$C_{t}=\mathbb{E}^{*}\left[e^{-r(T-t)}I_{\{S_{T}-K>0\}}|F_{t}\right]=e^{-r(T-t)}\mathbb{E}^{*}\left[I_{\{S_{T}-K>0\}}|F_{t}\right]=\ldots $

so if i rewrite $S_{T}$ as $S_{T}=S_{t}e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma(W^{*}_{T}-W^{*}_{t})}$ i will get something like this:

$$S_{T}=Y\cdot e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma\cdot X}$$

where $Y$ is $F_{t}$ - measurable and $X$ is independent of $F_{t}$.

Moreover, we can define $\phi(x,y)=I_{\{y\cdot e^{C+D\cdot x}-K>0\}}$ which is bounded.

It means that using the above theorem we get:

$$\ldots=e^{-r(T-t)}\mathbb{E}^{*}\left[I_{\left\{S_{t}e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma(W^{*}_{T}-W^{*}_{t})}-K>0\right\}}|F_{t}\right]$$ $$=e^{-r(T-t)}\mathbb{E}^{*}\left[I_{\left\{S_{t}e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma(W^{*}_{T}-W^{*}_{t})}-K>0\right\}}\right]$$ $$=e^{-r(T-t)}Q\left(S_{t}e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma(W^{*}_{T}-W^{*}_{t})}-K>0\right)=\ldots$$

and this is what we want.


You are basically correct. For any two independent random variables $X$ and $Y$, the theorem says that \begin{align*} E\big(\phi(X, Y) \mid \mathcal{G} \big) = E\big(\phi(X, y) \,\big)\, |_{y=Y}. \end{align*} Here, for the conditional expectation, \begin{align*} \mathbb{E}^*\bigg(I_{\big\{S_{t}e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma(W^*_{T}-W^*_{t})}-K>0\big\}} \mid \mathcal{F}_{t}\bigg) &= \mathbb{E}^*\bigg( I_{\big\{s\,e^{(r-\frac{1}{2}\sigma^{2})(T-t)+\sigma(W^*_{T}-W^*_{t})}-K>0\big\}} \bigg)\,\Big|_{s=S_t}\\ &= N(d_2). \end{align*}


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