# Double knockout binary pricing?

I'm studying the pricing of a Double-Barrier binary option on the price of $S$. By this I mean an option that pays $X$ at maturity $T$ if the lower ($H1$) or upper barriers ($H2$) are not hit during the lifetime of the option.

I was told that the valuation could be done by subtracting an up-and-in-cash(at expiry)-or-nothing struck at $H2$ from a down-and-out cash or nothing struck at $H1$. That is:

\begin{align*} KO_{H1}- KI_{H2} = &\ (X\ \ \text{if}\ \ \forall\ \ t \le T: S_{t}>H1) - (X\ \ \text{if}\ \ \exists\ \ t \le T: S_{t}>H2) \end{align*}

This valuation kind of makes sense to me because we are considering all the paths that are above $H1$ and subtracting the paths that got above $H2$ which would only leave us the paths between the lower and upper barrier.

However I am doubtful about it since I can't find this way of doing it in any place. Is there a mistake in it?

I've seen a formula for this which involves some infinite series and $sin(x)$ functions, but it seems way too different to my approach.

Much help appreciated

• That must be incorrect, because a path that breaches both barriers would pay out -X in your solution, but the original contract pays out zero. – dm63 Feb 11 '18 at 4:49
• I was just thinking about it! the way I see it if a path goes below $H1$ and then all the way above $H2$ the down-and-out option would have accounted for it (by not considering it) but then if we substract an up-and-in it would again consider all the paths that are above $H2$ (but negative). Thanks – Aldo Shumway Feb 11 '18 at 5:19

Assume that $H_1 < S_0 < H_2$. let \begin{align*} \tau_1 = \inf\{t: \, t>0 \text{ and } S_t \le H_1 \}, \end{align*} and \begin{align*} \tau_2 = \inf\{t: \, t>0 \text{ and } S_t \ge H_2 \}. \end{align*} Then, the option payoff is defined by \begin{align*} X\, \mathbb{I}_{\{\tau_1 >T\}} \mathbb{I}_{\{\tau_2 >T\}} &= X\, \mathbb{I}_{\{\tau_1 >T\}} \left(1-\mathbb{I}_{\{\tau_2 \le T\}}\right)\\ &=X\, \mathbb{I}_{\{\tau_1 >T\}} -X \, \mathbb{I}_{\{\tau_1 >T\}} \mathbb{I}_{\{\tau_2 \le T\}}\\ &=X\, \mathbb{I}_{\{\tau_1 >T\}} -X \, \left(1-\mathbb{I}_{\{\tau_1 \le T\}}\right) \mathbb{I}_{\{\tau_2 \le T\}}\\ &=X\, \mathbb{I}_{\{\tau_1 >T\}} -X \,\mathbb{I}_{\{\tau_2 \le T\}}+ X\,\mathbb{I}_{\{\tau_1 \le T\}}\mathbb{I}_{\{\tau_2 \le T\}}\\ &= (X\ \ \text{if}\ \ \forall\ \ t \le T: S_{t}>H_1)- (X\ \ \text{if}\ \ \exists\ \ t \le T: S_{t}>H_2)\\ &\quad + (X\ \ \text{if}\ \ \exists\ \ t_1 \le T \text{ and } t_2 \le T: S_{t_1}\le H_1 \text{ and } S_{t_2}\ge H_2). \end{align*}