# 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, 2018 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 Feb 11, 2018 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*}

• You put into math what my intuition was thinking of. Cheers! Feb 13, 2018 at 6:29

If you go short a knock-in call on the upper barrier and at the same time go long a knock-out call on the lower barrier, you basically replicate the lower barrier in the double barrier option scenario. But if your long short portfolio hits the upper barrier, profit is locked in at a constant level similiar to a bull spread. On the contrary if the upper barrier is hit by the spot in a double barrier option, the option is completely extinguished without profit.

Following Espen's great book "The complete guide to option pricing formulas", we discover in chapter 4.17.3, that a double knock-out barrier option is the same as going long a plain vanilla call and going short a double knock-in call with same strike, expiry and barrier levels. Valuation can be carried out in the way described in the book.

Regards,

• I think your answer is useful because you added a reference to the formula I've seen before. However I was asking for the mistake in what I was doing which it's clear once one considers a path that breaches both barriers (as it was mentioned in the comments).Thanks Feb 12, 2018 at 3:01

I cannot comment on answer of Gordon, hence comment here. He assumes that: \begin{align*} \mathbb{I}_{\{\tau_{1} >T\}} + \mathbb{I}_{\{\tau_{1} \le T\}} &= 1\\ \mathbb{I}_{\{\tau_{2} >T\}} + \mathbb{I}_{\{\tau_{2} \le T\}} &= 1 \end{align*} However, these are normalized not to 1 but to \begin{align*} \mathbb{I}_{\{\tau_{1} >T\}} + \mathbb{I}_{\{\tau_{1} \le T\}} =\mathbb{I}_{\{S_t < H_1\}}\\ \mathbb{I}_{\{\tau_{2} >T\}} + \mathbb{I}_{\{\tau_{2} \le T\}} = \mathbb{I}_{\{S_t > H_2\}}\\ \end{align*}.

Apparently, the full set is: \begin{align*} \mathbb{I}_{\{S_t < H_1\}} + \mathbb{I}_{\{S_t > H_2\}} + \mathbb{I}_{\{H_1 \le S_t \le H_2\}}=1\\ \end{align*}

or

\begin{align*} \mathbb{I}_{\{\tau_{1} >T\}} + \mathbb{I}_{\{\tau_{1} \le T\}} + \mathbb{I}_{\{\tau_{2} >T\}} + \mathbb{I}_{\{\tau_{2} \le T\}} + \mathbb{I}_{\{H_1 \le S_t \le H_2\}}=1\\ \end{align*}

From the other side, by looking at the "corridor" defined by "floor" and "ceiling" as $$(H_1, H_2)$$ and spanning till "wall" at $$T$$ we can define another full set: \begin{align*} \mathbb{I}_{\{\tau_{1} >T\}} \cdot \mathbb{I}_{\{\tau_{2} > T\}} + \mathbb{I}_{\{\tau_{1} \le T\}} + \mathbb{I}_{\{\tau_{2} \le T\}} = 1\\ \end{align*}

Hence we can expand one more: \begin{align*} \mathbb{I}_{\{\tau_{1} >T\}} \cdot \mathbb{I}_{\{\tau_{2} > T\}} = \mathbb{I}_{\{\tau_{1} > T\}} + \mathbb{I}_{\{\tau_{2} > T\}} + \mathbb{I}_{\{H_1 \le S_t \le H_2\}}\\ \end{align*}

Note, that by setting $$T \to \infty$$ it becomes

\begin{align*} \mathbb{I}_{\{H_1 \le S_t \le H_2\}} \to 0 \\ \end{align*}