Why a Knock-Out Option Value Vanishes at Its Barrier

# Does the Knock-out option price go to $0$ when the stock price goes to the barrier $B$?


# Does the Knock-out option price go to $0$ when the stock price goes to the barrier $B$?












I am reading Steven Shreve's book "Stochastic Calculus for Finance 2 Continuous-Time Models", page 304. My intuition is that when the stock price gets closer to the barrier, it will be more and more likely that the price will exceed the barrier in a near future, hence it has a large probability to become worthless. This leads to the consequence that the price of the option should be closer and closer to zero. But I can not justify this intuition from the formula on page 304. Can someone explain this? Thanks a lot.

The formula is $$V(0)=S(0)I_1-KI_2-S(0)I_3+KI_4$$ where $$\quad I_1=\frac{1}{\sqrt{2\pi T}}\displaystyle\int_{k}^be^{\sigma w-rT+\alpha w-\frac{1}{2}\alpha^2T-\frac{1}{2T}w^2}dw$$

$$I_2=\frac{1}{\sqrt{2\pi T}}\displaystyle\int_{k}^be^{-rT+\alpha w-\frac{1}{2}\alpha^2T-\frac{1}{2T}w^2}dw$$ and $$\quad I_3=\frac{1}{\sqrt{2\pi T}}\displaystyle\int_{k}^be^{\sigma w-rT+\alpha w-\frac{1}{2}\alpha^2T-\frac{2}{T}b^2+\frac{2}{T}bw-\frac{1}{2T}w^2}dw$$

$$I_4=\frac{1}{\sqrt{2\pi T}}\displaystyle\int_{k}^be^{-rT+\alpha w-\frac{1}{2}\alpha^2T-\frac{2}{T}b^2+\frac{2}{T}bw-\frac{1}{2T}w^2}dw$$

## Answer by Dom (score 0, accepted)

https://quant.stackexchange.com/a/29756

The reflection principle for a driftless Brownian motion $W_t$ starting at $W_0=0$ tells us that the probability of crossing a barrier $B>0$ before option expiry time $T$ is twice the probability of $W_T$ being above the barrier at expiry.

We now use this result in reverse.

We set $t=0$ and put the barrier just above the current value of $W_0=0$ by setting $B=\epsilon$ where $\epsilon$ is small. This is analogous to the stock price being just below the barrier.

It is clear from symmetry that the probability of W_T being above or below zero at expiry time $T$ will be approximately $1/2$, and exactly $1/2$ as $\epsilon \rightarrow 0$.

Hence the probability of the barrier being crossed at any time before expiry will be twice this number i.e. it will tend to 1.

If the barrier is a knock-out then the option price will tend to zero.

In practice the stock price process has a drift $r$ and is lognormal so the numbers are a little bit different. However the same intuition applies.

To see this in equation in Shreve's book you should note that he defines in (7.3.3) \begin{eqnarray} b=\frac{1}{\sigma} \log \frac{B}{S(0)} \end{eqnarray} So if $S(0) = B$ then $b=0$. Using the identities on page 305 you will see that this means that $I_1=I_3$ and $I_2=I_4$. Substituting these values into the formula on page 304 which is given by \begin{eqnarray} V(0)=S(0)I_1 - K I_2 -S(0)I_3+K I_4, \end{eqnarray} gives \begin{eqnarray} V(0)=0. \end{eqnarray} And so the knock out option price goes to zero when the stock price goes to the barrier.

## Answer by M. Jeunesse (score 0)

https://quant.stackexchange.com/a/29755

Let $X_t $ be a stochastic process. Your question looks like to prove that starting from x and for any t the probability that the time to be above b be smaller than t tends to 1 when $x\uparrow b$.

Let $t>0$. Indeed you want to prove: $$\mathbb{P}_x(\sup_{s\leq t}X_s \geq b)\to_{x\to b}=1$$

If no more assumption on $X $, then it can be true or false.

- Case 1

Let $X $ be a Brownian motion $$\mathbb{P}_x(\sup_{s\leq t}X_s \geq b)=2\mathbb{P}(W_t\geq b-x)$$ By the reflection principle. And you conclude that it tends to 1.

- Case 2

Let $X $ be a Poisson process compensated. I.e $X_t= N_t - \lambda t $ Then by conditioning on $N_t=0$ and $N_t>0$ you can prove that the limit when $x\to b $ tends to $1- P (N_t=0)<1$