User:Barrieroption/sandbox

There may exist situations where someone believes that for whatever reason the price of a stock will not exceed a certain threshold upon expiration of the option. This could be due to bearish technical analysis indicators, poor economy, a pending buyout, or other causes that would prevent a stock from rising. Conventional option pricing formulas assume there is no underling positive or negative bias, except for interest rates. This has applications such as allowing traders to better manage risk because it cannot always be assumed that the probability of a stock going higher or lower is equal. By setting a barrier, one can compute prices and probabilities under more conservative conditions.

Define barrier to be $$p_b$$

Strike $$p_s$$

Present price $$p_1$$

Where:

$$p_1 \le p_b$$

$$p_s \le p_b$$

But the barrier itself is not Dirichlet condition meaning that the option doesn't become worthless if it's crossed.

The new option pricing formula must meet the criteria:

$$C \to 0$$ as $$p_b-p_s \to 0$$ (assuming $$p_s>p_1$$)

$$C=p_1$$ if $$p_s=0$$

$$C=0$$ if $$p_1=0$$

As $$p_b \to \infty$$ you have the classic Black-Scholes.

$$t=0, C=p_1-p_s$$; or 0 if negative

We'll also make the barrier rise with interest $$b=be^{rt}$$ rate so that $$C=p_1$$ as $$r \to \infty$$

Rather than modeling a reflection barrier, we can meet the criteria of the option by restricting the payoff to:

$$x-p_s$$ for $$p_b > x > p_s$$ and $$p_b-p_s$$ for $$x>p_b$$

The following formula suffices:

$$C=\left[p_1N(d_1)-p_se^{-rt}N(d_2)-(mN(d_3)-p_be^{rt}N(d_4))\right]H(e^{rt}p_b-p_s)$$

$$m=p_1-(p_1-p_s)H(p_1-p_s-1)$$

Where $$H$$ is the heaviside function and:

$$\begin{align} d_1 &= \frac{1}{\alpha\sqrt{t}}\left[\ln\left(\frac{p_1}{p_s}\right) + \left(r + \frac{\alpha^{2}}{2}\right)t\right] \\ d_2 &= \frac{1}{\alpha\sqrt{t}}\left[\ln\left(\frac{p_1}{p_s}\right) + \left(r - \frac{\alpha^{2}}{2}\right)t\right] \\ d_3 &= \frac{1}{\alpha\sqrt{t}}\left[\ln\left(\frac{m}{p_be^{rt}}\right) + \left(r + \frac{\alpha^{2}}{2}\right)t\right] \\ d_4 &= \frac{1}{\alpha\sqrt{t}}\left[\ln\left(\frac{m}{p_be^{rt}}\right) + \left(r - \frac{\alpha^{2}}{2}\right)t\right] \\ \end{align} $$

Put options are the same as Black-Scholes. Obviously, this barrier assumption violates put-call parity.

With $$r=0$$, the barrier pricing formula also generates calls that aren't monotonically increasing with time and volatility, but contain an extrema point:

For $$p_1>p_s$$ the extrema for the call is $$\alpha\sqrt{t}=\sqrt{\ln(p_b p_1/p_s^2)}$$

For $$p_1<p_s$$ the extrema for the call is $$\alpha\sqrt{t}=\sqrt{\ln(p_b p_s/p_1^2)}$$

With the exception of buyouts, quoted options are always rising for each successive monthly expiration interval, so the barrier isn't always static, but could rise with time. For example: $$p_b=u t+p_1$$ for some constant $$u$$ where $$t$$ is the time until expiration. Or exponential $$p_b=p_1e^{ut} $$

If someone sells a put, the probability of assignment is $$p=1-(N(d_2)-N(d_4))$$ versus $$1-N(d_2)$$ for black scholes. Since the new probability is higher, one must sell further out of the money puts to prevent assignment.

For example, let's assume a stock has $$\alpha= .4,r=0,p_1=30$$ and it's expected that the price will not exceed $32 upon expiration. Assuming $$t=.1$$, what strike should the put be sold so that the probability of assignment is no higher than .2? Black-Scholes gives a strike of $26.8 versus $26.2 with the barrier.

The introduction of a barrier can also generate a volatility skew if Black-Scholes prices are forced to match those of barrier pricing. For example, consider a stock where $$\alpha= .6,p_1=30,r=0,p_b=35$$. Using the barrier formula, we can produce a table of calls for various strikes and then solve for the Black-Scholes volatility $$\alpha_2$$ so that we obtain the same call prices as done in this example:

As $$p_s \to 0 $$ it's evident $$\alpha_2=\alpha$$. Hence, options with skews may have barriers implicitly priced into them.