# Black and scholes option pricing

I have to solve the following problem in the Black and scholes model: find the price at anty $$t\in[0,T)$$ for an option whose payoff at the maturity is: $$$$0 \ \ \ \text{if} \ S_T

SOLUTION I have rewritten the payoff as: $$$$Payoff_T=(K_2-S_T)\textbf{1}_{\{K_1K_2\}}$$$$ Since $$S$$ evolves under martingale measure $$\mathbb{Q}$$ as a geometric Brownian Motion whit dynamic: $$$$S_t=S_se^{(R-\frac{\sigma^2}{2})(t-s)+\sigma Y\sqrt{t-s}}$$$$ where $$Y\sim N(0,1)$$ then I whant to compute for which values of $$Y$$: $$$$K_1 similarly I get: $$$$S_T Now applying the formula for the price in B$$\&$$S market: $$$$price_t=e^{-R(T-t)}E^{\mathbb{Q}}(Payoff_T|\mathcal{F}_t)\\ =e^{-R(T-t)}\bigg(\int_{y_1}^{y_2}(K_2-S_te^{(R-\frac{\sigma^2}{2})(T-t)+\sigma y\sqrt{T-t}})\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy+\int_{y_2}^{\infty}(K_2-K_1)\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy\bigg)$$$$ Now i omit the computation of this integrals (not difficult) and I have the final formula where I denote with $$\Phi(x)=P(X\leq x)$$ with $$X\sim N(0,1)$$: $$$$price_t=e^{-R(T-t)}(K_2-K_1)(1-\Phi(y_2))+K_2e^{-R(T-t)}(\Phi(y_2)-\Phi(y_1))-S_t(\Phi(y_2-\sigma\sqrt{T-t})-\Phi(y_1-\sigma\sqrt{T-t}))$$$$ At this point my questions are:

1. is this computation fine?
2. Since the second question of the exercise is to compute the delta of the contract (Derivative w.r.t the underlying S) is it possible to express the payoff in terms of Call/put options for which I know an explicit expression of the delta?
• There is a typo in your second indicator function, it should be $(K_1-K_2)\mathbf{1}_{\{S_T>K_2\}}$ not $(K_2-K_1)\mathbf{1}_{\{S_T<K_2\}}$ $-$ presuming the first description of the payoff is the correct one. Aug 4 at 11:34

Let us start by considering a bear spread strategy, consisting on long a European put with strike $$K_2$$ and short another European put with strike $$K_1$$. Then the payoff of this portfolio at expiry $$T$$ is: \begin{align} &(K_2-S_T)\textbf{1}_{\{S_T\leq K_2\}}-(K_1-S_T)\textbf{1}_{\{S_T\leq K_1\}} \\ &\qquad=(K_2-S_T)\left(\textbf{1}_{\{K_1< S_T\leq K_2\}}+\textbf{1}_{\{S_T\leq K_1\}}\right) -(K_1-S_T)\textbf{1}_{\{S_T\leq K_1\}} \\ &\qquad=(K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +((K_2-S_T)-(K_1-S_T))\textbf{1}_{\{S_T\leq K_1\}} \\ &\qquad=(K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}} \end{align} We've matched the first term in your payoff. To match your second, we actually need to subtract $$(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}}$$ and add $$(K_2-K_1)\textbf{1}_{\{S_T> K_2\}}$$: \begin{align} &(K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}} -(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}} +(K_2-K_1)\textbf{1}_{\{S_T> K_2\}} \\ \qquad&= (K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +(K_2-K_1)\textbf{1}_{\{S_T> K_2\}} \end{align} Yet, the subtracted term corresponds to the payoff of a cash-or-nothing put option with strike $$K_1$$ and cash payoff $$K_2-K_1$$, whereas the added term is equal to the payoff of a cash-or-nothing call option with strike $$K_2$$ and same cash payment. All these options have known prices and Greeks under the Black-Scholes model.
Therefore, you can price your payoff under a Black-Scholes setting by summing the Black-Scholes prices of 1) a European vanilla put with strike $$K_2$$, and 2) a European cash-or-nothing call with strike $$K_2$$ and cash payment $$C:=K_2-K_1$$, to which you subtract the prices of both 3) a European vanilla put with strike $$K_1$$, and 4) a European cash-or-nothing put with strike $$K_1$$ and cash payment $$C$$.
• @Pefok generally try to express the generic payoff using indicator functions, then refactor this expression so that only indicators of form $\textbf{1}_{\{S_T>K\}}$, $\textbf{1}_{\{S_T\leq K\}}$, etc. are involved as then you can express the payoff as a combination of vanilla options. Aug 16 at 16:37