# Analytical soluton to the Black-Scholes equation with a modified European Call Option

Please consider the following modified European Call Option

where $0 < a \leq 1$. When $a = 1$ the modified European call option is reduced to the standard European call option.

Transforming the Black-Scholes equation in the standard heat equation and using Fourier transform, I am obtaining the following analytical solution for such modified European call option (please do right click on the image to enlarge it)

When $a=1$ the standard Black-Scholes formula for the usual European call option is recovered, namely

My questions are:

1. Do you know another method to derive the solution for the modified European Cal option.

2. Do you know real life cases on which a modified European Call options are applied.

## 3 Answers

When a pay-off is piecewise linear plus jumps, it the same as the portfolio of calls and digital calls. Its price must agree with that of the portfolio by no arbitrage. Every time there is a jump we add in a digital call and every time there is a change in gradient we add in calls equal to the gradient change.

Here we have a call struck at $K$. Just below $2K$ the option pays $K$ and above it pays $a K.$ We have $(a-1)$ digital calls struck at $aK$ and $(a-1)$ calls as well to get the gradient to change to $a.$

Just add the BS formulas for the three contracts to get the price in the BS model.

Using the usual arbitrage arguments, we can write option prices as discounted expectations of future values under risk-neutral probabilities. That is $$V(S,0) = B(0,T) E\left[ V(S,T) \right]$$

Start by re-writing your particular payoff as the following sum $$C_K+aC_{2K}+KD_K$$ where $C_x$ is a call struck at $x$ and $D_x$ is a digital option struck at $x$.

Since expectation is a linear operator (i.e. $E[aX+bY] = aE[X]+bE[Y]$) we can then write the value of your option using the standard Black-Scholes formulas $BS_{Call}(\cdot)$ and $Digital_{Call}(\cdot)$ as

$$V(S,0) = BS_{Call}(K) + a\,BS_{Call}(2K) + K\,Digital_{Call}(K)$$

More generally, $any$ non-pathological terminal payoff $P(S_T)$ can be written in terms of a (finite or converging infinite) sum of standard calls and puts, or of digital calls and puts.

Linearity of the expectation operator then tells us that the present value $V(S_0)$ of that payoff is also a converging infinite sum.

From a measure theory point of view, this happens because linear functions (and step functions) provide a basis for the space of (nonpathological) payoff functions.

• Many thanks for you answer but I do not understand how $C_K+aC_{2K}+KD_K$ is equivalent to my payoff. Please can you give an explanation? Aug 18 '15 at 19:34

I will try to derive from the scratch as follows: Define the three indicator random variables for the three events: $1_A$ equals 1 if the event ${2K<S_T}$ happens,$1_B$ equals 1 if the event ${K\leq S_T \leq 2K}$, and $1_C$ as the indicator for any other events.

If there exists a stochastic discount factor $M_t$ that prices both the stock and bond, then we have

\begin{align} p_t &= E_t[M_T\{a(S_T-K) \cdot 1_A+(S_T-K) \cdot 1_B\}] \\ &= aE_t[M_TS_T1_A]-aE_t[M_TK1_A]+ E_t[M_TS_T1_B]-E_t[M_TK1_B]\\ &=a(Prob^s(A)-Prob^R(A)) + Prob^s(B)-Prob^R(B) \end{align}

Where $Prob^S$ and $Prob^R$ are the probability of these events under the measures with numeriare of the stock (assuming non-dividend-paying) and the money market account respectively. These probability are pretty easy to calculate.