1
$\begingroup$

I would like to consider a slight generalisation of this question, which I recall here:

At date of maturity $T_2$ the holder of a financial contract will obtain the amount: $$ \frac{1}{T_2−T_1}\int^{T_2}_{T_1}S(u)du $$ where $T_1$ is some time point before $T_2$. Determine the arbitrage free price of the contract at time $t$. Assume you live in a Black-Scholes world and that $t<T_1$.

Let's call $Z(t)$ the following quantity: $$ Z(t)=\int_{T_1}^t S(u)du, $$ defined only for $t>T_1$, otherwise $Z(t)=0$.

The price of the derivative should be a function of the form $F(t,S(t),Z(t))$.

I would like to dfind the Black-Scholes equation for this derivative.

Attempt

In the Black-Scholes world, we can use a bond and a stock of shares to replicate our derivative. The prices of the bond $B$ and the stock $S$ follow respectively these stochastic processes: $$ \begin{align} dB &= rBdt \\ dS &= \alpha Sdt + \sigma SdW. \end{align} $$ The payoff follows the process: $dZ = Sdt$ only for $t>T_1$, otherwise $Z$ is zero.

To replicate the derivative, we set up a portfolio $V$ whose relative weights are $u_B$ for the bond and $u_S$ for the stock.

The stochastic process for the price of the portfolio is: $$ dV = V\left(u_B r + u_S \alpha\right)dt + u_S \sigma V dW. $$ The stochastic process for the option price is, from Ito's lemma: $$ dF = \left(F_t + \alpha S F_S + S F_Z + \frac12 \sigma^2S^2 F_{SS}\right)dt + \sigma S F_S dW. $$ Since $V$ is a replicating portfolio, it must follow the same stochastic process as $F$, so we can solve for the relative weights of bond and stock: $$ \begin{align} u_S &= \frac{S F_S}{F} \\ u_B &= \frac{F_t + SF_Z + \frac12 \sigma^2 S^2 F_{SS}}{rF}. \end{align} $$ Imposing the obvious constraint $u_B + u_S = 1$, we get the Black-Scholes equation for this derivative: $$ F_t + rs F_s + H(t-T_1) sF_z + \frac12 \sigma^2s^2F_{ss} - rF = 0, $$ where $H(t-T_1)$ is the unit step which is zero for $t<T_1$ and one for $t>T_1$.

Question

  1. Is the equation above correct?
  2. If so, how does one solve it? The step function does not seem a big deal, as one can solve the equation with the term $sF_Z$ in the interval $ [T_1,T_2]$, and without it in the interval $[t,T_1]$. I am more concerned by the derivative $F_Z$ itself, as this would be solved trivially by separation of variables: $F(t,s,z)=A(t,s)\cdot(mz+q)$, $m$ and $q$ deterministic constants, but maybe the equation for $F$ should be recast as an integro-differential equation in the variables $(t,s)$ only?
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.