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.


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 $t>T_1$.


  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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.