I understand how to derive the black scholes solution if $dS_t$ = $\mu S_tdt$ + $\sigma S_tdW_t$ and r is constant. The solution is c(t, x) = $xN(d_{+}(T - t), x))$ - K$e^{-r(T - t)}N(d\_(T - t), x))$ where $d_{+}(\tau, x)$ = $\frac{1}{\sigma\sqrt{\tau}}$ * $[log\frac{x}{K} + (r + \frac{1}{2}\sigma^2)\tau]$, $d\_(\tau, x) = d_{+}(\tau, x) - \sigma \sqrt{\tau}$

However, I need to find the solution when, $dS_t = \mu_{t}S_tdt + \sigma_{t}S_tdW_t$ and $r_t$ are deterministic functions of t. I was asked to guess the solution, so it must be a very close analogue to the solution above. I thought about integrating over time, but I haven't been able to verify that this works, and I do need to verify the solution.

Any help in figuring out what the form and how to go about verifying that it is a solution would be appreciated.

Update: Someone asked to see some extra work, here is my guess of what the solution should be: c(t, x) = $xN(d_{+}(T - t), x))$ - $Ke^{-\int_0^{T - t}r_udu}$N(d_(T - t), x)) where $d_{+}(\tau, x) = \frac{1}{\int_0^\tau \sigma_udu}$ * $[log\frac{x}{K} + \int_0^\tau (r + \frac{1}{2}\sigma^2)]$, $d\_(\tau, x) = d_{+}(\tau, x) - \int_0^\sqrt{\tau} \sigma_udu$.

I don't know if this guess is even correct, and if it is I need to verify that it is a solution the Black-Scholes PDE.

  • $\begingroup$ Just plug in the deterministic function of t and solve the equation. It is not that hard. $\endgroup$
    – Matt Wolf
    Jan 26, 2015 at 4:49
  • $\begingroup$ I don't understand your comment, the functions are arbitrary functions of t. What is the solution then? $\endgroup$
    – user7348
    Jan 26, 2015 at 5:00
  • $\begingroup$ Can you show some of the work you have done and where you are stuck? This is not a homework site, please show a bit of effort and I am happy to try to help and I am sure others as well...(I am saying that because your equation shown looks awfully similar than the notation in Shreve's book "Stochastic Calculus for Finance II". Except, one time you attempt denote tau with "T-t" the other with τ. $\endgroup$
    – Matt Wolf
    Jan 26, 2015 at 5:07
  • $\begingroup$ @MattWolf You are correct that I was following the Shreve example. But, this is a modification of the problem from Shreve, and yes it is from a course I'm taking. I will try to talk to the professor, but they're not always available. I updated my question to show some progress. $\endgroup$
    – user7348
    Jan 26, 2015 at 5:24
  • $\begingroup$ Try to derive d(ln S(t)) and see where that gets you... $\endgroup$
    – Matt Wolf
    Jan 26, 2015 at 6:02

1 Answer 1


What you need is to identify the distribution of the asset price $S_T$, conditional on the information set $\mathcal{F}_{t}$ at time $t$, for $0\leq t < T$. Note that \begin{align*} S_T &= S_t \exp\bigg(\int_{t}^T \Big(r_s-\frac{\sigma_s^2}{2}\Big)ds + \int_t^T\sigma_s dW_s \bigg). \end{align*} Let \begin{align*} P(t, T) = \exp\bigg(-\int_t^T r_s ds \bigg), \end{align*} and \begin{align*} \hat{\sigma} = \sqrt{\frac{1}{T-t}\int_t^T\sigma_s^2 ds}. \end{align*} Then \begin{align*} S_T &= F(t, T)e^{-\frac{\hat{\sigma}^2}{2}(T-t) + \hat{\sigma}\sqrt{T-t} Z}, \end{align*} where $F(t, T)=S_t/P(t, T)$ is the forward price, and $Z$ is a standard normal random variable independent of $\mathcal{F}_t$. Consequently, the value at time $t$ of the option payoff $[\psi(S_T-K)]^+$, where $\psi = \pm 1$, is given by \begin{align*} P(t, T) E\Big([\psi(S_T-K)]^+ \mid \mathcal{F}_t \Big) &= P(t, T)\psi\big[F(t, T) N(\psi d_1) - KN(\psi d_2) \big], \end{align*} where \begin{align*} d_{1} = \frac{\ln F(t, T)/K + \frac{\hat{\sigma}^2}{2}(T-t)}{\hat{\sigma}\sqrt{T-t}}, \end{align*} and \begin{align*} d_2 = d_1 - \hat{\sigma}\sqrt{T-t}. \end{align*}


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