Pricing Arbitrary European Payoffs in Time-Varying Black–Scholes

# Solving the Black-Scholes for any arbitrary payoff


# Solving the Black-Scholes for any arbitrary payoff












Good evening,

I'm currently working on the following problem and I would like an opinion on it,

Let's consider the Black-Scholes model with (time-varying) volatility, $\sigma = \sigma(t)$, and (time varying) risk free return rate,$r=r(t)$.

$$ V_t + \frac{\sigma^2(t)}{2}S^2 V_{SS} + r(t)V_S-r(t)V = 0 \space, \space S>0,\space 0<t<T $$

And the following final condition: $$V(S,T) = \phi(S)\space , \space S>0$$ where $\phi$ represents the option's payoff.

I started by considering the following variable change, $$ S = e^x$$ $$ t = T - \theta $$ This allowed me to consider the following functions:

$$ U(x,\theta) = V(e^x,T-\theta) \space,\space \hat\sigma(\theta) = \sigma(T-\theta) \space,\space \hat r(\theta) = r(T-\theta) $$ This also turned my final condition into an initial condition, $U(x,0) = \phi(e^x) $, and I derived the following transformation.

$$ U_{\theta} = \frac{\hat\sigma^2(\theta)}{2}U_{xx} + \Big(\hat r(\theta) - \frac{\hat\sigma^2(\theta)}{2}\Big)U_x - \hat r(\theta)U \space,\space x \in \mathbb{R} \space,\space 0 < \theta < T $$

Then, I introduced a new time variable, $$ \tau(\theta) = \frac{1}{2} \int_{0}^{\theta} \hat\sigma^2(\xi)d\xi$$ I managed to prove that this function is a bijection from an interval $[0,T]$ to an interval $[0,\Upsilon]$. Therefore, $\tau$ is invertible and we can have $\theta = \theta(\tau)$

With this new time variable, I defined the following functions, $$ R(\tau) = \hat r(\theta(\tau)) \space,\space \Sigma(\tau) = \hat\sigma(\theta(\tau))$$ Which then allowed me to to define $$ k(\tau) = 2 \frac{R(\tau)}{\Sigma^2(\tau)} $$

Given $u(x,\tau) = U(x,\theta(\tau))$, I derived the following new equation, $$u_{\tau} = u_{xx} + (k(t)-1)u_x -k(t)u \space,\space x \in \mathbb{R} \space,\space 0 < \tau < \Upsilon $$

I then defined the following "updating factor", $$d(\tau) = e^{{\int_{0}^{\tau}k(\xi)d\xi}} $$ and a new function $$ v(x,\tau) = d(\tau)u(x,\tau) $$ This new function allowed me to derive the following transformation, $$v_\tau = v_{xx} + (k(t)-1)v_x \space,\space x \in \mathbb{R} \space,\space 0 < \tau < \Upsilon $$

I then solved the following PDE problem,

$$ \psi_\tau = (k(t)-1)\psi_x \space,\space x \in \mathbb{R} \space,\space 0<\tau<\Upsilon $$ $$ \psi(x,0) = x $$

This problem has the following solution, $$\psi(x,\tau) = x + \int_{0}^{\tau} k(\xi)-1 d\xi $$

With this $\psi$ solution, with $\psi = y$, I made a new transform with the following function, $$v(x,\tau) = w(\psi(x,\tau),\tau) $$ This transformation allowed me to achieve the heat equation, $$w_\tau = w_{yy} $$ With the initial condition, $$w(y,0) = \phi(e^y)$$

Having all of these transforms and functions, my main goal is to solve the first problem, given all this information above.

$$ V_t + \frac{\sigma^2(t)}{2}S^2 V_{SS} + r(t)V_S-r(t)V = 0 \space, \space S>0,\space 0<t<T $$ $$V(S,T) = \phi(S)\space , \space S>0$$

My question here is the following: should I start by solving the heat equation and reversing each transform one by one? Or is there a simpler way to solve this Black-Scholes equation?

In order not to have this post being twice as long as it is, I won't explicit any reasoning behind these proofs.

I was looking forward into having some kind of clue in order to have a starting point, because I'm really lost in all of this "mess". I really appreciate if you have read this far, and I apologize for the long post.

Thank you!

## Answer by Gabriele Pompa (score 2, accepted)

https://quant.stackexchange.com/a/63436

Yours is the (backward Kolmogorov) PDE of a Black-Scholes model with time-varying short rate and volatility. Now, have you considered at all risk-neutral evaluation and Feynman-Kač representation? See e.g. Bjork chap 5.

Because, the infinitesimal generator of that PDE is the same of the following SDE:

$$ dS(t) = r(t) S(t)dt + \sigma(t) S(t) dW(t) $$

where $W(t)$ is standard Brownian motion. If your $\sigma(t)$ is deterministic (or at least adapted to the $W(t)$-filtration), you still get that $S(T)$ is conditionally (to $S(t)$) log-normally distributed with an effective volatility factor proportional to the integrated rate of variance

$$ \int^T_t\sigma^2(s) ds $$

See e.g. bjork eq. (26.33). Something similar for the time-varying interest rate contribution. Basically, I think you are trying to solve a PDE which in fact you don’t really need to solve, since once you know the terminal conditional distribution of the underlying $S(T)$, say $p(S)$, you can effectively price any (European exercised) payoff integrating it

$$ \int^{\infty}_0 \phi(S) p(S) dS $$

and the solution of that integral is guaranteed to be (a) solution of your PDE. Am I missing something here? Personally I still like your deduction of the heat equation.

## Answer by Animesh Saxena (score 1)

https://quant.stackexchange.com/a/63430

What do you mean by solving it? A heat equation can be solved by a simple sin(x) exp(-x t) function as it will satisfy the equation.

I think what you really mean is satisfy the boundary conditions. Without the boundary condition there are various solutions possible. Assuming you are looking to solve for call option where boundary values are determined using Max(S-K,0) then you have to transfer this also to the heat equation form.

Heat equation solution can of course be taken from fourier series and N number of terms will satisfy it but you will have to calibrate it at the boundary points.