What's the intuition behind the transformation of Black-Scholes into Heat equation?

A sequence of transformations can be used to turn the Black-Scholes PDE into the heat equation.

Let $C(S, t)$ be the price of a vanilla European option at time $t$, maturing at time $T$, where the underlying stock's price is $S$.

$C(S, t)$ satisfies the Black-Scholes equation:

$$\frac{\partial C}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} + (r-q)S\frac{\partial C}{\partial S} - rC = 0$$

By introducing new variables $\tau = \frac{\sigma^2}{2}(T-t)$ and $x = \ln(S/K)$ and a suitable choice of constants $\alpha$ and $\beta$ we can ensure that $e^{\alpha x + \beta \tau} C(S, t)$ satisfies the heat equation (in $x$ and $\tau$).

As someone who has no intuition about PDEs, this last step is quite confusing for me.

The heat equation corresponds to Brownian motion so I was wondering if it's possible to carry out this transformation on the level of stochastic processes and only passing to PDEs once you somehow got Brownian motion.

Multiplication by such an $e^{\alpha x + \beta \tau}$ is a little reminiscent of the Girsanov transformation. I've played around with it, but not getting anywhere.

References: the particular coordinate transformation I give is described in full detail in Wilmott's Mathematics of Financial Derivatives

Another, similar one can be found here https://quant.stackexchange.com/a/110/23872

• Interesting question. What I was told was that "BS PDE is parabolic so not surprising that it can be reduced to Heat Equation the protoype of all parabolic PDE" , but I have never looked at the issue more closely, and in a stochastic process context. – Alex C Nov 13 '16 at 16:12
• there are actually two ways to reduce to the heat equation. One is a multiplication by an exponential. The other is by shifting coords. The former is more like the Girsanov. – Mark Joshi Nov 14 '16 at 0:56

First we should consider the Cauchy-Euler type of ordinary differential equation: $$t^2\frac{d^2 y}{dt^2}+at\frac{dy}{dt}+by=0\tag 1$$ Indeed we should set $\color{red}{t=e^x}$.Please note $$\frac{dy}{dx}=\frac{dt}{dx}\frac{dy}{dt}=e^x\frac{dy}{dt}=t\frac{dy}{dt}\tag 2$$ $$\frac{d^2y}{dx^2}=\frac{dt}{dx}\frac{dy}{dt}+t\frac{dt}{dx}\frac{d^2y}{dt^2}=t\frac{dy}{dt}+t^2\frac{d^2y}{dt^2}=\frac{dy}{dx}+t^2\frac{d^2y}{dt^2}$$ In other words $$\frac{d^2y}{dx^2}-\frac{dy}{dx}=t^2\frac{d^2y}{dt^2}\tag 3$$ Insert $(2)$ and $(3)$ in $(1)$ $$\color{red}{\frac{d^2 y}{dx^2}+(a-1)\frac{dy}{dx}+by=0}\tag 4$$

Second Indeed, the Black-scholes equation looks a little like the heat equation on the infinite interval in that it has a first derivative of the unknown with respect to time and the second derivative of the unknown with respect to the other (space) variable.On the other hand, notice:

• Each time the unknown is differentiated with respect to $S$, it also multiplied by the independent variable $S$, so the equation is not a constant coefficient equation.
• There is a first derivative of $C$ with respect to $S$ in the equation.
• The sign on the second derivative is the opposite of the heat equation form, so the equation is of backward parabolic form.

We eliminate each objection with a suitable change of variables. The plan is to change variables to reduce the Black-Scholes terminal value problem to the heat equation, then to use the known solution of the heat equation to represent the solution, and finally change variables back.

Also we know $$\color{red}{S_t=S_0e^{(r-\frac12 \sigma^2)t+\sigma B_t}}$$ On the other hand, set $$P(y,t,x,s)=P(B(t)\le y\,|\,B(s)=x)=\frac{1}{\sqrt{2\pi(t-s)}}\int_{-\infty}^{y}e^{-\frac{(u-x)^2}{2(t-s)}}dy$$ define $$V(t,x,y)=:\frac{d}{dy}P(y,t,x,0)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}$$ we have $$\frac{\partial V}{\partial t}=-\frac{1}{2t\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}+\frac{(y-x)^2}{2t^2\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}\tag 5$$ and $$\frac{\partial V}{\partial x}=\frac{y-x}{t\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}$$ and $$\frac{\partial^2 V}{\partial x^2}=-\frac{1}{t\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}+\frac{(y-x)^2}{t^2\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}\tag 6$$ Combine $(5)$ and $(6)$ $$\color{red}{\frac{\partial V}{\partial t}=\frac 12\frac{\partial^2 V}{\partial x^2}}\tag 7$$