# Why do we perform change of variable for Black Scholes equation

As an entry level financial engineer, I'm studying the Black Scholes equation, which looks like follows:

$${\frac {\partial V}{\partial t}}+{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+rS{\frac {\partial V}{\partial S}}-rV=0$$

But what is the practical reason for changing variable such as: $$y = lnS$$, $$\frac{\partial V}{\partial S}=\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial y}$$ and $$\frac{\partial^2 V}{\partial S^2}=\frac{\partial }{\partial S}(\frac{\partial V}{\partial S})=\frac{\partial }{\partial S}(\frac{1}{S}\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S}\frac{\partial }{\partial S}(\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2}$$

• We are trying to reduce the Black Scholes equation to another equation, the Heat Equation, for which we already know the solution. That is how PDEs are often solved, reduce them to an already known case by change of variables. – noob2 Jan 23 '20 at 22:41

# To make life easier

What is the practical reason for changing variables?

In short, because if we naively tried to find an analytic solution to the Black-Scholes equation as it stands, then it would be very messy, if at all possible. Only the brave, inexperienced, or over-confident would attempt to solve a PDE without trying to simplify it a bit. One of the nice things about maths, and even more so with science/finance, is that you typically find only a handful of differential equations typically pop if they're "organically derived". (Of course there are many nasty ones which do occasionally crop up). Thankfully most of these were solved a century or two ago, and so if we can transform our problem into a standard known form, then we can proceed both analytically/numerically/asymptotically on the shoulders of giants.

# Motivation for the change in variables

When faced with solving a PDE analytically (and numerically) there are typically a few approaches I would try. My typical order of preference goes as:

2. Change variables based on experience to mangle the PDE into a nicer form, ideally one that is well known already.
3. Cautiously run some numeric simulations to try and gain some insight.
4. Can I transform the PDE, e.g. periodic solutions suggest a Fourier Transform. Others might be the Legendre, Laplace, etc.
5. Try separating variables and reduce to systems of ODEs.
6. Look for Green's functions.
7. Magically recognise Feynman-Kac and use this.
8. Know some favourable basis, and try a series solution using this basis, e.g. polynomials.
9. ...

So let's try some of this on the Black-Scholes equation.

Much of the derivation below has been taken from this answer to this question: Transformation from the Black-Scholes differential equation to the diffusion equation - and back

Based on the above, we can see that changing variables would be one of the first things to try. Now the question is: out of the infinite number of possible change of variables I could try, which should I try? There are sometimes some obvious choices, and sometimes not, and of course if you've derived the PDE from a system, you already have some insight which you can utilise. So for the Black-Scholes equation we have:

$$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}+ rS\frac{\partial V}{\partial S}-rV=0$$

Now given in our derivation, if we follow the same line of reasoning as Black-Scholes, then we assumed $$S$$ followed a geometric Brownian motion, and hence is log-normal. If we took a simple contract whose value was the stock is would also be log-normal, which would suggest some sort of exponential process. Similarly we might hope the value of an option would be a positive process, again suggesting that a positive function, perhaps an exponential. Lastly, we might notice that to every $$\tfrac{\partial}{\partial S}$$ there is a corresponding $$S$$ term. While this is required for the dimensions to work, it might also suggest that each time we differentiate we get the same thing plus an $$S$$ term. This sounds reminiscent of the exponential function. With all this being said, there is no guarantee that plugging in an exponential process will make life any easier, but without the gift of hindsight, it doesn't sound like a bad idea, and would be one of the first things to try.

Picking up on the last observation, a second thing to notice before we start is that the PDE is linear and seems to have pairs of $$S\tfrac{\partial}{\partial S}$$ operators. Re-arranging the second order operator into this seems like a sensible first step prior to a change of variables. This gives: $$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2\left(S\frac{\partial }{\partial S}\right)^2V+\left(r-\frac{1}{2}\sigma^2\right)S\frac{\partial V}{\partial S}-rV=0.$$

Now trying the change of variables $$S=\exp(y)$$ based on our previous intuition we get $$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial y^2}+\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial y}-rV=0.$$

If we note that we are trying to determine the value earlier in time compared to the maturity then this suggests a transformation $$\tau = T - t$$, giving: $$\frac{\partial V}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial y}+rV=0.$$

If we now look at the last term, and compare it to the temporal derivative, we see that this would be solved if we had $$V = \exp(rt)$$. While this is not the full solution, perhaps it suggest we substitute in the a drifting term, which based on the Black-Scholes framework is simply just ensuring that all values are properly discounted/normalised to today's money. Hence this suggests $$u = \exp(r\tau)V$$, giving: $$\frac{\partial u}{\partial \tau}-\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2}-\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial u}{\partial y}=0.$$

Lastely, if we consider $$\tau$$ as being time-like and $$y$$ as being space-like, then we see that the coefficient $$(r-\tfrac{1}{2}\sigma^2)$$ corresponds to a velocity-like quantity, which then suggest we transform into a "stationary frame" using $$x=y+(r-\tfrac{1}{2}\sigma^2)\tau$$. Doing this gives the familiar heat equation, which is well understood and has known solutions, and so we stop at: $$\frac{\partial u}{\partial \tau}=\frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}.$$

# Conclusion

There are many ways to change variables, and to someone with a strong background in physics, PDEs, finance, probability, and the rest, several of these may seem obvious after years of experience. I certainly don't think they all are, but hopefully the above serves to help motivate sensible starting points for changing variables. In this situation I've only presented a case where everything worked out great. In reality there would be a substantial phase of trial and error, and just because something might sound sensible, doesn't mean it will reduce the complexity of the PDE.