# Put-Call parity and the log-transformed Black-Scholes PDE

We will assume zero interest rates and no dividends on the asset $$S$$ for clarity. The results can be easily generalized to the case with non-zero interest rates and dividends. Under those assumptions, the Black-Scholes PDE is: $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} = 0.$$

An implicit Euler discretisation on a uniform grid in $$S$$ of width $$h$$ with linear boundary conditions (zero Gamma) leads to:

$$V^{k+1}_i - V^{k}_i = \frac{1}{2}\sigma^2 \Delta t S_i^2 \frac{V^{k}_{i+1}-2V^k_{i}+V^{k}_{i-1}}{h^2}.$$ for $$i=1,…,m-1$$ with boundaries $$V^{k+1}_i - V^{k}_i = 0.$$ for $$i=0,m$$.

This forms a linear system $$M \cdot V^k = V^{k+1}$$ with $$M$$ is a tridiagonal matrix where each of its rows sums to 1 exactly. Furthermore, the payoff corresponding to the forward price $$V_i = S_i$$ is exactly preserved as well by such a system as the discretized second derivative will be exactly zero. The scheme can be seen as preserving the zero-th and first moments.

As a consequence, by linearity, the put-call parity relationship will hold exactly (note that in between nodes, any interpolation used should also be consistent with the put-call parity for the result to be more general).

This result stays true for a non-uniform discretisation, and with other finite difference schemes as shown in this paper.

It is common to consider the log-transformed problem in $$X = \ln(S)$$ as the diffusion is constant then, and a uniform grid much more adapted to the process. $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 \frac{\partial^2 V}{\partial X^2}-\frac{1}{2}\sigma^2 \frac{\partial V}{\partial X} = 0.$$

An implicit Euler discretisation on a uniform grid in $$X$$ of width $$h$$ with linear boundary conditions (zero Gamma in $$S$$ ) leads to:

$$V^{k+1}_i - V^{k}_i = \frac{1}{2}\sigma^2 \Delta t \frac{V^{k}_{i+1}-2V^k_{i}+V^{k}_{i-1}}{h^2}-\frac{1}{2}\sigma^2 \Delta t \frac{V^{k}_{i+1}-V^{k}_{i-1}}{2h}.$$ for $$i=1,…,m-1$$ with boundaries $$V^{k+1}_i - V^{k}_i = 0.$$ for $$i=0,m$$.

Such a scheme will not preserve the forward price anymore. This is because now, the forward price is $$V_i = e^{X_i}$$. In particular, it is not linear in $$X$$.

It is possible to preserve the forward by changing slightly the diffusion coefficient, very much as in the exponential fitting idea. The difference is that, here, we are not interested in handling a large drift (when compared to the diffusion) without oscillations, but merely to preserve the forward exactly. We want the adjusted volatility $$\bar{\sigma}$$ to solve $$\frac{1}{2}\bar{\sigma}^2 \frac{e^{h}-2+e^{-h}}{h^2}-\frac{1}{2}\sigma^2 \frac{e^{h}-e^{-h}}{2h}=0.$$ Note that the discretised drift does not change, only the discretised diffusion term. The solution is: $$\bar{\sigma}^2 = \frac{\sigma^2 h}{2} \coth\left(\frac{h}{2} \right) .$$ This needs to be applied only for $$i=1,…,m-1$$.

This is actually the same adjustment as the exponential fitting technique with a drift of zero. For a non-zero drift, the two adjustments would differ, as the exact forward adjustment will stay the same, along with an adjusted discrete drift.