Why Explicit Black–Scholes Grids Produce Stair Steps for Digital Options

# Numerical Solution to BS PDE - Digital Option


# Numerical Solution to BS PDE - Digital Option












Here is a relatively simple question about PDE's pricing.

Assume that we are within the BS framework and moreover that interest rate is zero. The price $V(t,S_t)$ of the digital is known to be $\Phi(d2)$.

Now consider the BS PDE and solve it backwards using the explicit method with the straightforward boundary conditions:

$$V(T,S_T) = payoff(S_T)$$ $$V(t,0) =0, \qquad \text{ for } 0\leq t \leq T$$ $$V(t,S_t) = 1, \qquad \text{ for } S_t \text{ large}$$

The numerical solution yields a price that is close enough to BS, but when I plotted the price against the spot at time zero I get this picture, where the PDE solution is the one in black and the red one is a numerical approximation to the SDE:

Some details: (Volatility $=25\%$, Strike $=100$, r $=0$)

Question: How can the staircase-like behaviour of the PDE, instead of a roughly strictly increasing one, be explained mathematically ?

I am assuming it is because the payoff is discontinuous (also pricing other derivatives that were continuous worked just fined), but I would like a semi-rigorous mathematical explanation.

Thanks in advance.

## Answer by Brian B (score 1, accepted)

https://quant.stackexchange.com/a/19283

Rather than thinking about the steps, think about the piecewise regions where your value is constant.

When using the explicit scheme, time zero option value at any stock price for your simple digital option is basically just a function of which antecedent nodes (accounting for backwards timestepping) were above or below the strike.

Slight modifications of the initial stock price are not affecting the value of the antecedent nodes, leading to no change in option value.

It's worth noting that an implicit scheme does not have the same problem, due its greater "smoothing" power -- from allowing neighboring points at a given timestep to influence each others' option values. This is related to, but not the same thing as, implicit schemes' greater stability under altered grid proportions. Implicit schemes are well-known for handling discontinuous payoffs much better than explicit schemes, and better than Crank-Nicholson schemes as well.

## Answer by user16891 (score 0)

https://quant.stackexchange.com/a/19305

We denote by $V_{i}^{n}=V(\tau_n,S_i)$ the value of a Digital call at time $\tau_n$ when the stock price is $S_i$. We use $N_S+1$ points for the stock price,and $N_{\tau}+1$ points for the maturity.Using ${{S}_{\min }}={{t}_{\min }}=0$ uniform grid for $(S,\tau)$ can be constructed as \begin{align} & {{S}_{i}}=i\times ds\quad \quad ,\quad \quad i=0\,,\,1\,,\,2\,,\,\ldots \,,\,{{N}_{S}} \\ & {{\tau }_{n}}=n\times d\tau \quad \,\,\,\,,\quad \quad n=0\,,\,1\,,\,2\,,\,\ldots \,,\,{{N}_{\tau}} \\ \end{align} Where the increments are $ds=S_{max}/N_S$ and $d\tau=\tau_{max}/N_{\tau}$.The first-order derivatives approximated with central differences for an interior point $(S_i,\tau_n)$ \begin{align} & \frac{\partial V}{\partial S}({{S}_{i}},{{\tau}_{n}})=\frac{V_{i+1}^{n}-V_{i-1}^{n}}{2ds} \\ & \frac{\partial V}{\partial t}({{S}_{i}},{{\tau}_{n}})=\frac{V_{i}^{n+1}-V_{i}^{n}}{dt},\\ \end{align} and second-order derivatives is approximated by $$\frac{{{\partial }^{2}}V}{\partial {{S}^{2}}}({{S}_{i}},{{\tau }_{n}})=\frac{V_{i+1}^{n}-2V_{i}^{n}+V_{i-1}^{n}}{ds{{\,}^{2}}}$$ thus,we have $$V_{i}^{n+1}=V_{i}^{n}+\,\left[ \frac{1}{2}{{\sigma }^{2}}{{i}^{2}}(V_{i+1}^{n}-2V_{i}^{n}+V_{i-1}^{n}) \right.\left. +\frac{r}{2}i(V_{i+1}^{n}-V_{i-1}^{n})-r\,V_{i}^{n} \right]d\tau,$$

You assumed $r=0$,then $$V_{i}^{n+1}=V_{i}^{n}+\,\frac{1}{2}{{\sigma }^{2}}{{i}^{2}}(V_{i+1}^{n}-2V_{i}^{n}+V_{i-1}^{n})$$ Indeed,without any changing variable, you convert Black-Scholes PDE into Heat equation.

we continue our discussion!.let \begin{align} & a=1-({{\sigma }^{2}}{{i}^{2}}+r)dt \\ & b=\frac{1}{2}\,\,\,\,\,({{\sigma }^{2}}{{i}^{2}}-ri)dt \\ & c=\frac{1}{2}\,\,\,\,\,({{\sigma }^{2}}{{i}^{2}}+ri)dt \\ \end{align} then $$V_{i}^{n+1}=aV_{i}^{n}+bV_{i-1}^{n}+cV_{i+1}^{n} $$

### Conditions of stability issue

- For $i=1,2,...,N_S$ , $a>0$, then $$ dt<\frac{1}{\sigma^2\,N_S^{\,2}+r}$$

- For $i=1,2,...,N_S$ , $b>0$, then $\sigma^2>r$

- $a+b+c<1$

## Answer by Mark Joshi (score 0)

https://quant.stackexchange.com/a/19328

there are has been a lot of papers on the analysis of the convergence of binomial trees for European options. You can regard a tree as an explicit finite difference method. The conclusions are that the location of the nodes near the strike determine the error. So if $\kappa$ is the fraction of the distance between the strike and the lower node as opposed to the distance between them, you get an asymptotic expansion whose coefficients are functions of $\kappa. $ (see Diener-Diener).

You can avoid the issue by adapting the tree to the strike so that $\kappa$ is constant. Similarly for PDEs.