# 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.

• Non-stability!. Do you have selected conditions for Method of Stability ? – user16891 Aug 12 '15 at 18:44
• I tried many different combinations for the path-step and time-step but didn't actually implement the stability condition for the BS. I considered different kinds of volatilities (not just BS) so the condition wasn't true anymore. In any case, the conditions are stable for most of the other stuff I priced (i.e. Vanilla Call) and they are far more complicated (and expensive) than the binary itself. In fact if you join the midpoints of the "step" pieces of the plot you get a nice convex function I believe :) – Alex Apas Aug 13 '15 at 17:28
• $r=0$ ? . This assumption is not consistent with reality – user16891 Aug 13 '15 at 17:40
• Indeed, I was interested mostly in the mathematics behind it. – Alex Apas Aug 16 '15 at 11:44

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.

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

1. For $i=1,2,...,N_S$ , $a>0$, then $$dt<\frac{1}{\sigma^2\,N_S^{\,2}+r}$$
2. For $i=1,2,...,N_S$ , $b>0$, then $\sigma^2>r$
3. $a+b+c<1$
• The third condition is trivial. – user16891 Aug 13 '15 at 19:36
• The second condition is trivial in this case. The third doesn't hold unless you make it less or equal. – Alex Apas Aug 16 '15 at 11:47
• In any case, this doesn't answer the original question, which was how can the staircase behaviour be explained. This is not a case of instability. I was interested in the step-function behaviour of the solution. – Alex Apas Aug 16 '15 at 11:50
• You can think about my question as related to the Heat equation if you like then ^^ The question is still the same, why does it exhibit such behaviour ? – Alex Apas Aug 16 '15 at 11:58
• If $r=0$ thus we have $V_t=\frac{1}{2}\sigma^2 s^2 V_{SS}$ – user16891 Aug 16 '15 at 12:08

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.