# Implicit finite difference method always guarantees positive and stable price of derivative?

For the following black scholes pde $$f_t + rSf_S+\frac{1}{2}\sigma^2S^2f_{SS} = rf$$

By denoting $$f_{i}^{n} =$$ Price of derivative at price node $$i$$ and time node $$n$$ and assume uniform grid, the corresponding implicit scheme would be $$a_if_{i-1}^n + b_if_{i}^n + c_if_{i+1}^n = f_i^{n+1}$$ where $$a_i = -\frac{1}{2}\Delta t \left( \frac{\sigma^2S_i^2}{\Delta S^2} - \frac{rS_i}{\Delta S} \right) = -\frac{1}{2}\Delta t(\sigma^2i^2 - ri)\\ b_i = 1+\Delta t \left( \frac{\sigma^2S_i^2}{\Delta S^2}+r \right) = 1+\Delta t(\sigma^2i^2 + r) \\ c_i = -\frac{1}{2}\Delta t \left( \frac{\sigma^2S_i^2}{\Delta S^2} + \frac{rS_i}{\Delta S} \right) = -\frac{1}{2}\Delta t(\sigma^2i^2 + ri)$$

In matrix form, $$CF_n + K_n = F_{n+1} \\ F_n = C^{-1}\left( F_{n+1}-K_n \right)$$ where $$F_n= \begin{pmatrix} f_1^n \\ f_2^n \\ \vdots \\ f_{M-1}^{n} \end{pmatrix}\\ C = \begin{pmatrix} b_1 & c_1 & 0 & \cdots & 0 & 0 \\ a_2 & b_2 & c_2 & \cdots & 0 & 0 \\ 0 & a_3 & b_3 & \cdots & 0 & 0 \\ 0 & \vdots &\vdots &\ddots &\vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{M-1} & b_{M-1} \end{pmatrix}$$ $$K_n = \begin{pmatrix} a_1f_0^n \\ 0 \\ \vdots \\ 0 \\ c_{n-1}f_M^n \end{pmatrix}$$ where $$f_0$$ and $$f_M$$ are two ends of the price grid with some boundary conditions.

There are two questions to ask

1. All the cofficients should be greater than or equal to zero to guarentee that the pricing of derivative is alwasy positive, because referecnece I have read so far mentions that for explicit scheme the cofficients must be greater than equal to zero but not for implicit scheme. I guess it is not necessary, since $$a_i \geq 0$$ when $$\frac{\Delta S}{S_i} \geq \frac{\sigma^2}{r}$$ and this would hold for small $$S_i$$.
2. For stability, I think that $$\left\|C\right\|_{\infty} \geq 1$$ as we take inverse of $$C$$. When $$a_i < 0$$ and $$c_i \geq 0$$, \begin{align} |a_i|+|b_i|+|c_i| &= \frac{1}{2}\Delta t(\sigma^2i^2 - ri) + 1 + \Delta t(\sigma^2i^2 + r) - \frac{1}{2}\Delta t(\sigma^2i^2 + ri) \\ &= -\Delta t \frac{rS_i}{\Delta S} + 1 + \Delta t\frac{\sigma^2S_{i}^{2}}{\Delta S^2} + \Delta r \end{align} and it should be greater than or equal to 1. \begin{align} & -\Delta t \frac{rS_i}{\Delta S} + 1 + \Delta t\frac{\sigma^2S_{i}^{2}}{\Delta S^2} + \Delta r \geq 1 \\ \implies & -\frac{rS_i}{\Delta S} + \frac{\sigma^2S_{i}^{2}}{\Delta S^2}+r \geq 0 \\ \implies & -rS_i\Delta S + \sigma^2S_{i}^{2} + \Delta S^2r \geq 0 \end{align} By letting $$g(S_i, \Delta S) = -rS_i\Delta S + \sigma^2S_{i}^{2} + \Delta S^2r$$, it requires minimum of $$g$$ greater than or equal to 0. $$g_{S_i} = -r\Delta S + 2\sigma^2S_i = 0 \implies S_{i}^{*} = \frac{r\Delta S}{2\sigma^2}$$ and \begin{align} g(S_{i}^{*},\Delta S) &= -\frac{r^2\Delta S^2}{2\sigma^2} + \frac{r^2\Delta S^2}{4\sigma^2} + r\Delta S^2 \\ &= -\frac{2r}{4\sigma^2} + \frac{r}{4\sigma^2} + 1 \\ &= -\frac{r}{4\sigma^2} + 1 \geq 0 \\ \implies & \frac{\sigma^2}{r} \geq \frac{1}{4} \end{align} Therefore, I think that the iteration is not stable for $$\frac{\sigma^2}{r} < \frac{1}{4}$$.

I have tried to find references, but most of them used change of variables to transform black scholes pde into normal heat equation and used von-neumann stability analysis, so I could not find an answer. Thank you in advance.

Edit: $$c_i \geq 0$$ is impossible since $$c_i \geq 0 \implies \sigma^2i^2+ri \leq 0 \implies i \leq -\frac{r}{\sigma^2}$$ Hence, $$|a_i|+|b_i|+|c_i| > 1$$ for any $$a_i$$. Please ignore the second question.

• What is $K_n$? It will make your question much easier to read and spare the reader the guesswork to define the matrices especially $C$.
– Hans
Jul 3 '20 at 19:49
• @Hans Sorry, I should have put what C looks like. I just added more details on matrices and coefficients Jul 4 '20 at 4:19
• That looks slightly better. But what is $c^N_{M-1}$? Is it $c_{M-1}$ raised to the $N$'th power? I doubt that. The index for the boundary condition does not seem right. What is $N$? Is it the final time and the maximum of $n$? Why do you use $f^N_i$ as the entry of $F_n$? Should it not be $f^^n_i$? Your notations are so sloppy and confusing.
– Hans
Jul 4 '20 at 18:21
• @Hans It should be $c_{M-1}$ and $f_{i}^{n}$ indicates value of f at price node i and time node n. You can think it as $f_{i,n}$. Also $F^n$ would seem natural but this would be confusing when I take inverse of it, so instead I put it as $F_n$ Jul 5 '20 at 5:19
• I understand the reasoning. I asked you to check your definition of $F_n$ where you used $f_i^N$ instead of $f_i^n$. That is a mistake. You should be careful writing your question.
– Hans
Jul 5 '20 at 15:55

For the original PDE, the positivity can be deduced from the maximum principle for a parabolic operator. There is also a discrete version of the maximum principle for the finite difference parabolic operator as for example stated in Hung-Ju Kuo and N. S. Trudinger, On the discrete maximum principle for parabolic difference operators which can be applied to prove the positivity of the implicit finite difference scheme of the PDE.

It is easier to show the discrete maximal principle for the canonical heat equation which is obtained by transforming the original PDE through eliminating both $$r$$ and the $$S$$ in the coefficients.

• It would be nice for this answer to be expanded a bit Jul 10 '20 at 14:16
• @BrianB: OK. I will do that after a while. Kind of busy right now.
– Hans
Jul 12 '20 at 7:20
• For implicit scheme, it was straightforward to show that discrete maximum pricinple holds for $a_i < 0$ and $c_i < 0$. However, I could not figure out for the case when $a_i > 0$. Can we show that the maximum principal holds for $a_i >0$ or should we just avoid small $S_i$ when volatility is very small? Jul 17 '20 at 13:11
• @spar7453: Yes, the discrete maximum principle is easy to be shown to be true for, e.g. $r=0$, I thought you needed it for the most general setup. See my second paragraph just added.
– Hans
Jul 17 '20 at 16:46
• If we transform original pde to $u_\tau(\tau,x) = u_{xx}(\tau,x)$, where $x = lnS$ and $\tau = \frac{1}{2}(T-t)\sigma^2$ and show that the maximum principal holds, can we say that it also holds for $a_i>0$ in the original case? I think that it is possible to show by taking non-uniform grid on $x$ to match with $\Delta S$ Jul 18 '20 at 5:23

For positivty, I think it will also depend on the boundary condition. The idea is briefly stated as following.

Let say we have a boundary condition $$f(S_T,T) = min\{S_T - K,0\}$$, $$f(b,0) = 0$$ and $$f(a,t) = 0$$. By Feynman-Kac Theorem, the solution of the original PDE can be written as:

$$f = e^{-rT}\mathbb{E}[min\{S_T-k,0\}]$$ , where $$dS_t = rS_tdt + \sigma S_tdW_t$$

We know $$min\{S_T-k,0\} \leq 0, \forall S_T \in [0,\inf)$$. Therefore, we have $$f \leq 0$$.

If your numerical method converges, your solution, in this case, will be less than or equal to 0 too.

For stability, please see PDE methods for pricing barrier options.

• Your reasoning is incorrect. The OP is asking for the positiveness of each of the finite difference step. The Feynman-Kac formula only applies to the continuum. Also, your min should be max.
– Hans
Jul 10 '20 at 18:23
• Oh, I am sorry I misunderstood the question. I thought Spar was asking the positiveness of the solution of the PDE as "positive and stable price of derivative" is mentioned in the title. I was trying to give an example that a derivative satisfying the Black Scholes pde can be negative. Jul 11 '20 at 15:04