# Is it possible to replicate the payoff of a portfolio of options taken from a set of strikes {K1}, given another set {K2} with the same underlying?

Let's say I have 2 different and independent (can't be mixed) set of strikes, {K1} and {K2}. If I create a portfolio of options using the first set and calculate my payoff at expiration, would it be possible to replicate (or approximate) it with the other set of strikes {K2}?

I would appreciate too if you could share some papers on the topic.

I assume that you want to minimize some error function of your replication. For simplicity, I will focus on the squared error integral below.

Without loss of generality, let us assume that for some target strike level $$K$$ (from set 1), there exist $$2n$$ symmetrically (around $$K$$) spaced strikes in set 2, with spacing $$\pm h,\pm2h,\pm3h...,\pm nh$$. In order to minimize the squared distance of the replication, we want to minimize:

\begin{align} E(w_1,\ldots,w_n)&\equiv\int_L^U\left(\sum_j^n\frac{w_j}{2}\left(\left(x-(k+jh)\right)^++\left(x-(k-jh)\right)^+\right)-(x-k)^+\right)^2\mathrm{d}x\\ &=\int_L^U\left(\sum_j^n\frac{w_j}{2}\left(f_j^+(x)+f_j^-(x)\right)-f(x)\right)^2\mathrm{d}x \end{align} In order to control the global error level, we must make sure that the leading order of $$x$$ is zero, i.e. $$\sum w_j=1$$; else the hedge would explode. Then, we can set $$L=k-nh,U=k+nh$$ as our integration range.

The Lagrangean for our problem is

$$L=E-\lambda(w^Te-1)$$ (where $$e$$ is a vector of ones) with optimality condition \begin{align} E_w -\lambda e &= 0 \\ e^Tw&=1 \end{align} Let's look more closely at the gradient to $$E$$:

\begin{align} \frac{\partial E}{\partial w_i}&=\int_L^U\left(\sum_j^n\frac{w_j}{2}\left(f_j^+(x)+f_j^-(x)\right)-f(x)\right)\left(f_i^+(x)+f_i^-(x)\right)\mathrm{d}x\\ &=\int_L^U\sum_j^nw_j\left(f_j^+(x)+f_j^-(x)\right)\left(f_i^+(x)+f_i^-(x)\right)-f(x)\left(f_i^+(x)+f_i^-(x)\right)\mathrm{d}x\\ &\equiv w^TG_i-g_i \end{align} where the functions $$G_i, g_i$$ result from integration and are independent of the choice of $$w_i$$. The gradient of the error function is thus linear in $$w$$, i.e. $$E_w=Gw-g$$. The FOC becomes

$$\begin{pmatrix} G&e\\ e^T &0 \end{pmatrix} \begin{pmatrix} w\\ \lambda \end{pmatrix} = \begin{pmatrix} g\\1 \end{pmatrix}$$ which can be solved thru linear algebra. Please note that I have chosen the strike spacing solely for convenience, the method works with irregularly spaces strikes as well.

### An example.

Say we have an option with strike $$K=10$$, and the replication instruments have strikes $$7,8,9,11,12,13$$ (three pairs). After replacing $$w_3=1-w_1-w_2$$, the corresponding error integral computes to

$$E=\frac{1}{6}(27 + 20 w_1^2 + 23 w_1 (-2 + w_2) - 26 w_2 + 7 w_2^2)$$

with FOC: $$\begin{pmatrix} 40&23\\ 23 & 14 \end{pmatrix} \begin{pmatrix} w_1\\w_2 \end{pmatrix} = \begin{pmatrix} 46\\26 \end{pmatrix}$$

The optimal weights solve to $$w_1\approx 1.483871, w_2\approx -0.58065$$, $$w_3=1-w_1-w_2\approx 0.09677$$, with corresponding error $$E\approx 0.06989$$.

A naive approximation using only the two nearest neighbors (which was my first idea) would result in a (larger) error of $$E=1/6$$. For reference, here's a plot of the error integrands when using two, four, or six neighbouring strikes ($$n=1,2,3$$):

As user dm63 wrote in his answer, this is truly just an exercise in function approximation. Say you want to approximate function $$f$$ using $$n$$ test functions $$g_i$$ in a square error sense. Each test function has some contribution $$w_i$$. Let $$g=\left(g_1,\ldots,g_n\right)^T$$ and $$w=\left(w_1,\ldots,w_n\right)^T$$:
\begin{align} I &\equiv \int_D \left( w^Tg-f \right)^2\mathrm{d}x\\ &= \int_D w^Tgg^Tw -2w^Tgf+f^2 \mathrm{d}x\\ &=w^T\begin{pmatrix} &&\cdots&\\ &&\cdots&\\ \cdots&\cdots&\cdots&\cdots\\ &&\cdots& \end{pmatrix}w\\ &-2w^T\begin{pmatrix} \\ \\ \cdots\\ \end{pmatrix}+\\ &\equiv w^THw-2w^Th+ \end{align} where $$=\int_D f(x)g(x)\mathrm{d}x$$ - This is a quadratic problem and can be solved using corresponding methods.
• I think there is a typo in your first expression, $f_j^+(x)$ should be equal to $(x-(k\color{blue}{+}jh))^+$. Commented Feb 10, 2022 at 23:24