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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.

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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 ($n=1,2,3$):

enter image description here

Addendum

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 $w=\left(w_1,\ldots,w_n\right)^T$:

$$ \begin{align} I &\equiv \int_D \left( w^Tg-f \right)^2\mathrm{d}x\\ &= \int_D w^Tgg^Tw -2w^Tgf+f^2 \mathrm{d}x\\ &=w^T\begin{pmatrix} <g_1,g_1>&<g_1,g_2>&\cdots&<g_1,g_n>\\ <g_1,g_2>&<g_2,g_2>&\cdots&<g_2,g_n>\\ \cdots&\cdots&\cdots&\cdots\\ <g_1,g_n>&<g_2,g_n>&\cdots&<g_n,g_n> \end{pmatrix}w\\ &-2w^T\begin{pmatrix} <f,g_1f>\\ <f,g_2>\\ \cdots\\ <f,g_n> \end{pmatrix}+<f,f>\\ &\equiv w^THw-2w^Th+<f,f> \end{align} $$ where $<f,g>=\int_D f(x)g(x)\mathrm{d}x$ - This is a quadratic problem and can be solved using corresponding methods.

HTH?

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    $\begingroup$ I think there is a typo in your first expression, $f_j^+(x)$ should be equal to $(x-(k\color{blue}{+}jh))^+$. $\endgroup$ Feb 10, 2022 at 23:24
  • $\begingroup$ Thanks, I've updated accordingly. $\endgroup$ Feb 11, 2022 at 7:52
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This is more of a math question than a quant finance question. You’re asking if a piece wise linear function with “breaks” at {K1} can be approximated by a piece wise linear function with “breaks” at {K2}. Well if {K2} is sufficiently fine (strikes very close together) then you can get arbitrarily close, yes. But if it isn’t , you will be a long way off.

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