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Given a vanilla options market, i.e. $C(S,K,t, T)$ for all strikes $K$, is it possible to replicate $C^2 (S,K,t,T)$? So I am looking for a self-financing portfolio which has a price equal to $C^2(S,K,t,T)$ for a fixed strike $K$ and for all $t$.

Thanks.

2 Answers2

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I assume your trade $V(S,K,t,T)$ is European. Its payoff is: $$\begin{align} V(S,K,T,T)&=C^2(S,K,T,T) \\[3pt] &=\max(S_T-K,0)^2 \\[3pt] &=\boldsymbol{1}_{\{S_T\geq K\}}(S_T-K)^2 \\[3pt] &=\boldsymbol{1}_{\{S_T\geq K\}}f(S_T) \end{align}$$ where $f(x)=(x-K)^2$. By Carr-Madan's static replication formula (see this question or this paper), we have(1): $$\begin{align} f(S_T)&=f(K)+f'(K)(S_T-K)+\int_0^{K}f''(k)(k-S_T)^+\text{d}k+\int_{K}^{\infty}f''(k)(S_T-k)^+\text{d}k \\[3pt] &=2\int_0^{K}(k-S_T)^+\text{d}k+2\int_{K}^{\infty}(S_T-k)^+\text{d}k \end{align}$$ where $(x)^+=\max(x,0)$. Multiplying by $\boldsymbol{1}_{\{S_T\geq K\}}$: $$\boldsymbol{1}_{\{S_T\geq K\}}f(S_T)=2\int_{K}^{\infty}(S_T-k)^+\text{d}k$$ Multiplying by the discount factor $D(t,T)$ and taking the conditional expectation under the risk-neutral measure $Q$, we get the following theoretical replicating strategy: $$V(S,K,t,T)=2\int_{K}^{\infty}C(S,k,t,T)\text{d}k$$ Given in practice there is no availability of a continuum of call options, the following approximation is made: $$V(S,K,t,T)\approx2\sum_{i=0}^nC(S,k_i,t,T)\delta_i$$ where $\{k_i:i=0,\dots,n\}$ are the quoted strikes with $k_0=K$ and $\delta_i=k_{i+1}-k_i$.

(1) We have chosen as threshold value the strike $K$.

Daneel Olivaw
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  • Thanks, but I probably did not state my question as clearly as I should have. What I am looking for is the replicating portfolio for $\left( E_t \left[ (S_T-K)+ \right] \right)^2$. What you gave is an upper bound, namely $ E_t \left[ (S_T - K)^2+ \right]$. At maturity they will of course be the same. I may actually have found the answer myself, but still checking a few things before I dare post it. In the meanwhile of course if you or anyone else has the solution would be great to know. –  Mar 16 '19 at 18:27
  • @ilovevolatility What is the payoff of your trade? From your description it seems the only cash flow is $\max(S_T-K,0)^2$ at maturity, as you state in your comment, with no intermediate payment thus by no arbitrage the price of your “squared option” must be the same as the price of the claim I describe and hence the hedging strategy. Can you be more specific regarding the structure of your deal? – Daneel Olivaw Mar 16 '19 at 21:07
  • It's true that both have the same value at maturity but they are still different. Take for example the special case $K=0$, then the difference $E_t[ S_T^2] - (E_t[S_T])^2$ is related to the volatility of $S$. –  Mar 17 '19 at 07:18
  • So you are asking: is there a payoff whose price is the square of the regular option price. – dm63 Mar 17 '19 at 09:15
  • @dm63 Exactly, so the question is what is $g(S_{T'})$ such that for all $t < T' < T$ we have $E_t [g(S_{T'})] = \left( C(S_t,t,K,T) \right)^2$. I suspect I'd need options on options but would like to avoid that if there is any other way e.g. by dynamic trading in vanilla options. Since $dC^2 = 2CdC + (dC)^2$ I need to hold $2C$ options at any time, plus something that has a p/l of $(dC)^2$ over a time interval $dt$. It's the last part that I am struggling with, maybe I am missing something very simple though. –  Mar 17 '19 at 10:44
  • @ilovevolatility May I ask the origin of the question? The natural approach would be to start from the payoff and derive a price process, as in a client goes to a bank asking for a specific payoff (ie. to hedge against some risk) and the bank responds with a price and structures a hedging strategy. Here the problem is inversed, in that we want to derive a payoff from a price process. Is there any assurance the payoff will be unique? For example, in the absence of dividends, an EQ European option has the same price as an American one. – Daneel Olivaw Mar 18 '19 at 16:55
  • @DaneelOlivaw I was working on a problem where I thought I needed Volga P/L from an option that has a strike at the point where it's Vanna and Volga is initiallty zero (the $d_2=0$ point). Now I can get Volga p/l even though Volga is zero if I look at the option squared. Hence my question. But it turns out, after endless mucking about on this problem, that I don't need this C^2 'fake' Volga p/l afterall. That said the problem itself I find quite interesting in a more general setting, i.e. replicating Greeks of options using other options. –  Mar 18 '19 at 17:54
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There is no terminal $\mathcal{F}_T$ mesurable payoff $g$ such that $e^{-r(T-t)} E_t[g] = C(S_t, t, T, K)^2$, simply because $E_t[g]$ must be a martingale and $e^{r(T-t)} C(S_t, t, T, K)^2$ is not.

So any deal that has npv $C(S_t, t, T, K)^2$ must involve a stream of intermediary payoffs $ h(S_t,t) dt$, which you can solve for by plugging $V(S,t) = C(S, t, T, K)^2$ in the BS PDE $$ \frac{\partial V}{\partial t} + r S \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} -rV + h(S,t) = 0 $$ to obtain $$ h(S,t) = -\left(\frac{\partial V}{\partial t} + r S \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} - rV\right) $$ along with the terminal payoff $g(S) = \max(S-K,0)^2$

Antoine Conze
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  • Ok, I was wondering if there wasn't a $-rV$ term missing. But it's the discounted payoff $e^{-r(T-t)}E_t[g]$ that must be a payoff, why do you write $\color{blue}{e^{r(T-t)}}C(S_t,t,T,K)^2$ must be a martingale, and not $C(S_t,t,T,K)^2$? – Daneel Olivaw Mar 19 '19 at 13:39
  • Yes I fixed some typos, sorry about that. $e^{-rt} C(S_t, t, T, K)^2$, or equivalently $e^{r(T-t)} C(S_t, t, T, K)^2$, must be a martingale if $C(S_t, t, T, K)^2$ is to represent the npv of a single terminal payoff. – Antoine Conze Mar 19 '19 at 13:44
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    Ok. Just to add to the excellent answer, note that the expression for $h(S,t)$ can thus be decomposed in 1) the (Black-Scholes) hedging P&L for a vanilla option, 2) a funding cost $rV$, and 3) a third term $(\sigma S\Delta_t^{BS})^2$. Not sure how to interpret/hedge this last term... – Daneel Olivaw Mar 19 '19 at 13:59