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Assumption 2.1 - If the payoff $P$ of a financial instrument is non negative, then the price $p$ of the financial instrument is non negative.

Assume $C$ is just the price of the call option, and $C^{*}$ is also the price of a different call option with the same parameters as $C$ except that $\tau = T - t$ where $T$ is the maturity and $t$ is current time

Assume no dominance, Assumption 2.1. Show that the price of call option should satisfy $$(S - B_t(T)K)_+ \leq C(T,K,S)\leq C^{*}(T,K,S)\leq S$$ Therefore, for any price quote $C^{*}(\tau,K,S)$ of a call option with strike $K$ and time to maturity $\tau$, there exists a unique $\sigma^{imp}(\tau,K,S)$ such that $$C(\tau,K,S,\sigma^{imp},r) = C^{*}(\tau,K,S)$$ $\sigma^{imp}(\tau,K,S)$ is called implied volatility. See figure belowfigure

Attempted proof: Let the price of a call option with strike $K$ be denoted $C(K)$. Assume when we purchase a call option the stock price $S$ is equal to the strike price $K$. Hence we have $$C(T,K,S)\leq S$$ at time $T$.

I am not really sure where to go from here, any suggestions is greatly appreciated.

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    $\begingroup$ You may need to define the notations. For example, what are $L$, $C$, $C^*$, and what is the difference between $C^*(T, K,S)$ and $C^*(\tau, K,S)$? $\endgroup$
    – Gordon
    Mar 3, 2016 at 19:43
  • $\begingroup$ @Gordon Sorry some typos one second $\endgroup$
    – Wolfy
    Mar 3, 2016 at 20:34
  • $\begingroup$ @Gordon I fixed the typos to the question $\endgroup$
    – Wolfy
    Mar 3, 2016 at 20:38
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    $\begingroup$ In this sense, I will assume that $C(T,K,S)=C^*(T,K,S)$, while $C^*(\tau, K, S)$ is just a different notation for $C^*(T,K,S)$. That is, they are all the same. $\endgroup$
    – Gordon
    Mar 3, 2016 at 20:52
  • $\begingroup$ @Gordon could you see my new question I posted its right up your alley $\endgroup$
    – Wolfy
    Oct 11, 2016 at 19:33

1 Answer 1

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Note that $(S_T-K)^+ -S_T \le 0$, By the dominance principle, \begin{align*} 0 &\ge E\left(\frac{S_T-K)^+ -S_T}{e^{rT}}\right)\\ &= E\left(\frac{S_T-K)^+}{e^{rT}}\right) - E\left(\frac{S_T}{e^{rT}}\right)\\ &=C(T, K, S)-S. \end{align*} That is, \begin{align*} C(T, K, S) \le S. \tag{1} \end{align*} On the other hand, since \begin{align*} (S_T-K)^+ -(S_T-K)\ge 0, \end{align*} by the dominance principle, \begin{align*} E\left(\frac{(S_T-K)^+ -(S_T-K)}{e^{rT}}\right) \ge 0. \end{align*} That is, \begin{align*} C(T, K, S) &\ge S-e^{-rT}K.\tag{2} \end{align*} Moreover, since \begin{align*} (S_T-K)^+ \ge 0, \end{align*} by the dominance principle again, \begin{align*} C(T, K, S) &\ge 0.\tag{3} \end{align*} In summary, from (1)-(3), \begin{align*} \big(S-e^{-rT}K\big)^+ \le C(T, K, S) \le S. \end{align*} Here, for a given volatility $\sigma$, \begin{align*} C(T, K, S)(\sigma) &= S\Phi(d_1)-e^{-rT} K \Phi(d_2), \end{align*} where \begin{align*} d_1= \frac{\ln\frac{S}{e^{-rT}K }+\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}, \end{align*} and \begin{align*} d_2= \frac{\ln\frac{S}{e^{-rT}K}-\frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}, \end{align*} is a continuous function of $\sigma$. It is easy to see that \begin{align*} \lim_{\sigma \rightarrow +\infty}C(T, K, S)(\sigma) = S, \end{align*} and \begin{align*} \lim_{\sigma \rightarrow 0+}C(T, K, S)(\sigma) = \big(S-e^{-rT}K\big)^+. \end{align*} Therefore, for any value $C^*$ that satisfies \begin{align*} \big(S-e^{-rT}K\big)^+ < C^* < S, \end{align*} there is a volatility, which we denote by $\sigma^{imp}$, such that \begin{align*} C(T, K, S)(\sigma^{imp}) = C^*. \end{align*}

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  • $\begingroup$ Any chance you could revise this to include the no dominance principle $\endgroup$
    – Wolfy
    Mar 4, 2016 at 2:48
  • $\begingroup$ @MorganWeiss: I do not know the no dominance principle. You may need to provide some background information. $\endgroup$
    – Gordon
    Mar 4, 2016 at 13:40
  • $\begingroup$ I did I have the definition of it, it's called assumption 2.1. Do you see it? $\endgroup$
    – Wolfy
    Mar 4, 2016 at 15:17

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