2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 '16 at 19:43
  • $\begingroup$ @Gordon Sorry some typos one second $\endgroup$ – Wolfy Mar 3 '16 at 20:34
  • $\begingroup$ @Gordon I fixed the typos to the question $\endgroup$ – Wolfy Mar 3 '16 at 20:38
  • 1
    $\begingroup$ 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 '16 at 20:52
  • $\begingroup$ @Gordon could you see my new question I posted its right up your alley $\endgroup$ – Wolfy Oct 11 '16 at 19:33
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ Any chance you could revise this to include the no dominance principle $\endgroup$ – Wolfy Mar 4 '16 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 '16 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 '16 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.