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Any help with those queries is greatly appreciated

Question 1: How to understand call and put options?

So far, I know that call option is a contract with an expiry date $T > t$ and a strike $K > 0$ in which:

  1. the holder (who has the long position) has the right to buy the underlying share for the strike at the expiry date;

  2. the writer (who has the short position) is obliged to deliver the share for the strike if the holder exercises their right.

Does the long position have to pay something initially? If no, then isn't this a form of arbitrage as at maturity, if the price of the share is $>K$ it will just pay the strike and if not, then it will just pay the price, hence he is not going to lose anything?

asdf
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  • This isn't a math question, so far as I can see. – saulspatz Jan 30 '18 at 19:08
  • Of course -- the holder has to pay a premium for a contract that has the potential to be extremely valuable but never less than zero. Otherwise, there would never be a seller. Clearly the existing multi-trillion dollar options market is not one big free lunch. The math is about how to place a fair value on the option. Look up Black-Scholes option pricing formula. – RRL Jan 30 '18 at 19:26
  • Thanks! Is the price that the long position has to pay equal to this $c(S_t, t, T,K)$ expression, i.e. the price of the call option? If you make this whole thing an answer, I'd be more than happy to upvote it and acccept it – asdf Jan 30 '18 at 19:40

1 Answers1

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An arbitrage opportunity exists if an asset price $V_t$ (considered as a stochastic process on a filtered probability space) satisfies at present time $0$ and future time $T$

$$V_0 = 0, \quad P(V_T \geqslant 0) = 1, \quad P(V_T \neq 0) > 0.$$

In other words, we could buy the asset for nothing and the future value is almost surely nonnegative with a non-zero probability of a positive value.

In the absence of arbitrage, there exists a probability measure $Q$ called the risk-neutral measure such that the price of an asset $V_t$ at any time $0 \leqslant t \leqslant T$ can be obtained as an expected value. In particular, it is the expectation of the future price under conditioned on all information known at time $t$ and discounted at the risk-free rate $r$:

$$V_t \,= \,e^{-r(T-t)}\,\,E^{Q}(V_T \,| \,\mathcal{F}_t).$$

Turning to a call option, the future value or payoff at expiration time $T$ is given in terms of an underlying asset (for example, a stock price). At expiry, the holder of the call option receives the underlying asset with price $S_T$ in exchange for the payment of the strike price $K$ so the value of the option at expiry should be $C_T = \max(S_T- K,0)$. The maximum appears here because a rational holder would not choose to exercise if the value of the asset were less than the strike price. Furthermore, in practice, many options do not require a physical exchange but are, rather, settled in cash.

Since a stock price cannot have a value below zero (shareholders have limited liability) the value of the call option at time $t$ can be expressed as

$$C_t = e^{-r(T-t)} \int_0^\infty \max(S_T-K,0) f(S_T |S_t)\, dS_T \\= e^{-r(T-t)} \int_K^\infty (S_T-K) f(S_T|S_t) \, dS_T$$

where $f(S_T|S_t)$ is the conditional probability density of the future stock price $S_T$ (given the current known price $S_t$).

The usual starting point for option pricing is to assume that, under the risk-neutral measure, $S_t$ follows a stochastic process of the form

$$\frac{dS_t}{S_t} = r \, dt + \sigma \, dZ_t$$

where $Z_t$ is a Brownian motion and the parameter $\sigma$ is called the volatility. In this case, the probability density $f$ will be a normal PDF and a closed-form solution can be found depending on the parameters already discussed:

$$C_t = C(t; T,S_t,K,r, \sigma) = \text{Black-Scholes formula}$$

$$$$

RRL
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