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Suppose I buy certain number of at-the-money call or put options, and if I want to exercise those options before the expiry date, then do I need to have the cash needed to buy the entitled amount of shares of the underlying stock?

like if I pay a certain amount for a premium on ATM call options entitling me the right to buy Y number of shares of the underlying, then upon exercise, do I need the cash required to buy Y number of the shares?

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This would depend on whether the option you bought is cash- or physically-settled.

Let $V_t$ be the intrinsic value of your option at time $t$, $T$ its maturity and $y$ the number of shares it gives right to. For example, for a call option of strike $K$ written on an underlying $S$ which price process is $(S_t)_{t \geq 0}$, the intrinsic value is $-$ independently on whether the call is European or American:

$$ \forall \, t \in [0,T], \: V_t = y\max(S_t-K,0)$$

Letting $\tau$ be the time of exercise $-$ for a European call $\tau \in \{T,\infty\}$ and for an American one $\tau \in [0,T] \, \cup \, \infty$, $\tau=\infty$ meaning that the option is not exercised $-$ we have:

  • For a cash-settled option, you will receive the cash amount $V_{\tau}\$$ at exercise time $\tau$;
  • For a physically-settled option, you will receive $y$ shares of the underlying $S$ in exchange for the cash amount $yK\$$.

Hence for a physically-settled option you would need indeed the cash amount $yK\$$ to buy the $y$ shares of $S$. In practice, I believe most exchange-traded vanilla options are cash-settled although I am clearly not sure about this.

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  • $\begingroup$ Great answer. If I want to use ATM calls or puts as a leverage, then given predicted expectation and variance of the underlying asset, do you know how the use of the options would transform that expectation and variance? $\endgroup$ – Kevvy Kim May 29 '17 at 20:09
  • $\begingroup$ @KevvyKim That would make a good additional question. On top of my head, taking a call as an example I would say that the expectation of your cash profit $\pi$, i.e. $\pi=S_T-S_0$ with $t=0$ being today and $K=S_0$, would increase (ignoring the option's premium), but the volatility would also probably increase: deep ITM options behave similarly to the underlying hence similar volatility, whereas deep OTM price is highly convex on the underlying price, so I would guess that overall both effects would yield a higher volatility than the underlying. $\endgroup$ – Daneel Olivaw May 29 '17 at 20:21
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    $\begingroup$ Most stock options are physically settled. $\endgroup$ – RRG May 30 '17 at 6:36

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