If a call may expire OTM, why won't a \$1 increase in the underlying's price necessarily increase the call's price by \$1? Zvi Bodie, Alex Kane, Alan J. Marcus. Investments (2018 11 edn). p 723 scanned.


  &nbsp; &nbsp; &nbsp;  Figure 21.9 verifies that the slope of the call option valuation function is less than 1.0,
  approaching 1.0 only as the stock price becomes much greater than the exercise price. This
  tells us that option values change less than one-for-one with changes in stock prices. Why
  should this be? Suppose an option is so far in the money that you are absolutely certain it
  will be exercised. In that case, every dollar increase in the stock price would increase the
  option value by \$1. But if there is a reasonable chance the call option will expire out of the
  money, even after a moderate stock price gain, a \$1 increase in the stock price will not necessarily
  increase the ultimate payoff to the call; therefore, the call price will not respond
  by a full dollar.   


Pls elaborate the embolding? The authors didn't spell out all the reasons. If a call has a chance of expiring OTM


why won't a \$1 increase in the underlying asset's price "necessarily increase the ultimate payoff to the call"?
why won't the call price increase by a full $1?
The reason that a call option price does not increase one-to-one with spot price of the underlying stock is simple: A 1 dollar increase in stock price only increases the likelihood that the option will be ITM. It is not a guarantee of 1 dollar increase in payoff at maturity. Naturally the price of call does not increase by 1 dollar. 

More precisely, the price of the call should increase by the probability that the option will be ITM, which is less than 1. This fact is model-independent. (Try buying calls at one dollar plus current price whenever the underlying increases by one dollar, and see what happens.)

Accordingly, any model of option prices will have this property. Consider, for example, two basic models---the binomial tree and the Black-Scholes model.

Binomial tree

Suppose the stock price today is $S$. Tomorrow the stock price can be $S+\Delta S$ with probability $q$ and $S-\Delta S$ with probability $1-q$.

The price $C$ of a European call option with strike $K &gt; S - \Delta S$ is therefore equal to the expected payoff $(S + \Delta S - K)q$.

(To be correct, assuming risk-free rate $r$, $q$ should be the risk-neutral probability given by 
$$
q e^{-r} (S + \Delta S) + (1- q) e^{-r} (S - \Delta S) = S.
$$
But this is not central to the discussion. 
)

If $S$ increases to $S + 1$, the price of the call becomes
$C = (S + 1 + \Delta S - K)q$.
The increase in the price of the call is exactly the probability $q$ of the call being ITM.

Black-Scholes

In the context of the Black-Scholes, 
the sensitivity of option price with respect to the underlying is called the delta of the option. It is a standard quantity that arises in delta hedging.

Let $C(S_0, K)$ denote the Black-Scholes formula for price of European call with strike $K$ maturing at time $T$.
The delta of the call is the partial derivative $\frac{\partial C}{\partial S_0}$. The delta is the increase in $C(S_0, K)$ when $S_0$ increases by 1 dollar. 

As you would expect, it's immediate from the Black-Scholes formula that the delta $\frac{\partial C}{\partial S_0}$ is equal to the (risk-neutral) probability that the option will be ITM at maturity.

(In Black-Scholes, under risk-neutral dynamics, the price of the stock $S_T$ at maturity $T$ is given by
$$
S_0 e^{( r + \frac12 \sigma^2 )T + \sigma \sqrt{T} N} 
$$
where $N$ is a standard normal random variable. 
So $\frac{\partial C}{\partial S_0}$ is equal to 
$$
P( S_0 e^{( r + \frac12 \sigma^2 )T + \sigma \sqrt{T} N} &gt; K)
$$
where $N$ is a standard normal random variable. 
)