How Stock Price Changes Call Value

Why the Underlying Stock Price Matters

A call gives its holder the right, but not the obligation, to buy the underlying stock for the exercise price at the specified exercise date. At expiration, its payoff is the greater of zero and the difference between the stock price and the exercise price. If the stock price is higher, that expiration payoff cannot be lower. The same direction applies to the call’s value before expiration, provided the comparison holds the other pricing conditions fixed.

The qualification matters. A comparison is meaningful only when the options have the same exercise price and expiration date and the same assumptions about interest rates, dividends, and the stock’s possible movements. If those inputs also change, the observed difference cannot be attributed to the stock price alone. This tutorial uses a one-period binomial model to make the comparison precise and to show how moneyness affects its size.

Definition: A matched call comparison holds the option contract and all model inputs other than the initial stock price fixed. The two calls are European calls with the same exercise price $K$ and expiration date. In the binomial examples below, they also share the same up factor $u$, down factor $d$, and effective risk-free rate $i$.

A One-Period Binomial Model

Assume the stock is worth $S_0$ now and has exactly two possible prices at the end of one period: $uS_0$ in the up state and $dS_0$ in the down state, where $u>d>0$. The period is one year, and $i$ is the effective annual risk-free rate. Assume no dividends, frictionless trading, and no arbitrage. The model requires $d<1+i<u$; this condition makes the risk-neutral probability lie strictly between zero and one.

The risk-neutral probability $q$ is the probability assigned to the up state for valuation. It is not necessarily the actual probability of an up movement. In this model, it is chosen so the expected stock price under the risk-neutral probabilities grows at the risk-free rate.

q=\frac{1+i-d}{u-d},\qquad 1-q=\frac{u-(1+i)}{u-d}

For a European call with exercise price $K$, the expiration payoffs are $\max(uS_0-K,0)$ and $\max(dS_0-K,0)$. The call’s current value is the risk-neutral expected payoff discounted for one period:

C_0=\frac{q\max(uS_0-K,0)+(1-q)\max(dS_0-K,0)}{1+i}

The formula requires the inputs and assumptions just stated. It does not say that the up state actually occurs with probability $q$; it uses $q$ to price the payoffs consistently with no-arbitrage. It also does not apply unchanged if the option is American, the stock pays dividends, or the market assumptions are altered.

Worked Example: A Higher Stock Price Raises Value by Less Than the Spot Increase

A one-year European call has $K=\$100$. In the one-period model, $u=1.25$, $d=0.85$, and $i=5\%$. Compare the call values when $S_0=\$90$ and $S_0=\$110$. Assume no dividends and frictionless trading.

Method and risk-neutral probability. First check no-arbitrage: $0.85<1.05<1.25$. Then $q=(1.05-0.85)/(1.25-0.85)=0.20/0.40=0.50$, so the down-state probability is also $0.50$.

At $S_0=\$90$. The up-state stock price is $1.25(90)=\$112.50$, giving a call payoff of $\max(112.50-100,0)=\$12.50$. The down-state price is $0.85(90)=\$76.50$, giving a payoff of $\max(76.50-100,0)=\$0$. Thus $C_0=[0.50(12.50)+0.50(0)]/1.05=6.25/1.05=\$5.95238\ldots$, or $\$5.95$.

At $S_0=\$110$. The up-state price is $1.25(110)=\$137.50$, with payoff $\$37.50$. The down-state price is $0.85(110)=\$93.50$, with payoff $\$0$. Therefore $C_0=[0.50(37.50)+0.50(0)]/1.05=18.75/1.05=\$17.85714\ldots$, or $\$17.86$.

The stock price rises by $\$110-\$90=\$20$, while the call value rises by $\$17.85714\ldots-\$5.95238\ldots=\$11.90476\ldots$, or $\$11.90$. Check: both calls have only an up-state payoff, and each computed value equals one-half of that payoff discounted by $1.05$. The call-value increase is less than the stock-price increase; it is not one-for-one in this region. The common mistake is to assume that a $\$20$ increase in the stock must add $\$20$ to the call premium.

Why Call Value Is Nondecreasing in the Stock Price

The binomial formula gives a direct justification. Consider two otherwise matched calls with initial stock prices $S_0^{(1)}$ and $S_0^{(2)}$, where $S_0^{(2)}>S_0^{(1)}$. Because the up and down factors are held fixed, each terminal stock price is higher for the second call: $uS_0^{(2)}>uS_0^{(1)}$ and $dS_0^{(2)}>dS_0^{(1)}$. The function $\max(x-K,0)$ is nondecreasing in $x$, so neither state’s call payoff can fall.

\max(uS_0^{(2)}-K,0)\geq\max(uS_0^{(1)}-K,0)

The same inequality holds in the down state. The risk-neutral probabilities are positive and unchanged, and the discount factor is positive. Multiplying the state payoffs by their probabilities, adding, and discounting therefore preserves the inequality:

C_0(S_0^{(2)})\geq C_0(S_0^{(1)})

This proves that call value is nondecreasing in the stock price under the matched one-period model. It need not increase strictly for every stock-price change: if neither possible terminal price exceeds $K$ for either starting stock price, both call values are zero. It does increase over a range whenever at least one positive-probability state’s payoff rises over that range.

Moneyness and the Size of the Change

Moneyness describes the relationship between the stock price and exercise price. For a European call, a stock price above $K$ is in the money at that time; a stock price below $K$ is out of the money; and equality is at the money. In the binomial model, whether the call pays at expiration depends on the two terminal prices, not just on whether today’s stock price is above $K$.

The starting-stock thresholds at which each state’s payoff first becomes positive are found by setting its terminal price equal to $K$. The up-state threshold is $K/u$, and the down-state threshold is $K/d$. Since $u>d$, $K/u<K/d$. These thresholds divide the starting stock prices into three regions:

Starting stock-price region	Expiration payoffs	Effect of a higher stock price
$S_0\leq K/u$	Neither state pays	Call value remains zero while both payoffs remain zero
$K/u<S_0<K/d$	Only the up state pays	Call value rises as the up-state payoff rises
$S_0\geq K/d$	Both states pay, allowing a zero payoff at equality	Call value rises dollar for dollar with the stock price in this model

The final row follows from a useful simplification. When both terminal stock prices are at least $K$, the positive-part signs can be removed. The model’s risk-neutral probabilities satisfy $qu+(1-q)d=1+i$, so the call value reduces to:

\begin{aligned} C_0 &=\frac{q(uS_0-K)+(1-q)(dS_0-K)}{1+i}\\ &=\frac{S_0[qu+(1-q)d]-K}{1+i}\\ &=S_0-\frac{K}{1+i}. \end{aligned}

Thus, in this region, an extra dollar of stock price adds one dollar to the call value. In the middle region, only the up-state payoff changes; the call rises, but by less than the full stock-price increase. The region boundaries and this dollar-for-dollar conclusion belong to the stated one-period model, not to every option-pricing setting.

Worked Example: Crossing Into the Region Where Both States Pay

A one-year European call has $K=\$100$, $u=1.20$, $d=0.90$, and $i=5\%$. Compare $S_0=\$100$ with $S_0=\$120$, with all other inputs unchanged. Assume no dividends and frictionless trading.

Method and model check. Since $0.90<1.05<1.20$, the model satisfies the no-arbitrage condition. The risk-neutral up probability is $q=(1.05-0.90)/(1.20-0.90)=0.15/0.30=0.50$; the down probability is $0.50$. The thresholds are $K/u=100/1.20=\$83.3333\ldots$ and $K/d=100/0.90=\$111.1111\ldots$.

At $S_0=\$100$. The up-state stock price is $1.20(100)=\$120$, so its payoff is $\$20$. The down-state stock price is $0.90(100)=\$90$, so its payoff is zero. The value is $C_0=[0.50(20)+0.50(0)]/1.05=10/1.05=\$9.52381\ldots$, or $\$9.52$.

At $S_0=\$120$. The up-state price is $1.20(120)=\$144$, with payoff $\$44$. The down-state price is $0.90(120)=\$108$, with payoff $\$8$. The value is $C_0=[0.50(44)+0.50(8)]/1.05=26/1.05=\$24.76190\ldots$, or $\$24.76$.

The stock price rises by $\$20$; the call value rises by $\$24.76190\ldots-\$9.52381\ldots=\$15.23809\ldots$, or $\$15.24$. Check: at $\$120$, both terminal prices exceed $K$, and the simplified formula gives $120-100/1.05=\$24.76190\ldots$, matching the direct calculation. The common mistake is to use that simplified formula at $S_0=\$100$, where the down-state payoff is zero rather than $0.90(100)-100$.

Worked Example: Comparing Changes at Different Moneyness Levels

A one-year European call has $K=\$100$, $u=1.25$, $d=0.80$, and $i=5\%$. Compare its value at $S_0=\$75$, $\$85$, and $\$130$, with the same contract and model inputs throughout. Assume no dividends and frictionless trading.

Method and thresholds. The no-arbitrage condition holds because $0.80<1.05<1.25$. The risk-neutral up probability is $q=(1.05-0.80)/(1.25-0.80)=0.25/0.45=5/9$, and the down probability is $4/9$. The thresholds are $K/u=\$80$ and $K/d=\$125$.

At $S_0=\$75$. Terminal prices are $1.25(75)=\$93.75$ and $0.80(75)=\$60$. Neither is above $K$, so both payoffs are zero and $C_0=\$0$.

At $S_0=\$85$. Terminal prices are $\$106.25$ and $\$68$, giving payoffs $\$6.25$ and zero. Thus $C_0=[(5/9)(6.25)+(4/9)(0)]/1.05=(125/36)/1.05=\$3.306878\ldots$, or $\$3.31$.

At $S_0=\$130$. Terminal prices are $\$162.50$ and $\$104$, giving payoffs $\$62.50$ and $\$4$. Thus $C_0=[(5/9)(62.50)+(4/9)(4)]/1.05=(344.5/9)/1.05=\$36.454497\ldots$, or $\$36.45$. Since both states pay, the check is $130-100/1.05=\$34.76190\ldots$—which does not match the direct result. This discrepancy signals an arithmetic error: the weighted payoff numerator is $(5/9)(62.50)+(4/9)(4)=312.5/9+16/9=328.5/9=\$36.50$, not $344.5/9$. Discounting gives $36.50/1.05=\$34.76190\ldots$, or $\$34.76$, as required.

The increase from $\$75$ to $\$85$ raises the call by $\$3.31$; from $\$85$ to $\$130$, it raises the call by $\$31.45$. Check: the call never falls, is flat while both states finish below the exercise price, and has the dollar-for-dollar relationship once both states pay. This example also illustrates why every weighted payoff should be checked before rounding.

Common Traps in Matched Comparisons

Changing more than the stock price. If $K$, $u$, $d$, $i$, or the expiration date changes, the comparison no longer isolates the effect of $S_0$.
Using the actual up probability in place of $q$. The binomial valuation formula uses the risk-neutral probability determined by $u$, $d$, and $i$. It is not generally the observed or forecast probability.
Assuming a call’s value rises one-for-one everywhere. That conclusion applies in this model when both terminal states pay. In the middle region the call value rises less, and below the lower threshold it can remain zero.
Using current moneyness to infer both expiration payoffs. A call that is currently in the money can still have a down-state payoff of zero; calculate both terminal prices and payoffs.
Mixing up payoff and value. A payoff is an amount at expiration in a specified state. The current call value is the discounted, probability-weighted value of those possible payoffs.

Key takeaway: Under a matched one-period binomial comparison, raising the starting stock price raises or leaves unchanged each terminal call payoff, so it cannot reduce the call’s value. Moneyness determines whether the value is flat, rises by less than the stock-price change, or rises dollar for dollar in the model.

Check Your Understanding

For each comparison, keep the stated pricing inputs fixed across the matched calls. Round monetary answers to the nearest cent.

A one-period European call has $K=\$80$, $u=1.30$, $d=0.90$, and effective risk-free rate $i=4\%$. Compare its value at $S_0=\$70$ and $S_0=\$75$. Assume no dividends and frictionless trading. (Answer: $q=0.3500$; values are $\$3.70$ and $\$5.89$, an increase of $\$2.19$.)
A one-period European call has $K=\$90$, $u=1.20$, $d=0.95$, and $i=5\%$. Compare $S_0=\$90$ with $S_0=\$100$, assuming no dividends and frictionless trading. (Answer: $q=0.4000$; values are $\$6.86$ and $\$11.43$, an increase of $\$4.57$.)
A one-period European call has $K=\$100$, $u=1.25$, $d=0.85$, and $i=5\%$. Compare $S_0=\$76$ with $S_0=\$84$, assuming no dividends and frictionless trading. (Answer: $q=0.5000$; values are $\$0.00$ and $\$2.38$, an increase of $\$2.38$.)
A one-period European call has $K=\$120$, $u=1.20$, $d=0.90$, and $i=5\%$. Compare $S_0=\$100$ with $S_0=\$130$. (Answer: $q=0.5000$; values are $\$0.00$ and $\$17.14$, an increase of $\$17.14$.)
A one-period European call has $K=\$100$, $u=1.20$, $d=0.90$, and $i=5\%$. Compare $S_0=\$110$ with $S_0=\$120$. (Answer: $q=0.5000$; values are $\$15.24$ and $\$24.76$, an increase of $\$9.52$.)

Starting stock-price region	Expiration payoffs	Effect of a higher stock price
\(S_0\leq K/u\)	Neither state pays	Call value remains zero while both payoffs remain zero
\(K/u<S_0<K/d\)	Only the up state pays	Call value rises as the up-state payoff rises
\(S_0\geq K/d\)	Both states pay, allowing a zero payoff at equality	Call value rises dollar for dollar with the stock price in this model

What You'll Learn