Volatility and Option Value

Why Volatility Matters to Option Holders

An option holder’s expiration payoff is asymmetric: the holder can benefit from favorable stock-price movements but does not have to accept a negative payoff on the option itself. A wider range of possible stock prices can therefore increase an option’s expected payoff, even when the expected stock price stays the same. This is the basic reason that greater volatility generally increases both call and put values.

The comparison needs a careful condition. A higher standard deviation alone does not guarantee that one distribution has higher expected value for every convex payoff. The clean result applies when the more dispersed distribution is a mean-preserving spread of the less dispersed one: it has the same mean, but introduces additional dispersion without changing that mean. In standard option-pricing models, increasing volatility while holding the other inputs fixed has this effect on call and put values.

Definition: The volatility of a stock price describes the scale of its fluctuations, often measured as a standard deviation or, in a model, as a standard deviation parameter. A mean-preserving spread replaces some outcomes with more dispersed outcomes while preserving their probability-weighted mean. A call or put payoff is convex if its value at a weighted average stock price is no greater than the correspondingly weighted average of its values at the component stock prices.

A One-Period Pricing Model

Use a one-year model to make the comparison precise. Assume a European option on a non-dividend-paying stock, with exercise price $K$, expiration in one year, and a known effective annual risk-free rate $i$. Assume frictionless trading, no arbitrage, and that the stated probabilities are risk-neutral pricing probabilities. Under these assumptions, the risk-neutral expected terminal stock price is $S_0(1+i)$, where $S_0$ is the current stock price. The option value is its risk-neutral expected expiration payoff discounted at the risk-free rate.

E_Q[S_T]=S_0(1+i), \qquad C_0=\frac{E_Q[\max(S_T-K,0)]}{1+i}, \qquad P_0=\frac{E_Q[\max(K-S_T,0)]}{1+i}.

Here $E_Q$ denotes expectation using risk-neutral probabilities, $C_0$ is the call value, and $P_0$ is the put value. These probabilities are pricing weights; they need not be forecasts of real-world frequencies. The equations apply to the stated one-year, European, no-dividend setting. Other maturities, dividends, or exercise rights require the relevant pricing assumptions and formulas.

The call payoff $\max(S_T-K,0)$ is convex: below $K$ it is flat at zero, and above $K$ it rises one-for-one with the stock price. The put payoff $\max(K-S_T,0)$ is also convex: above $K$ it is zero, and below $K$ it rises as the stock price falls. That convexity is the key to the volatility comparison.

Worked Example: Comparing Two Dispersions for a Call and Put

A stock is worth $S_0=\$100$. The effective annual risk-free rate is $i=5\%$, so the one-year risk-neutral expected stock price is $\$100(1.05)=\$105$. Consider European options with $K=\$100$. In Distribution A, the terminal stock price is $\$90$ or $\$120$, each with probability $0.50$. In Distribution B, it is $\$70$ or $\$140$, each with probability $0.50$. Find both option values under each distribution.

Method and check of the assumptions. Both distributions have the required risk-neutral mean:

E_Q[S_T\mid A]=0.50(90)+0.50(120)=105,\qquad E_Q[S_T\mid B]=0.50(70)+0.50(140)=105.

The second distribution is more dispersed around the same mean. For the call under A, the payoffs at $\$90$ and $\$120$ are $\$0$ and $\$20$, so the expected payoff is $0.50(0)+0.50(20)=\$10$. Discounting gives $C_0=\$10/1.05=\$9.52$, to the nearest cent.

For the call under B, the payoffs are $\$0$ and $\$40$. The expected payoff is $0.50(0)+0.50(40)=\$20$, and the value is $C_0=\$20/1.05=\$19.05$.

For the put under A, the payoffs are $\$10$ and $\$0$. The expected payoff is $0.50(10)+0.50(0)=\$5$, so $P_0=\$5/1.05=\$4.76$. Under B, the put payoffs are $\$30$ and $\$0$, giving an expected payoff of $0.50(30)+0.50(0)=\$15$ and a value of $P_0=\$15/1.05=\$14.29$.

Result and trap avoided. Both options are more valuable under B, even though both distributions have the same expected terminal price. The comparison would not isolate volatility if the means differed; a change in mean would also affect expected option payoffs.

Why a Mean-Preserving Spread Raises Convex Payoffs

The basic justification uses the defining inequality for a convex payoff $g$. If a stock-price outcome $m$ is replaced by two outcomes $m-a$ and $m+a$, each with probability one-half, their average remains $m$. Convexity implies that the average payoff at the two new outcomes is at least the payoff at the original outcome:

g(m)\leq \frac{g(m-a)+g(m+a)}{2}.

This inequality is Jensen’s inequality for two equally weighted outcomes. If the outcome $m$ had probability $q$ in the original distribution, the replacement changes its contribution to expected payoff from $qg(m)$ to $q[g(m-a)+g(m+a)]/2$. The expected-payoff change is therefore

q\left(\frac{g(m-a)+g(m+a)}{2}-g(m)\right)\geq 0.

The replacement preserves the expected stock price because $q(m-a)/2+q(m+a)/2=qm$. Applying the inequality to the convex call and put payoffs shows that neither expected payoff falls. Discounting both by the same positive factor $1/(1+i)$ preserves the comparison, so neither option value falls either.

This argument establishes the result for a mean-preserving spread. It does not establish that any two distributions with the same mean are ordered by their variance. The full distribution matters: an increase in variance, by itself, is not a universal guarantee that expected value rises for every convex payoff.

Worked Example: Splitting One Outcome to Increase Dispersion

A stock has $S_0=\$100$, the one-year effective annual risk-free rate is $i=5\%$, and European options have $K=\$110$. Under Distribution A, $S_T=\$90$ or $\$120$, each with probability $0.50$. Under Distribution B, the $\$120$ outcome in A is replaced by $\$100$ and $\$140$, each with probability $0.25$; the $\$90$ outcome still has probability $0.50$. Calculate the call and put values and verify that this is a mean-preserving spread.

Check the pricing mean. The required risk-neutral mean is $S_0(1+i)=\$105$. Distribution A has mean $0.50(90)+0.50(120)=\$105$. Distribution B has mean $0.50(90)+0.25(100)+0.25(140)=45+25+35=\$105$. The replacement preserves the mean because $[100+140]/2=120$.

Call under A. The payoffs are $\max(90-110,0)=\$0$ and $\max(120-110,0)=\$10$. The expected payoff is $0.50(0)+0.50(10)=\$5$, and the value is $\$5/1.05=\$4.76$.

Call under B. The payoffs at $\$90,\$100,\$140$ are $\$0,\$0,\$30$. The expected payoff is $0.50(0)+0.25(0)+0.25(30)=\$7.50$, and the value is $\$7.50/1.05=\$7.14$.

Put under A. The payoffs are $\max(110-90,0)=\$20$ and $\max(110-120,0)=\$0$. The expected payoff is $0.50(20)+0.50(0)=\$10$, so the value is $\$10/1.05=\$9.52$.

Put under B. The payoffs at $\$90,\$100,\$140$ are $\$20,\$10,\$0$. The expected payoff is $0.50(20)+0.25(10)+0.25(0)=\$12.50$, and the value is $\$12.50/1.05=\$11.90$.

Check. Both distributions have risk-neutral mean $\$105$, while B replaces one outcome by two equally weighted outcomes around it. Both calculated option values rise, as convexity predicts. The common mistake is to compare payoff averages without discounting; the option values are the discounted amounts.

The Call and Put Payoff Shapes

The two option types benefit from different tails. A call gains from unusually high stock prices; a put gains from unusually low stock prices. A mean-preserving spread adds dispersion in both directions, so it can increase both option payoffs even though the options respond to opposite price movements. Which option is more valuable depends on its strike and the pricing inputs; volatility alone does not make their values equal.

Worked Example: A Put When the Strike Exceeds the Expected Price

A stock has $S_0=\$100$, the one-year effective annual risk-free rate is $i=2\%$, and the European put has $K=\$110$. Distribution A assigns probability $0.50$ to each of $\$92$ and $\$112$. Distribution B assigns probability $0.50$ to each of $\$72$ and $\$132$. Find the call and put values under each distribution.

Verify the risk-neutral means. The required mean is $\$100(1.02)=\$102$. For A, $0.50(92)+0.50(112)=\$102$. For B, $0.50(72)+0.50(132)=\$102$. Thus the comparison keeps the pricing mean fixed.

Calls. With $K=\$110$, the A call payoffs are $\$0$ and $\$2$, giving expected payoff $0.50(0)+0.50(2)=\$1$. Its value is $\$1/1.02=\$0.98$. The B call payoffs are $\$0$ and $\$22$, giving expected payoff $\$11$ and value $\$11/1.02=\$10.78$.

Puts. The A put payoffs are $\$18$ and $\$0$, giving expected payoff $\$9$ and value $\$9/1.02=\$8.82$. The B put payoffs are $\$38$ and $\$0$, giving expected payoff $\$19$ and value $\$19/1.02=\$18.63$.

Check. Both call and put values rise as the distribution spreads out, with the risk-neutral mean unchanged. The values are not equal: the strike and the shape of each payoff matter. Do not assume a call and put have equal values simply because they share an exercise price.

Common Exam Traps and Limits of the Result

A common error is to call a distribution “more volatile” and compare option values before checking its mean. For instance, suppose $S_0=\$80$, $i=2\%$, and a one-year call has $K=\$80$. A proposed risk-neutral distribution with terminal prices $\$60$ and $\$100$, each with probability $0.50$, has mean $0.50(60)+0.50(100)=\$80$. The tempting calculation is an expected payoff of $0.50(0)+0.50(20)=\$10$, then a value of $\$10/1.02=\$9.80$. But this is not a valid risk-neutral distribution for the stated stock price and rate: the required mean is $\$80(1.02)=\$81.60$, not $\$80$.

One possible repair is to use terminal prices $\$60$ and $\$103.20$, each with probability $0.50$, whose mean is $\$81.60$. The call payoffs are $\$0$ and $\$23.20$, so the expected payoff is $\$11.60$ and the value is $\$11.60/1.02=\$11.37$. This repair makes a valid pricing example, but it is not by itself a volatility comparison: a valid comparison must also hold the risk-neutral mean fixed across the distributions being compared.

Other checks that prevent errors include:

Use the payoff, not the stock-price change. A call payoff is $\max(S_T-K,0)$; a put payoff is $\max(K-S_T,0)$. Neither payoff can be negative.
Use the probabilities with their outcomes. Compute each payoff at its stated terminal price before taking the probability-weighted average.
Discount once. In this one-year model, divide the expected payoff by $1+i$ to obtain value today.
Check what is held fixed. The volatility comparison assumes the same strike, maturity, risk-free rate, and risk-neutral expected stock price.
Do not infer a universal ordering from variance alone. The convexity argument applies to a mean-preserving spread, not every possible pair of distributions with different variances.

Key takeaway: Under risk-neutral pricing, a mean-preserving spread raises the expected payoff of a convex option payoff. Because European call and put payoffs are convex, greater volatility generally raises both values when the other pricing inputs and the risk-neutral expected stock price are held fixed.

Check Your Understanding

For each problem, use risk-neutral probabilities and discount the expected payoff as stated.

A stock has $S_0=\$50$, $i=4\%$, and a one-year European call with $K=\$52$. Its terminal price is $\$40$ or $\$64$, each with probability $0.50$. Find the call value and verify the risk-neutral mean. (Answer: Mean $=\$52$; call value $=\$5.77$.)
A stock has $S_0=\$80$ and $i=0\%$. A call with $K=\$75$ is priced under either terminal distribution: A has outcomes $\$60,\$100$ with equal probabilities; B has outcomes $\$40,\$120$ with equal probabilities. Find both values. (Answer: A $=\$12.50$; B $=\$22.50$.)
A stock has $S_0=\$100$, $i=0\%$, and options with $K=\$100$. Compare equally likely outcomes $\$80,\$120$ with equally likely outcomes $\$60,\$140$. Find the call and put values under each distribution. (Answer: First distribution: call $=\$10$, put $=\$10$; second distribution: call $=\$20$, put $=\$20$.)
Under a risk-neutral distribution, an outcome of $\$120$ with probability $0.50$ is replaced by outcomes $\$100$ and $\$140$, each with probability $0.25$. Does this replacement change the expected stock price, and what does convexity imply for expected call and put payoffs? (Answer: The expected stock price is unchanged; neither expected payoff can decrease.)
A stock has $S_0=\$100$, $i=5\%$, and a one-year call with $K=\$110$. Distribution A has outcomes $\$90$ and $\$120$, each with probability $0.50$. Distribution B has outcomes $\$90,\$100,\$140$ with probabilities $0.50,0.25,0.25$. Find both call values. (Answer: A $=\$4.76$; B $=\$7.14$.)

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