Time to Expiration for European Puts

Why Put Value and Time Can Move in Either Direction

A longer expiration gives the underlying asset more time to fall below a put’s exercise price. That added opportunity can increase the expected payoff. But a put pays its exercise price only at expiration, and the present value of that future payment is lower when it is farther away under a positive interest rate. These effects work in opposite directions, so a longer expiration does not guarantee a higher European put value.

The relevant comparison holds the current stock price, exercise price, interest-rate environment, and volatility assumptions fixed while changing expiration. This is a comparison of model values, not a claim that any individual put’s market price must move in a particular direction over time.

Definition: A European put with exercise price $K$ and expiration $T$ pays $\max(K-S_T,0)$ at expiration, where $S_T$ is the stock price then. Its value today is the present value of that contingent payoff under a pricing model. “Longer expiration” means a larger $T$, with the other stated inputs held fixed.

Risk-Neutral Valuation and the Two Competing Effects

Under risk-neutral valuation, a derivative’s price is the discounted expected value of its payoff using risk-neutral probabilities. In a discrete model with a constant effective rate $i$ per period, a European put expiring after $n$ periods has value

P_n=\frac{E^Q[\max(K-S_n,0)]}{(1+i)^n}.

Here $E^Q$ denotes expectation under the risk-neutral probability measure. The numerator can change with expiration because the distribution of the future stock price changes. The denominator also changes: for a positive rate, discounting reduces the present value of a fixed future dollar as its payment date moves farther away. The actual payoff is not fixed, however, so the numerator may rise or fall as well.

For continuous compounding at a constant continuously compounded risk-free rate $r$, the corresponding expression is

P(T)=e^{-rT}E^Q[\max(K-S_T,0)].

This expression makes the balance clear. A longer expiration can create more opportunities for a low $S_T$, but the strike payment embedded in the payoff is discounted over a longer period. Neither effect alone determines the result.

Worked Example: A Longer Expiration Lowers the Put Value

Consider a hypothetical, no-dividend stock in a recombining binomial model. Each one-year period has an up factor $u=1.20$, a down factor $d=0.80$, and a risk-free gross return $R=1.10$. The one-year risk-neutral probability of an up move is

p=\frac{R-d}{u-d}=\frac{1.10-0.80}{1.20-0.80}=0.75.

The down probability is $1-p=0.25$. Since $0.80<1.10<1.20$, the risk-neutral probability lies between zero and one. Let $S_0=\$100$, $K=\$100$, and compare expiration after one year with expiration after two years. The same annual tree parameters apply in each period.

One-year put. The up-state stock price is $100(1.20)=\$120$, giving payoff $\max(100-120,0)=\$0$. The down-state stock price is $100(0.80)=\$80$, giving payoff $\max(100-80,0)=\$20$. The expected payoff is

0.75(0)+0.25(20)=\$5.

Discount one year at the gross return $1.10$:

P_1=\frac{5}{1.10}=\$4.54545\ldots\approx \$4.55.

Two-year put. The possible terminal stock prices are $100(1.20)^2=\$144$, $100(1.20)(0.80)=\$96$, and $100(0.80)^2=\$64$. Their probabilities are $0.75^2=0.5625$, $2(0.75)(0.25)=0.3750$, and $0.25^2=0.0625$, respectively. The corresponding put payoffs are $\$0$, $\$4$, and $\$36$. Thus,

E^Q[\text{payoff}]=0.5625(0)+0.3750(4)+0.0625(36)=0+1.50+2.25=\$3.75,

and discounting for two years gives

P_2=\frac{3.75}{(1.10)^2}=\frac{3.75}{1.21}=\$3.09917\ldots\approx \$3.10.

The longer put is worth less in this model: $\$3.10<\$4.55$. The added time does produce a possible $\$36$ payoff in the lowest state, but the risk-neutral expected payoff falls from $\$5.00$ to $\$3.75$, and the two-year payoff is discounted more heavily. The result is checked by comparing the computed values at both maturities using the same tree assumptions.

Black–Scholes Maturity Sensitivity

A standard continuous-time benchmark makes the competing effects explicit. Assume the stock follows geometric Brownian motion with constant volatility $\sigma>0$, the risk-free rate $r$ is constant and continuously compounded, the stock pays no dividends, and the option is European. Under these assumptions, the Black–Scholes put value is

P(T)=K e^{-rT}\Phi(-d_2)-S_0\Phi(-d_1),

where $S_0$ is the current stock price, $\Phi$ is the standard normal cumulative distribution function, and

d_1=\frac{\ln(S_0/K)+(r+\sigma^2/2)T}{\sigma\sqrt{T}}, \qquad d_2=d_1-\sigma\sqrt{T}.

In this formula, $K e^{-rT}\Phi(-d_2)$ is the discounted exercise-price component, while $S_0\Phi(-d_1)$ is the stock-price component subtracted from it. Changing $T$ changes both components as well as the probabilities represented by $d_1$ and $d_2$.

Theorem: Under the no-dividend Black–Scholes assumptions above, the derivative of put value with respect to expiration is

\frac{\partial P}{\partial T} = \frac{S_0\phi(d_1)\sigma}{2\sqrt{T}} -rK e^{-rT}\Phi(-d_2),

where $\phi$ is the standard normal probability density function. A positive derivative means that a small increase in expiration raises the put value locally; a negative derivative means that it lowers the value locally. This is a local sensitivity, not by itself a guarantee of monotonicity over a large change in expiration.

The formula’s two terms explain the result. The first is positive and reflects the added value associated with uncertainty and the possibility of more downside. When $r>0$, the second term is negative: extending the time to the strike payment lowers its present value. If the positive term is larger, the put locally increases with expiration; if the negative term is larger, it locally decreases. If $r=0$, the second term vanishes, so the derivative is positive under these assumptions. If $r<0$, the second term contributes positively rather than negatively.

Why the Maturity-Sensitivity Formula Holds

The key cancellation in the derivation comes from the relationship between the two normal densities in the Black–Scholes formula. Differentiate the put price with respect to $T$, treating $S_0$, $K$, $r$, and $\sigma$ as fixed. Differentiating the discounted strike term produces a direct discounting term and a term involving the derivative of $d_2$. Differentiating the stock term produces a term involving the derivative of $d_1$.

Differentiate both price components.
The direct derivative of $K e^{-rT}$ is $-rK e^{-rT}$. The derivatives of $\Phi(-d_2)$ and $\Phi(-d_1)$ introduce $-\phi(d_2)\partial d_2/\partial T$ and $-\phi(d_1)\partial d_1/\partial T$, respectively.

Use the density identity.
For the stated $d_1,d_2$, $S_0\phi(d_1)=K e^{-rT}\phi(d_2)$. Therefore the terms involving derivatives of $d_1$ and $d_2$ combine into $K e^{-rT}\phi(d_2)(\partial d_1/\partial T-\partial d_2/\partial T)$.

Differentiate their difference.
Since $d_1-d_2=\sigma\sqrt{T}$, its derivative is $\sigma/(2\sqrt{T})$. The combined density term becomes $K e^{-rT}\phi(d_2)\sigma/(2\sqrt{T})=S_0\phi(d_1)\sigma/(2\sqrt{T})$. Adding the direct discounting term gives the stated formula.

Thus the formula is not simply a statement that “more time helps” or “discounting hurts.” It quantifies both effects under a specific model. The assumptions matter: the expression above is for a no-dividend stock and constant Black–Scholes inputs. It should not be applied unchanged if those assumptions are different.

Worked Example: A Longer Expiration Locally Raises the Put Value

Use the Black–Scholes model with $K=\$100$, $T=1$, $r=0.04$, and $\sigma=0.20$. Choose $S_0=100e^{-0.06}\approx\$94.1765$. These inputs make $d_1=0$, because $\ln(S_0/K)=-0.06$ and $(r+\sigma^2/2)T=0.04+0.02=0.06$. Hence $d_2=-0.20$.

Use $\phi(0)=0.3989423$, $\Phi(0.20)=0.5792597$, and $e^{-0.04}=0.9607894$. The positive term in the maturity derivative is

\frac{S_0\phi(d_1)\sigma}{2\sqrt{T}} = \frac{94.1765(0.3989423)(0.20)}{2} \approx \$3.7571.

The discounting term is

rK e^{-rT}\Phi(-d_2) = 0.04(100)(0.9607894)(0.5792597) \approx \$2.2262.

Therefore,

\frac{\partial P}{\partial T} \approx 3.7571-2.2262 =\$1.5309\text{ per year}.

At these inputs, a small increase in expiration raises the put value locally because the positive term exceeds the negative discounting term. This is a sensitivity calculation, not a prediction that the put must rise by exactly $\$1.5309$ for a full additional year; the derivative and model inputs can change as $T$ changes. The calculation is checked by subtracting the two displayed components with the same precision.

When Downside Opportunity Outweighs Discounting

The next binomial comparison uses the same stock tree and risk-free return as the first example, but a different exercise price. It shows that the direction can reverse without changing the market model.

Worked Example: A Longer Expiration Raises the Put Value

Use the one-year binomial parameters $u=1.20$, $d=0.80$, $R=1.10$, $p=0.75$, and $S_0=\$100$, but set $K=\$120$. Compare one-year and two-year European puts.

One year. The up-state price is $\$120$, with payoff $\max(120-120,0)=\$0$. The down-state price is $\$80$, with payoff $\$40$. The risk-neutral expected payoff and present value are

0.75(0)+0.25(40)=\$10, \qquad P_1=\frac{10}{1.10}=\$9.0909\ldots\approx \$9.09.

Two years. The terminal prices are $\$144$, $\$96$, and $\$64$, with probabilities $0.5625$, $0.3750$, and $0.0625$. The payoffs are $\$0$, $\$24$, and $\$56$, respectively. Thus,

E^Q[\text{payoff}] =0.5625(0)+0.3750(24)+0.0625(56) =0+9.00+3.50 =\$12.50.

Discounting for two years gives

P_2=\frac{12.50}{1.21}=\$10.33058\ldots\approx \$10.33.

Here the longer put is worth more: $\$10.33>\$9.09$. The two-year risk-neutral expected payoff rises from $\$10.00$ to $\$12.50$, enough to outweigh the additional discounting. As a check, both present values use the same $R=1.10$ and the appropriate number of years. The common mistake is to conclude that a longer expiration must always help; the first example shows why that conclusion fails.

A Pure Discounting Illustration

A limiting deterministic example isolates the discounting effect. Suppose, hypothetically, that a stock starts at $\$80$ and grows with certainty at the risk-free gross return $R=1.10$ each year. Let $K=\$100$. After one year, $S_1=\$88$, so the payoff is $\$12$ and its present value is $12/1.10=\$10.9091$. After two years, $S_2=\$96.80$, so the payoff is $\$3.20$ and its present value is $3.20/(1.10)^2=\$2.6446$. There is no added uncertainty in this illustration; the payoff itself shrinks as the stock grows, and discounting further lowers its present value. It is a model illustration, not an assumption that real stock prices are deterministic.

Common Mistakes and Practical Checks

Assuming more time automatically raises a put. More possible downside is relevant, but it must be balanced against discounting and the full change in the payoff distribution.
Discounting only once for a multi-period option. In the binomial examples, the two-year expected payoff is divided by $1.10^2=1.21$, not by $1.10$.
Using ordinary probabilities in place of risk-neutral probabilities. The valuation calculation uses probabilities consistent with the risk-free return and the tree’s up and down factors.
Treating a local sensitivity as a full-maturity comparison. The sign of $\partial P/\partial T$ describes the local direction at the stated inputs. The derivative can change as expiration changes.
Changing several inputs at once. To isolate the expiration effect, hold $S_0$, $K$, the model, and all other assumptions fixed.

For an exam-style comparison, first name the model and its assumptions, then compute or interpret the value at each expiration using the same inputs. In a binomial model, calculate each terminal payoff, its risk-neutral expected value, and the correct discount factor. Under Black–Scholes, use the maturity-sensitivity formula to determine the local direction and explain which of its two terms dominates. This disciplined comparison is aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

Use the stated model assumptions in each problem. Round dollar values to the nearest cent.

In the binomial model with $u=1.20$, $d=0.80$, $R=1.10$, and $S_0=\$100$, find the values of a European put with $K=\$90$ expiring in one year and two years. (Answer: $P_1\approx\$2.27$, $P_2\approx\$1.34$; the longer put is lower.)
Use the same binomial model to compare a put with $K=\$130$ at one and two years. Show the two-year expected payoff before discounting. (Answer: $P_1\approx\$18.18$, two-year expected payoff $=\$16.875$, $P_2\approx\$13.95$; the longer put is lower.)
A hypothetical stock grows deterministically at a risk-free gross return of $1.05$ per year. If $S_0=\$75$ and $K=\$100$, calculate the present value of the European put payoff at one and two years. (Answer: $P_1=100/1.05-75\approx\$20.24$, $P_2=100/(1.05)^2-75\approx\$15.70$.)
In the no-dividend Black–Scholes model, at a particular maturity suppose $S_0=\$100$, $K=\$100$, $T=1$, $r=0$, $\sigma=0.20$, and $\phi(d_1)=0.3989423$. Use the maturity-sensitivity formula to find the local direction and rate of change. (Answer: $\partial P/\partial T=100(0.3989423)(0.20)/2\approx+\$3.9894$ per year; the put locally increases.)
For the same one-year binomial tree as in question 1, calculate the put values for $K=\$120$ and explain why its direction differs from that of the $K=\$90$ put. (Answer: $P_1\approx\$9.09$, $P_2\approx\$10.33$; the two-year expected payoff rises to $\$12.50$, enough to outweigh additional discounting.)

What You'll Learn