Why Long-Dated Calls Approach the Stock Price in Black–Scholes
Summary
The document examines the Black–Scholes value of a European call as its expiration moves far into the future, assuming no dividends. In the formula, the call value depends on two probabilities expressed under different measures. As maturity increases, the Black–Scholes terms approach values that make the call price approach the current stock price, even though the lognormal density’s mode moves toward zero.
The responses explain this apparent tension through the distribution’s increasingly long right tail: under the risk-neutral measure, the expected stock price grows with time, and discounting offsets that growth in the option valuation. The density’s peak near zero does not determine its mean. The discussion also clarifies that equivalent probability measures need only agree on which events have zero probability; they need not assign the same probability to an in-the-money event. The treatment is qualitative and relies on the stated no-dividend Black–Scholes setup, without exploring other models or dividend assumptions.
Key ideas
- With no dividends, the Black–Scholes call value tends toward the current stock price as maturity becomes very large.
- The lognormal distribution’s mode can move toward zero while its right tail sustains a rising risk-neutral mean.
- Discounting offsets the risk-neutral growth in the underlying when valuing the call.
- Equivalent measures can assign different probabilities to an event while agreeing on which events have zero probability.
- The explanation is specific to the assumptions described and does not establish behavior under other pricing models.
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# Intepreting European call option when expiration approaches to infinity
# Intepreting European call option when expiration approaches to infinity
Assume that dividend = 0, then the price of call option is
$$ C = S\cdot P_{s}[S(T) > K] - e^{-rT}K\cdot P_F[S(T) > K] = SN(d_1)-e^{-rT}KN(d_2) $$ where
$P_s[S(T) > K]$ = Probability of ITM when $S(t)$ is set to be a numeraire and
$P_F[S(T) > K]$ = Probability of ITM under Forward measure
When $T \rightarrow \infty$ , $N(d_1) \rightarrow 1 $ and $N(d_2) \rightarrow 0$ regardless of strike price $K$ and therefore $C = S$.
However, when $T \rightarrow \infty$, then this will squeeze probability density function of stock price at $0$.
My questions are
- Why the price of call option equals to $S$, when the probability density function of stock price spikes at 0.
- If probability measure under Stock price numeraire and forward measure are equivalent, then the probability $P[S(T) > K]$ shouldn't agree? or they are not equivalent in this case? or is it just $P_s[S(T)>K] \rightarrow 1$ not $P_s[S(T)>K] = 1$?
## Answer by Soumirai (score 1, accepted)
https://quant.stackexchange.com/a/60604
- The mode of the lognormal tends towards 0 when T increases, but the risk-neutral mean (assuming no dividends) is S0*exp(rT) which increases with time
Then in Black Scholes, discounting cancels out the risk neutral drift of the stock (S0*exp(rT)*DF = S0)
- Not sure I get the second point, could you please re-formulate?
## Answer by Kermittfrog (score 1)
https://quant.stackexchange.com/a/60617
In the very long run, expected stock prices will diverge to infinity under the risk neutral measure and at the distribution will spread more and more.
Thus, in the very long run, we will have $E(S(t)) \to \infty$ and the fraction of the cdf that covers the range $0\ldots X$ becomes smaller and smaller. Thus, a very (very)long termed call option is effectively an investment in the underlying itself.
HTH?
## Answer by dm63 (score 1)
https://quant.stackexchange.com/a/60626
- Even though the density spikes at zero, it also develops a very long right hand tail, which creates the required expectation.
- If the probability under stock numeraire and forward measure are equivalent , there is no reason to believe that $P[S>K]$ should agree. The only requirement is that the zero probability regions agree, and since both measures have positive probability on $(0,\infty)$ this is satisfied.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.