Intepreting European call option when expiration approaches to infinity

Question

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

1. Why the price of call option equals to $S$, when the probability density function of stock price spikes at 0.
2. 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 1

1. 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)

2. Not sure I get the second point, could you please re-formulate?

Answer 2

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 3

1. Even though the density spikes at zero, it also develops a very long right hand tail, which creates the required expectation.

2. 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.