I am looking at a perpetual put option where the strike price is initially the stock price $K(0)=S(0)$ (i.e. at the money), but the strike price grows at the constant risk-free rate $r$ [i.e. $K(t)=S(0)\exp(rt)$]. The stock price S(t) follows GBM with no dividend, so that $dS(t)/S(t)=rdt+\sigma dW(t)$.

If I use the money market account as a numeraire so that $Z(t)=S(t)/B(t)$ where $B(t)=\exp(rt)$, isn't this then just the price of a perpetual put on $Z(t)$ with constant strike price $K=(0)=Z(0)=S(0)$ and $dZ(t)=\sigma dW(t)$? If yes, I thought I should be able to price this with Merton's perpetual put formula (for instance, https://www.ma.utexas.edu/users/mcudina/Lecture14_1and2.pdf) by setting $r=q$ (i.e. zero drift).

But the Black-Scholes price of a European put with strike $S(0)*\exp(rT)$ and maturity $T$ is $S(0)*[N(\sigma \sqrt{T}/2)-N(-\sigma \sqrt{T}/2)]$ which approaches $S(0)$ when $T \to \infty$. So doesn't that mean that after some maturity $T^*$ the Black-Scholes price of an European put will be higher than the price of a perpetual put (same strike and volatility)?


1 Answer 1


No, your first assumption isn't quite right: you forgot a scaling factor, and in fact we have $P(S_0,S_0.B_T) = P(S_0/B_T,S_0) * B_T$.

This is turn will also approach $S_0$ when $T \to \infty$.


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