Perpetual Put Valuation with a Growing Strike

Summary

The document considers a put whose strike grows at the risk-free rate while the underlying stock follows geometric Brownian motion with no dividends. The question reframes the stock relative to the money market account and asks whether this produces a standard perpetual put with a constant strike and zero drift. It also compares the proposed perpetual-option valuation with the limiting price of a European put as maturity becomes very long.

The response identifies a scaling factor omitted in the proposed change of numeraire: the original option price equals the transformed option value multiplied by the money market account. With this adjustment, the transformed valuation also tends toward the initial stock price as maturity grows, addressing the apparent contradiction. The exchange gives a concise pricing identity rather than a full derivation or a broader discussion of assumptions, so it does not supply a complete treatment of perpetual-option valuation.

Key ideas

• A strike growing at the risk-free rate can be related to a constant-strike payoff after scaling the underlying by the money market account.
• Changing numeraire does not remove the need to scale the option value back into the original units.
• The response gives a pricing identity that reconciles the long-maturity comparison with the transformed option.
• The explanation is brief and does not derive the identity or explore its assumptions in detail.

Tags

# Perpetual Put vs European Put


# Perpetual Put vs European Put












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)?

## Answer by Ivan (score 1)

https://quant.stackexchange.com/a/41986

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.