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Let us consider perpetual American put option with interest rate: $r = 0$.

The Black-Scholes equation in this case has the form: $$ \frac{1}{2} \sigma^2 S^2 \frac{d^2 V(t, S)}{dS^2} + (r-d)S \frac{dV(t, S)}{dS} - rV(t, S) = 0. $$

After applying PDE method I obtained that: $$ V(S) = A S^{\frac{2d+\sigma^2}{\sigma^2}}+ B, $$ when $2d+\sigma^2\geq0$.

Then using the fact that $V(S)\rightarrow 0$ when $S\rightarrow +\infty$ I have to set $A = 0$ and $B = 0$.

Finally, I obtained: $$ V(S) = 0. $$

I interpret it in the way that this option is worthless.

Hence, the stopping region is an empty set, and continuation region has the form: $C = \mathbb{R}^{+}$, because there is no optimal moment to exercise this option.

My question is:

  1. Is this solution correct?

  2. If I obtain that an option is worthless, which means that: $V(S) = 0$ can I say that continuation region is $\mathbb{R}^{+}$?

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  • $\begingroup$ An European put does not have value $0$ when the interest rate is null, I don't see why a perpetual American put would have a lower value. $\endgroup$ – Daneel Olivaw Apr 30 '18 at 21:55
  • $\begingroup$ You are right, so it is a mistake in my solution. $\endgroup$ – MathMen Apr 30 '18 at 22:07
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The PDE doesn't equal zero when I replace with the expression of $V(S)$ you gave. Threre is an issue in your PDE's the boundary conditions and its solution.

Let's start with the regular american put PDE, and deduce the perpetual american put ones, then solve for the price:

Regular american put

  • Black scholes PDE is satisfied by the price for $S(t) > S^*(t), t < T$
  • On $S(t) = S^*(t)$, $V(S(t), t, T) = K - S^*(t)$
  • With a final condition: $V(S, T, T) = (K - S(T))^+$
  • And the additional condition $\frac{\partial V}{\partial S}(S^*(t), t) = -1$ (we need this an additional equation to be able to solve for $V$ and $S^*$ at the same time).

Perpetual american put

Now, for the perpetual american, the value shouldn't depend on time, only on the level of the underlying: $$V_\infty(S, t) = V_\infty(S) = \lim_{(T - t) \rightarrow \infty} V(S,t,T)$$

Doing a change of variable $\theta = T - t$, and denoting $S^*_\infty = \lim_{\theta \rightarrow \infty} S^*(\theta)$ in the equations above gives:

  • Black Scholes PDE for $S > S^*_\infty$
  • $V_\infty(S^*_\infty) = K - S^*_\infty$
  • $\frac{\partial V_\infty}{\partial S}(S^*_\infty) = -1$

Seeking a solution of the form $A S^\alpha + B S^\beta$, we have:

  • $\alpha, \beta = \frac{-(r-d-0.5\sigma^2) +/-\sqrt{(r-d-0.5\sigma^2)^2 + 2r\sigma^2}}{\sigma^2} $.
    • if $r = 0$ and $d + 0.5 \sigma^2 \geq 0$, then $\alpha >0$ and $\beta = 0$, in this case we have no solution (the put price should be between $K$ and $K-S$)
    • if $r = 0$, and $d + 0.5 \sigma^2 < 0$, then $\alpha = 0$ and $\beta < 0$. In this case, $A = 0$ because $V(S)$ should be zero when $S \rightarrow \infty$ (argument that you made)

The last two equations above give in this case $B{S^*_\infty}^\beta = K - S^*_\infty$ and $\beta B {S^*_\infty}^{-1} = -1$

Which leads to: $$ V_\infty(S) = (K - S^*_\infty) \left( \frac{S}{S^*_\infty} \right)^\beta $$ where: $S^*_\infty = \frac{\beta}{\beta - 1} K$

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  • $\begingroup$ I am sorry, I wanted to write: "Then using the fact that $V(S)\rightarrow 0$ when $S\rightarrow \infty$ I have to set $A=0$ and $B=0$. I edited now my main post. $\endgroup$ – MathMen May 2 '18 at 20:03
  • $\begingroup$ Ah ok, it's clearer now indeed. However, I think there's something off with the solution of the PDE when r = 0. I substituted your function but didn't get zero :/ Conceptually, if r = 0 and d > 0 your stock price will be eventually reach zero, so I'd expect the price of the perpetual put to be equal to K rather than 0. $\endgroup$ – byouness May 2 '18 at 21:43
  • $\begingroup$ Edited my answer accordingly. $\endgroup$ – byouness May 2 '18 at 22:36
  • $\begingroup$ I understand what you mean, but if I assume that $V(S) = K$ for all S, then the condition which states that $V(S)\rightarrow 0$ when $S\rightarrow\infty$ will not hold. The second thing which seems to be strange for me is the fact that the maximum possible payoff for this option is equal to $K$, so selling this option for K creates a situation in which the seller cannot lose money. In the worst situation he/she will have $0$ at the end. $\endgroup$ – MathMen May 3 '18 at 8:24
  • $\begingroup$ I had a look, and have rewritten a clearer solution. You were close but some bits were missing. $\endgroup$ – byouness May 3 '18 at 10:10

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