According this answer, https://quant.stackexchange.com/a/39298/29108, the European put price (with maturity $T$) at time $t$ for a stock whose current price is $0$ should be the strike $K$ discounted from $T$ to $t$. So $P(t,T)K = p$.

Is this exactly true? I had thought that the value $P(t,T)K$ should be an upper bound on the put price, as there is still a chance the stock price increases in the time before maturity. And if the European put price is basically growing at the risk free rate, doesn't this remove the probability of stock price movements?

Any help would be appreciated!

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    $\begingroup$ For GBM zero is an "absorbing barrier". The price of the stock is going to remain at zero if it ever hits that level, If the option is European all you can do is wait until maturity to receive K. That is worth $K e^{r(T-t)}$ today. $\endgroup$ – Alex C Feb 5 '19 at 1:35
  • $\begingroup$ So for a stock price of $0.01$, the expression becomes an inequality? $\endgroup$ – Slade Feb 5 '19 at 1:37
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    $\begingroup$ Yes, for a price of 0.01 there is till some chance of appreciation for the stock and therefore still some "time value" for the option. $\endgroup$ – Alex C Feb 5 '19 at 1:40
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    $\begingroup$ Okay, nice. Thanks a lot. If you put your comments as an answer, I can accept it! $\endgroup$ – Slade Feb 5 '19 at 1:40

Thanks to @Alex C for clarifying. His answer is in the comments, but I wanted to have the question answered for others in the future.

Assuming a geometric brownian motion as the stock price process (like in Black-Scholes), when the stock price becomes zero, it will stay at zero 'forever'. So basically the stock price is known in the future to be $0$, and thus an European Put option on the stock with strike $K$ and maturity date $T$, is essentially a zero-coupon bond delivering $K$ on date $T$, since both are riskless investments. So discounting the known, riskless payoff of the option, $K - 0$, to the present, the option value (just like the value of the zero-coupon bond) is $P(t,T)K$.

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