First of all, I realise this question might border on `meta-finance', so I'd be totally OK if it gets closed.

Having said that, the question itself:

Given a stock $S$, in the absence of default it is quite clear that the probability the stock price at time $T$ takes values in $(0,\infty)$ is $1$ for all $T$.

In the presence of default, however, the price at $T$ isn't even defined if the company defaulted at $t<T$. Can we express this mathematically as $P(0<S_T < \infty) = f(T) \leq 1$?

To illustrate what I mean with a simple example, consider the probability density of the stock price (zero interest rate and dividend yield for now) given by $$ q(S_T,T | S_t,t) = e^{-c(T,t)} p(S_T,T | S_t,t) $$ where $p$ is the usual lognormal density with constant volatility $\sigma$ and $c(t,T)$ is a positive deterministic function of $t$ and $T$. In this simple example let's assume that we only observe a term structure of IVs and no skew/smile.

Then, $$ E\left[(S_T-K)_+ \right] = e^{-c(T,t)} C^{BS} (S_t,K,\sigma,T) $$ We can calibrate $c(t,T)$ to the IV TS through $$ C^{BS} (S_t,K,IV(T),T) = e^{-c(T,t)} C^{BS} (S_t,K,\sigma,T) $$

I am working on generalising this to stochastic $c$ and/or a local $c$ to get skew/smile.

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    $\begingroup$ We express this usually as $S_T=0$ when $T$ is on or after the time of default. This defines $S_T$ for all $T$. $\endgroup$
    – Kurt G.
    Oct 7, 2022 at 9:10
  • $\begingroup$ @KurtG. Yes that's the usual definition / interpretation. I am trying to look at it another way. See my edit, I saw your comment after my edit. $\endgroup$
    – user34971
    Oct 7, 2022 at 9:15
  • $\begingroup$ Besides, $S_T = 0$ isn't even an allowed value in many models. It doesn't make sense. It's like saying the empty set is zero. It isn't. $\endgroup$
    – user34971
    Oct 7, 2022 at 9:23
  • $\begingroup$ If I get this correcty, you are looking for the density of a defaultable stock price process? Isn't that something like $p(S_T=0)+(1-p(S_T=0))q(S_T)$, with $q$ the density under no-default? $\endgroup$ Oct 7, 2022 at 10:11
  • $\begingroup$ @Kermittfrog I see two close votes under the reason of 'basic question' (I was hoping at least for off-topic :)), but the `basic question' I have is: if the price of a stock is the result of measurement (i.e. trading, or at the very least bid-ask), it's easy to say its price is zero after it has defaulted, been de-listed, migrated to a parallel universe, but how about thinking about it as that its probability of having a price is not 1 anymore (since it could happen that at $T$ the stock just doesn't exist anymore), what happens then? That's the gist of what I'm trying to do. $\endgroup$
    – user34971
    Oct 7, 2022 at 10:17