Pricing PDE of Asian option by Shreve

Question

I am currently working on "Stochastic Calculus for finance II, continuous time model" from Shreve. In chapter 7.5 Theo 7.5.1 he derives a pricing PDE with boundary conditions for an Asian call option and i do not understand his derivation of the first boundary condition. So we have \begin{align*} dS_t=rS_tdt+S_tdW_t\\ Y_t=\int_{0}^{t}S_udu\ \end{align*} The boundary condtion is given by \begin{align*} v(t,0,y)=e^{-r(T-t)}\max(\frac{y}{T}-K,0). \end{align*} His derivation: "IF $S_t=0$ and $Y_t=y$ for some variable $t$, then $S_u=0, \ \forall u \in[t,T]$ and so $Y_u$ is constant on $[t,T]$ and therefore $Y_T=y$ and the value of the Asian call option at time t is $e^{-r(T-t)}\max(\frac{y}{T}-K,0)$".

Problem: I do not understand why $S_t=0$ implies $S_u=0, \ \forall u \in[t,T]$.
Thank you very much for your help!

Answer 1

Intuitively, once a stock hits value zero the underlying company is bankrupt and the value remains zero (this might not be true in the real world but it's a common assumption). So, once $S_t$ is zero it remains zero. Mathematically, if $S_t = 0$:

$$ \begin{align} dS_t &= rS_tdt+S_tdW_t \\ &= r \times 0 \times dt + 0 \times dW_t \\ &= 0. \end{align} $$