Why Zero Stock Price Is an Absorbing Boundary in an Asian Option PDE

Summary

The note explains the zero-stock-price boundary condition for a continuously monitored Asian call in a Black–Scholes setting. The average-price state is the accumulated stock value, so if the stock reaches zero and stays there, the accumulated value remains fixed through expiry. The payoff is then determined by that fixed accumulated amount divided by the full averaging period, discounted back to the boundary time.

The answer supports this reasoning by substituting zero into the stock’s drift and diffusion terms: both vanish, leaving no change in the stock value. This relies on the specified stochastic model and its assumption that zero is absorbing. It is a model property rather than a universal statement about real companies or all asset-price processes; the note itself acknowledges that real-world bankruptcy outcomes can differ.

Key ideas

• In the stated stock-price model, both drift and diffusion vanish when the stock price is zero.
• Zero is therefore an absorbing state, so the stock remains at zero after reaching it.
• The accumulated stock-price integral stays constant from that time until expiry.
• The Asian call boundary value is the discounted payoff based on the fixed accumulated value and averaging period.

Tags

# Pricing PDE of Asian option by Shreve


# Pricing PDE of Asian option by Shreve












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 by Bob Jansen (score 4, accepted)

https://quant.stackexchange.com/a/78207

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 $S_t = 0$:

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

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