Small-Strike Put Prices and the Probability of a Zero Stock Value

Summary

The question examines the limit of a put price divided by its strike as the strike approaches zero, under zero interest rates. It rewrites the put payoff expectation as the probability that the terminal stock value is below the strike, minus a scaled truncated expectation of the stock value. The proposed result is that the limit equals the probability of the stock ending at zero.

The accepted response reaches that conclusion by asserting that the scaled expectation vanishes, but its final step is not fully justified as written. In particular, the document assumes a density representation, which excludes a possible point mass at zero, and it does not establish the limiting behavior of the second term. For a nonnegative stock value, that term can instead be bounded using the probability of a positive value below the strike, which tends to zero as the strike decreases to zero. The result is useful for connecting deep out-of-the-money put prices to a zero-value probability, subject to the stated pricing assumptions and a sound treatment of atoms.

Key ideas

• With zero interest rates, a put price can be expressed as an expectation of its terminal payoff.
• Dividing by strike separates a low-tail probability term from a scaled truncated expectation.
• The proposed limiting quantity is the probability that the terminal stock value is zero.
• A density-only derivation needs care because the distribution may have an atom at zero.

Tags

# How do I prove that $\lim_{K\searrow0}\frac{P(K,T)}{K} = \mathbb P(S_T=0)$?


# How do I prove that $\lim_{K\searrow0}\frac{P(K,T)}{K} = \mathbb P(S_T=0)$?












I am trying to prove that $$\lim_{K\searrow0}\frac{P(K,T)}{K} = \mathbb P(S_T=0)$$ where $P(K,T)$ denotes the put option price with maturity $T$ and strike $K$ for some stock $S$. Assuming interest rates $r=0$ we write $$\lim_{K\searrow0}\frac{P(K,T)}{K} = \lim_{K\to0}\frac{\int_0^K(K-S)f(S)dS}{K}$$ $$= \lim_{K\searrow0} \left(\int_0^Kf(S)dS - \frac1K\int_0^KSf(S)dS\right)$$ $$= \lim_{K\searrow0} \left(\mathbb P(S_T\le K) - \frac1K\mathbb E[S_T1_{S_T<K}]\right)$$ where $f(S)$ is the density of $S_T$.

Now $\lim_{K\searrow0} \mathbb P(S_T\le K) = \mathbb P (S_T=0)$, but I am not sure whether it is "obvious" that $\frac1K\mathbb E[S_T1_{S_T<K}]$ tends to zero as $K$ tends to zero.

## Answer by Hans (score 4, accepted)

https://quant.stackexchange.com/a/12813

$$\lim_{K\searrow0}\frac{P(K,T)}{K} = \lim_{K\to0}\frac{\int_0^K(K-S)f(S)dS}{K}$$ $$= \lim_{K\searrow 0} \left(\int_0^Kf(S)dS - \frac1K\int_0^KSf(S)dS\right)$$ $$= \lim_{K\searrow 0} \big(\mathbb P(S_T\le K)\big) - Sf(S)\big\vert_{S=K=0}$$ $$= \mathbb P(S_T=0),$$ where $f(S)$ is the density of $S_T$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.