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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.

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  • $\begingroup$ Which model do you consider? Black Scholes or something different? $\endgroup$ – Ric Jun 27 '14 at 6:39
  • $\begingroup$ @Richard I am only assuming that $P(K,T)=\mathbb E[(K-S_T)^+]$ for some process $S_T$ that has a desnity $f(S)$ (one might write $f_T(S)$ to indicate that this density is in fact associated with $S_T$ and not with any other $S_t$). $\endgroup$ – Phil-ZXX Jun 27 '14 at 10:12
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$$\lim_{K\searrow0}\frac{P(K,T)}{K} = \lim_{K\to0}\frac{\int_0^K(K-S)f(S)dS}{K}$$ $$= \lim_{K\searrow 0} \left(\int_0^Kf(S)dS - \frac1K\int_0^KSf(S)dS\right)$$ $$= \lim_{K\searrow 0} \big(\mathbb P(S_T\le K)\big) - Sf(S)\big\vert_{S=K=0}$$ $$= \mathbb P(S_T=0),$$ where $f(S)$ is the density of $S_T$.

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  • $\begingroup$ Why is the last step valid? Could you mention a rule from calculus? $\endgroup$ – Ric Jun 27 '14 at 6:07
  • $\begingroup$ @Richard I think it's as follows: If $g(x)$ is a function, then $\frac1\epsilon\int_a^{a+\epsilon} g(x)dx \to g(a)$ as $\epsilon\to 0$ (assuming $g(x)$ is smooth "enough"). $\endgroup$ – Phil-ZXX Jun 27 '14 at 10:15
  • $\begingroup$ yes ... I thought so too. I wonder how this theorem is called and what the conditions are ... $\endgroup$ – Ric Jun 27 '14 at 10:55
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    $\begingroup$ @Richard: This is the Fundamental Theorem of Calculus en.wikipedia.org/wiki/…. It is valid, almost everywhere (in measure), so long as g in Tom's comment is integrable. It is true everywhere, if it is continuous (no need for it to be smooth as Tom states) and it is a simple application of the mean value theorem. $\endgroup$ – Hans Jun 27 '14 at 16:30
  • $\begingroup$ So we have $\lim_{K\rightarrow 0} \frac{1}{K} \int_0^K x dF(x) = 0$ right? A reference would be what they call the second fundamental theorem of calculus as described here. The main point is that the limit $K\rightarrow 0$ leads to taking the derivative and we can take the function inside the integral and evaluate it at $K$ ... Sorry but we can not let $K$ tend to zero, use this as an agrument for the theorem and then evaluate at $K=0$. This is still not 100% rigorous, in my mind ... $\endgroup$ – Ric Jun 30 '14 at 6:48

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