I have an exercise where I need to show that the prices of call options $ C(t,K)=E((S_t-K)^+),t \in [0,T]$ with Strike $K$ for fixed $t$: $$\frac{\partial ^+C(t,K)}{\partial K}=-P(S_t>K).$$ We havent discussed Black Scholes model yet. I guess this will be the introduction exercises for the BS formulas. With: $$\frac{\partial ^+}{\partial K}=\lim_{h↓0}\frac{C(t,K+h)-C(t,K)}{h}$$ I get: $\frac{\partial ^+C(t,K)}{\partial K}=\lim_{h↓0}\frac{C(t,K+h)-C(t,K)}{h}=\lim_{h↓0}\frac{E((S_t-(K+h))^+)-E((S_t-K)^+)}{h}=\lim_{h↓0}\frac{P(S_t>K+h)(E(S_t|S_t>K+h)-(K+h))-P(S_t>K)(E(S_t|S_t>K)-K)}{h}...$

From there I dont know how to proceed further. Using L'Hospital b/c we have $"\frac{0}{0}"$ or left term could be 0. Please help.


First note that delta is the derivative w.r.t. to the spot and not the strike. The latter is often called "dual delta". Also, you don't need any knowledge of Black-Scholes as this is a model-independent result.

The result follows from the general expression of the call price

\begin{equation} C_0 = e^{-r T} \mathbb{E}_\mathbb{Q} \left[ \left( S_T - K \right)^+ \right] = e^{-r T} \int_K^\infty (x - K) \mathrm{d}F(x), \end{equation}

where $F$ is the risk-neutral distribution function of $S_T$. Differentiating w.r.t. $K$ yields

\begin{equation} \frac{\partial C_0}{\partial K} = -e^{-r T} \int_K^\infty \mathrm{d}F(x) = -e^{-r T} \mathbb{Q} \left\{ S_T > K \right\}. \end{equation}

This is probably one of the most common questions here; search for "Breeden-Litzenberger" for related answers.

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  • $\begingroup$ Thanks for your quick answere. It's not needed to know that F is a distribution of S discrete or continuous with given density f? Thats why I tried to avoid integration. $\endgroup$ – TheDude Jan 8 '18 at 16:54
  • $\begingroup$ That's why I am integrating w.r.t. the distribution function $\mathrm{d}F(x)$ instead of w.r.t. to the density $f(x) \mathrm{d}x$, which might not always be defined. I.e. the above to integrals are Riemann-Stieltjes integrals. $\endgroup$ – LocalVolatility Jan 8 '18 at 16:58
  • $\begingroup$ Thank you so much. I researched about it for 2 days straight and then finally set the question on exchange. $\endgroup$ – TheDude Jan 8 '18 at 17:03

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