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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.

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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. $\endgroup$ Commented Jan 8, 2018 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. $\endgroup$ Commented Jan 8, 2018 at 16:58
  • $\begingroup$ Thank you so much. I researched about it for 2 days straight and then finally set the question on exchange. $\endgroup$ Commented Jan 8, 2018 at 17:03

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