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How can I prove that under the risk-neutral probability:

$\mathbb{P}[S_{t}<K]=-\frac{\partial{C}}{\partial{K}}(K,T)$

where

$S_{t}$ is the stock price, K is the strike price, C is the call option price

Thank you !

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Your posting has an error, that is, the identity should be \begin{align*} -P(0, T) \mathbb{P}(S_T > K) = \frac{\partial C}{\partial K}. \end{align*} The derivation below is based on this assumption. We denote by $f(x)$ the density function for $S_T$. Then \begin{align*} \mathbb{P}(S_T > K) = \int_K^{\infty} f(x) dx, \end{align*} and \begin{align*} C(K, T) &= P(0, T) \mathbb{E}\big((S_T-K)^+ \big)\\ &=P(0, T) \int_K^{\infty}(x-K) f(x) dx\\ &= P(0, T)\bigg[\int_K^{\infty} x f(x) dx - K \int_K^{\infty}f(x) dx \bigg]. \end{align*} Then, \begin{align*} \frac{\partial C}{\partial K} = -P(0, T)\int_K^{\infty} f(x) dx. \end{align*} That is, \begin{align*} -P(0, T) \mathbb{P}(S_T > K) = \frac{\partial C}{\partial K}. \end{align*}

We can additionally obtain that \begin{align*} P(0, T) f(K) = \frac{\partial^2 C}{\partial K^2}. \end{align*}

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  • $\begingroup$ Okay thank you very much . Now I understand where this formula comes from. In my formula, it was supposed that r = 0 that's why there is not the term P(0,T) but I think there is no effect about the false formula I wrote ... $\endgroup$ – glork May 5 '15 at 7:42
  • $\begingroup$ Just an interrogation: $\mathbb{E}[(S_{T}-K)+]P(0,T)=C(K,T)$ is this a formula that we can prove or is this the definition of the price of the option ? $\endgroup$ – glork May 5 '15 at 7:53
  • $\begingroup$ @glork: The formula $\mathbb{E}[(S_T-K)^+]P(0, T) = C(K, T)$ is obtained by assuming deterministic interest rates, or by treating $\mathbb{E}$ as the expectation operator under the $T$-forward measure. $\endgroup$ – Gordon May 5 '15 at 12:31

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