How can I prove that under the risk-neutral probability:
$\mathbb{P}[S_{t} where $S_{t}$ is the stock price, K is the strike price, C is the call option price Thank you !
Answer
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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