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1

This has been asked many times already. Volatility always refers to a model. And unless stated otherwise this model is the Black-Scholes model. In this model the volatility is the standard deviation of the log-returns divided by the square-root of time: $$ \log(\frac{S_{t}}{S_0}) = (r - \frac{1}{2}\sigma^2)t + \sigma W_t \sim \mathcal{N}\left( (r - ...


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One reason is that implied volatility measures the relative value of the option as the price of an option depends on various parameters. As everyone has its own pricing model, it's insane to quote all parameters. This little simple IV tells you everything you'd need to know for valuation.


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The most common use for implied volatility in valuation is for asseing options or option like postions. A volatile instrument is likley to activate or put an option postion in the money just on the basis of its volatility rather than any fundamental change in the intrinsic or fair market value of the underlying. This needs to be taken into account when ...


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Since $W_{2t}-W_{t}$ is independent of $W_t$ and has the same law as $W_{2t-t}=W_t$ we only have to compute $$P(X(X+Y)<0)$$ where $(X,Y)$ follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either $X<0$ and $X+Y>0$ or the opposite. The two events have the same probability since ...


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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, ...


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Your decomposition is correct. I will show here the computation for one term: \begin{align*} P(W_t < 0, W_{2t} >0) &= E(W_t < 0, W_{2t}-W_t > -W_t)\\ &= E\Big(E\big(\mathbb{1}_{\{W_t < 0\}}\mathbb{1}_{\{W_{2t}-W_t > -W_t\}}\mid W_t\big)\Big)\\ &= E\Big(\mathbb{1}_{\{W_t < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ ...



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