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Greeks are named quantities representing sensitivity of option price to change in underlying parameters. Use of [greeks] tag should relate to one more named quantities, such as delta or gamma.

Something went wrong in the third equality of the equation where you compute $\partial C_0 / \partial K$. Starting from the second equality, you can use that \begin{equation} S_0 \mathcal{N}' \left( …
answered Jun 13 '17 by LocalVolatility
Find the topic of model-independent properties of option prices very interesting as well. Here are some results that I am aware of and the respective references in the literature. Some are already con …
answered Sep 17 '16 by LocalVolatility
I think you nearly got there but made a few mistakes in the application of l'Hopital's rule. First Limit In the first case, you got \begin{eqnarray} \lim_{S_0 \rightarrow \infty} \Omega & = & \lim_ …
answered Mar 11 '17 by LocalVolatility
Assume that the time $t$ forward for the maturity $T > t$ is given by \begin{equation} F_t(T) = \left( S_t - D_t(T) \right) e^{r (T - t)}, \end{equation} where $D_t(T)$ is the time $t$ value of all …
answered Mar 7 '17 by LocalVolatility
The reason is that in many common models including geometric Brownian motion, the variance of the logarithmic returns is proportional to time. Thus, their standard deviation/volatility is proportional …
answered Sep 18 '16 by LocalVolatility
The risk exposures/sensitivities of long and short positions always have different signs. This has to hold since derivatives are zero sum games. Vega is always positive for a long position in a Europ …
answered Sep 18 '16 by LocalVolatility
It seems like he is assuming that the shorter term volatilities change more than the longer term ones and the relatively sensitivity is proportional to $1 / \sqrt{T}$. Thus, this hedge is not against …
answered Sep 19 '18 by LocalVolatility