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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.

5
votes
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
16
votes
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
2
votes
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
2
votes
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
3
votes
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
5
votes
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
4
votes
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