Pricing and valuation of contingent claims

When it comes to the option Greeks, it is really easy to imagine delta - delta(option)/delta(stock price), then if deep-in-the-money delta moves to 1 for call option and vice versa -1 for put option then when you look at the gamma -delta(delta)/delta(stock price) so

1. why they say gamma is non negative and when it comes to put option it is positively correlated with delta, however in gamma it is negatively correlated with delta?
2. Why gamma and vega of put and call options are alike, so what are the main differences to differentiate them?
Could anyone please explain for me, quite difficult to understandâ€‹:joy:

the Greeks are just the derivatives (I know, confusing because derivative means 2 things):
if V is the option price, S the stock price, and \sigma the volatility, then
Delta =\frac{\partial V}{\partial S}
Gamma =\frac{\partial\;\textrm{Delta}}{\partial S}=\frac{\partial^{2}V}{\partial S^{2}}
Vega =\frac{\partial V}{\partial\sigma}

If you remember put-call parity,
long call and short put = long stock and short bond
call - put =S-Xe^{-r(T-t)}
If you differentiate put-call parity with respect to stock price, you find
Delta(call)-Delta(put)=1.
If you differentiate a second time with respect to stock price, you find
Gamma(call)=Gamma(put)
Similarly, if you differentiate put-call parity with respect to volatility, you find
Vega(call)=Vega(put).

In the Black-Scholes model, you have a formula for the value of a call and of a put, and you can differentiate those formulae to find the Greeks.
Gamma =\frac{1}{\sigma S\sqrt{2\pi(T-t)}}\exp\left[-\frac{1}{2\sigma^{2}(T-t)}\left(\log(S/X)+(r+\frac{1}{2}\sigma^2)(T-t)\right)^2\right]
Gamma is the same for the call and the put.
r is interest rate, \sigma is volatility, t is time, with option expiry at t=T

When stock price is zero, Gamma is zero.
For stock price S>0, Gamma is >0.
Hence `non-negativeâ€™ which means it can be either positive or zero but not negative,
Under the Black-Scholes model, if the stock price is zero, it remains zero. But if the stock price is >0 it will never become zero.
Gamma is the derivative of Delta with respect to stock price
For stock price >0, Gamma will be positive which means Delta will always increase as stock price increases for S>0.
As you say, for the call, Delta goes from 0 (when S=0) to 1 (in the limit S\to\infty)
and for the put, Delta goes from -1 (when S=0) to 0 (in the limit S\to\infty)

If you work out Vega (derivative with respect to volatility),
Vega =\frac{S\sqrt{(T-t)}}{\sqrt{2\pi}}\exp\left[-\frac{1}{2\sigma^{2}(T-t)}\left(\log(S/X)+(r+\frac{1}{2}\sigma^2)(T-t)\right)^2\right]
so that Vega=Gamma \times S^{2}\sigma(T-t)
Yes, the formulae are alike, but Vega and Gamma are different things.
If you graph them, Vega and Gamma both look like skewed Normal Distributions.
Vega will be larger than Gamma as S\to\infty
Gamma will be larger than Vega as S\to 0

1 Like