4

If $\sigma=0$, the stock price is deterministic and grows at rate $r$. In one year, it is thus worth $100\cdot e^{0.05}\approx 105.13$. The strike is $K=100$. Your payoff is thus $5.13$. Discounting at rate $r$, you get as today’s fair option price $5.13\cdot e^{-0.05}\approx4.88$. Note that there is no randomness and the stock price is perfectly predictable....


3

Let $C\left(S,t\right) $ represent the price of the call option when the underlying price is S at time t. Now if S changes by h instantaneously, the call price becomes $C\left(S+h, t\right) $. So the change in the call option price is: $C\left(S+h, t\right) - C\left(S,t\right) $ Which you can approximate via first order Talyor series: $C\left(S+h, t\right)...


2

I will answer in two parts: 1): Assuming this is meant in the Black Scholes sense (one dimensional diffusion), here is an alternative explanation motivated by the convexity of the option price. In simple terms, if you plot the Back Scholes Call price as a function of S, you get this convex curve (blue curve): Now if you draw the tangents at two different ...


2

As @AlexC shows this is true in the Black Scholes model. However as the late Mark Joshi comments in the link, the result is not true in all models. For example , in the stochastic vol model , suppose implied vols fall if the underlying stock moves away from at the money in either direction. Perhaps the losses on implied vol exceed the gamma profits from the ...


2

Here is the reasoning I use. (Assuming familiar facts from the standard BSM setup). The Delta of a Call is positive, as S goes up the Delta goes up towards 1. (Why? The call has more moneyness, more probability of exercise, so the option behaves more like the stock). The Delta of a Put is negative, as S goes up the Delta goes towards zero (less chance of ...


2

To add a bit to Will Gu's answer: Compute $\mathbb{E} \left[ \left. S_T \right| S_T > K \right]$ using the fact that $S_T$ is lognormally distributed with mean $ln(S_0) + (r - \sigma^2/2)T$ and variance $\sigma^2 T$. Then find the pdf of the lognormal distribution on, e.g., Wikipedia, and compute the expectation integral. You may find the following ...


1

In the following, I am assuming the BS73 model and I assume that "ATM" means $$ S = Xe^{-r\tau} $$ The pricing formula for a European call then becomes $$ \tag{1} O\propto N\left(+\frac{1}{2}\sigma\sqrt{\tau}\right)-N\left(-\frac{1}{2}\sigma\sqrt{\tau}\right) $$ times some scaling factor which is irrelevant for our purpose. Clearly, $$ Vega\equiv\frac{\...


1

Based on your computation, you can observe that the $N’$ term is always positive, between 0 and 0.4. As $\sigma$ is always positive, you can focus on the $-d_2$ term. When $d_2 > 0$, i.e. call is ITM, delta has a negative sensitivity to volatility ; conversely for OTM call. That is in line with your remark.


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