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5

In the Black-Scholes-Merton model, with model option price $V$ as a function of underlying price $S_t$, strike price $X$, continuously compounded risk-free rate $r$, continuously compounded dividend yield $y$, time-to-maturity (in year fractions) $\tau$ and implied volatility $\sigma$, our $\Delta$ is defined as $$ \Delta\equiv \frac{\partial V}{\partial S_t}...


5

In addition to the good answers and charts already given: Assume a Black-Scholes world to start with (no skew). Then taking $r=q=0$ for simplicity but without loss of generality, first note that $$ \frac{\partial P^{BS}}{\partial K} = \frac{1}{K} \left(P^{BS} - S \frac{\partial P^{BS}}{\partial S} \right) $$ where $P^{BS}$ denotes the put option with strike $...


4

VIX almost always only spikes when SPX goes down as @Jan Stuller also mentions in a comment. Insofar the question is a bit counterfactual. I frequently use twin axis in the charts that follow. The position of the label corresponds to the axes the ticker belongs to. These are essentially two question in real world scenarios. 1 ) VIX and Vega: VIX up, IVOL up,...


4

Vega is the option's price sensitivity to the volatility (i.e. IV). In the graph below, vega is shown to be a strictly positive function in volatility, which means that at any point in the graph (i.e. for any value of IV, irrespective of whether the option is OTM, ATM or ITM), the option price: will increase in value if IV goes up (because Vega is positive) ...


4

An actual option with an independent existence cannot have a negative price. But we are talking here about 'embedded options' that are part of another security (in this case a USTR bond) and cannot be separated from their parent. Their price is not quoted in the marketplace but is found by a calculation. The problem is that this calculation comes up with a ...


3

This is quite a brain teaser, at least it was for me. The way I thought about this initially was based on statistics. This lead me to believe that higher IVOL should always decrease the probability of exercise, no matter if ITM, ATM or OTM. If the probability decreases for ITM call options ($S_0 > K$), there is no intuitive reason why it should increase ...


3

The best way is to start with definitions (instantaneous and their finite difference versions) of Greeks. For a currency pair $(FOR,DOM)$ with FX rate $S$, the number of $[DOM]$ (domestic, numeraire, right-side) units needed to buy one $[FOR]$ (foreign, asset, left-side) unit, let $V(S)$ be an option's price in $[DOM]$ units. Note that the unit of $S$ is: $$ ...


3

Tricky right? The reason is that when dealing with stocks or options or any sort of spot market asset, there is an original cash settlement. For instance: You have an account with $4000 and you buy AMZN for 3500. 3500 leaves your account and 1 share of AMZN comes in. Your account is now 500 + 1 share AMZN. AMZN price goes to 4000. Your account is now worth ...


3

You are correct, in that you need not (explicitly) specify real world dynamics to calculate option prices. Indeed in many rates derivatives models, you simply assume a unique risk neutral measure exists (completeness), specify the dynamics under the risk neutral measure (risk neutral probabilities) and price your options. At the same time, it is important to ...


3

See the graph below. Let's define the PNL as the position's payoff at expiry plus accrued initial investment, i.e. collected / paid option premia. Assuming $K_1=95,K_2=100,K_3=105$ (i.e. $\lambda=0.5$), the orange payoff diagram below belongs to a setup where $C_2<\lambda C_1 + (1-\lambda) C_3$: You paid some net fee initially, and you obtain a position ...


3

Yes. The Options Clearing Corp handles this in a way that you would expect. 2 for 1 splits are easy: strike prices are halved and you get 2 for every 1 that you had before. Then everything continues trading as before. Non-standard splits, mergers, and other share dividends become more complex. On a 3 for 2, split, the share count is changed, the strikes are ...


2

Delta is not the probability of finishing in the money as suggested in another answer, N(d2) is. The foot note mentions this. See Understanding N(d1) and N(d2): Risk-Adjusted Probabilities in the Black-Scholes Model by Lars Tyge Nielsen for a detailed explanation. If time and vol is low, $d1 \approx d2$ and delta will be closer to the risk-adjusted ...


2

Agree with @Brian B. With BS, you cannot have the issue in (1). Tree, grid, Monte Carlo could all result in errors though. (2) is a likely reason. I just tried in Julia for ATM, 0 div and rates plus 0.2 vol and 1 year tenor. Shifts smaller than ~ 0.00008 result in an error for Gamma. Delta seems to be less sensitive for this, and it is fine for at least 1e-7 ...


2

Everything (warning: I have not checked 3rd order greeks) that is not delta is in terms of ccy2 in the standard Garman Kohlhagen model. Gamma is not in CCY1 by default either (some vendors like Bloomberg display it like that to be consistent with Delta). Why Let's start by not looking at FX but equity to help build intuition. The actual price of an option is ...


2

I guess generally what ATM means depends a lot on asset classes. FX vols are quoted as ATM DNS (delta neutral straddles). This in itself can be Spot, Forward, Spot premium adjusted, forward premium adjusted with the following formulas retrieved from the working paper FX volatility smile construction : However, based on your wording I assume you think 50D ...


2

Not sure if you meant only short puts with "when option is ITM increase in volatility will decrease the delta, whereas for OTM option increase in volatility will increase the delta". Either way, you cannot generalize it like that as you figured out yourself. Adding a few remarks to @Kermittfrog's excellent answer. If you plot delta and the ...


1

Copy pasting parts of an answer I did here as it illustrates the limits of call and put option premia. $N(d2)$ is the probability that a call option with an exercise price of $K$ is exercised in a risk-neutral world. Therefore, $(1− N(d2)$ or $N(-d2)$ is the probability that a put with the same exercise price will be exercised. Let's plot this as a function ...


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