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11

In my opinion, professionals mainly trade options if they want to trade the volatility. I believe there is a mathematical proof that shows that if the realized underlying volatility between the option inception and maturity exceeds the implied volatility of the option (priced in at inception of the option), the option seller would lose money if they delta-...


9

In addition to other reasons mentioned here, options tend to be expensive to trade (they have high bid-ask spreads). These do add up in institutional asset management, so best avoided. Further, if trading options at any significant scale, the underwriter of the option will end up shorting the stock anyway, to cover their own risk. As the price fluctuates, ...


8

Under the assumption that the underlying cannot have a negative value, then the value of a put option cannot be greater than the strike. The reason behind this doesn't require maths, it's fairly simple: the lowest possible value of the underlying is zero. At that price, you make the maximum possible profit, of K. The value of the option is the probability ...


7

If a european option value becomes lower than intrinsic value it gets negative time value. In this circumstance the theta becomes positive because as time approaches to expiry the option value has to converge to intrinsic value. For european options there are 2 circumstances that can lead to the option value being lower than intrinsic value deep ITM puts ...


7

So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. Yes indeed However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever exercise ...


6

American put-call symmetry relies on the observation that trading $S$ for $K$ is optimal when $\frac1K$ is optimally traded for $\frac1S$. So long as the dynamics of the inverse process $\frac1S$ are sufficiently tractable, you can derive the symmetry formula. You don't have to have a "pure" GBM for this to work. For example, non-constant (and even price-...


5

The PPUT strategy is an example of a "tail protection strategy". The objective is to have a return somewhat similar to the return of the S&P 500 but with better performance during "crashes" (sharp down moves in the S&P). The strategy buys puts, which cost money (i.e. detract from returns) but are helpful when the SP drops more than 5% (improving the ...


5

As you can see from the wiki page, the delta of a put is $$\Delta = -e^{-qT}N(-d_1)= -e^{-qT} \left(1-N(d_1)\right)$$ Recall that this $\Delta$ is the derivative of the value of the put $p$ with respect to the value of the underlying stock $S$: $\frac{\partial p}{\partial S}$. So this means that if the underlying goes up by 1, the price of the put change ...


5

Given one satisfies margin requirements anyone can short exchange traded options as long as local regulators permit (American retail investors at present are not permitted, for example, to trade futures options . As long as there is a market and one finds a willing counterpart nothing speaks against shorting options contracts. Some brokers might require a ...


5

If you could hedge continuously with zero transaction costs, the gamma would be irrelevant: you would perfectly replicate with delta hedging and be done. In practice, hedging is discrete and there is a certain amount of slippage giving a random outcome with mean zero. The larger the gamma, the bigger the variance of slippage. Trading more frequently ...


4

$$\lim_{K\searrow0}\frac{P(K,T)}{K} = \lim_{K\to0}\frac{\int_0^K(K-S)f(S)dS}{K}$$ $$= \lim_{K\searrow 0} \left(\int_0^Kf(S)dS - \frac1K\int_0^KSf(S)dS\right)$$ $$= \lim_{K\searrow 0} \big(\mathbb P(S_T\le K)\big) - Sf(S)\big\vert_{S=K=0}$$ $$= \mathbb P(S_T=0),$$ where $f(S)$ is the density of $S_T$.


4

If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way). Let now $K\gg S_t$,...


4

I think it is far easier to understand by just drawing the payoffs. You have two put options: A European put option on a non-dividend paying stock with strike price 80 is priced at 8 dollars, and a put option on the same stock with strike price 90 dollars is priced at 9 dollar The difference between the two payoffs is equal to 10 dollars (90 strike puts ...


4

Puts are not available on all names, or might only be available for a limited set of expiries. I'm sure there are other reasons but those are the two most obvious.


4

At the risk of making maybe three obvious points: 1- Many funds' investment theses are not predicated on a particular price point on a specific expiry date. They simply believe that X is too high relative to Y, which is too low relative to X. Expressing this view via an options position would implicitly require them to take an additional view about the ...


3

I am assuming you are short EUR and long USD based on your description of your hedges. I am also assuming the size of your hedges and your fx position are the same. In the first example of a hedge of this position, you cannot be delta neutral. A long at-the-money (ATM) EUR call will have a higher delta than the short out-of-the-money (OTM) call--you will ...


3

There is no contradiction at all here. $Ke^{-rT}$ goes to zero for $T$ going to $\infty$ so the relation you mention suggests that $-S\leq p \leq 0$ as $T$ gets bigger. If $r>0$. Put prices can (in theory) not be negative so $p$ goes also to zero according to the relation you mention. Is this consistent with the pricing function of Black-Scholes? Yes. ...


3

I think the chain of logic should be as follows: We have put value >= intrinsic. Therefore either put value > intrinsic or put value= intrinsic. If put value > intrinsic, then it is not optimal to exercise. If put value = intrinsic , it may be optimal to exercise.. Hence there is no contradiction.


3

It is my understanding that a replicating portfolio for a put involves short selling stock and lending money. You cannot statically replicate an option. So this is not true in general, you'll need to re-balance your replicating portfolio (underlying + cash) dynamically if you want to replicate the option. This will imply sometimes buying stock and borrowing ...


3

A butterfly in general has a payoff of the form \begin{align*} (X_T-K_c)^+ + (K_p-X_T)^+-(X_T-K_{atm})^+-(K_{atm}-X_T)^+, \end{align*} where $X_T$ is the asset value at maturity $T$, while $K_c$, $K_p$, and $K_{atm}$ are strike levels.


3

This can be due to various effects. I will list you 2 of them off the top of my head: Jumps/Crashes : assume you were to price a put option which expires in a few days from now. Your diffusion model tells you this option should be worth $\$3.25e^{-7}$ since it is very out of the money as of today. What price will you quote? Well, in order to be conservative,...


3

It depends on the derivatives exchange but e.g. Eurex exchange can also be used by retail investors as long as they are qualified (concerning their max. risk level) and their bank offers access to it (some at least do that).


3

Of course you can sell options and you can certainly sell options on most major indices. Thinkorswim (TDAmeritrade) offers and excellent platform. Moreover, one can short options without "full" account privileges provided a defined risk trade is entered (such as an iron condor or call spread)


3

Maybe it will help your intuition if you think in terms of log-moneyness $\ln S/K$ instead of $S/K$. Let's look at a `deep' in the money put $K=100, S=10$. That sounds really deep in the money, but the value of log-moneyness for this situation is only $-2.3$, which is not that much if you consider the possible range of $\ln S/K$ is $(-\infty,\infty)$. So ...


3

You need an implied volatility assumption in addition to the price drop assumption to compute that. With a higher implied volatility increase the "profitability peak" you have will gravitate towards lower strikes. Vega is a more important pnl factor in that situation than pure delta, it's not surprising that an option which is still OOM would be more ...


3

It's the interest rate component. That is more meaningful in the formula. Note that the call becomes more expensive. Think about it this way. You could buy the call and sell the put instead of being long the stock. This gives you a synthetic long position. You need to pay the market the cost of borrow (r). That makes the calls more expensive and the ...


3

Cost. And greed. They want to squeeze every penny that is possible out of their transaction. It costs much less, maybe nothing, to short stocks that do not even exist. However the risk is substantial in case someone tries the short squeeze on you, as you could lose your shirt. This is just another example of the black swan biting those who do not really ...


3

Two main reasons: cost/premium: there is upfront premium associated with purchase of any put option. If your option ends up out of money, your premium is lost. for example, if stock price remains where it is (or higher than strike of your option), you will still end up losing money via premium timing: Even if your bet is correct that eventually stock goes ...


2

Assuming zero interest, the put option has the price \begin{align*} KN(-d_2)-S_0N(-d_1), \end{align*} and delta $-N(-d_1)$. When $N(-d_1)$ units of stocks are shorted and invested in bonds, the total value in bonds is $KN(-d_2)$, which is indeed greater than the option price. However, as you have shorted $N(-d_1)$ units of stocks, your portfolio value is \...


2

This is not quite right. The covered call you are describing is equal to selling a Put with the same strike price (\$105) and holding ( \$105 / (1+r) ) in the bank. If you draw the Payoff diagram this will become apparent. Put call relationships are summarized as the Put-Call parity: $$ S - C = D \cdot K - P $$ Where $S$ the underlying, $D$ is the ...


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