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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

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 ...

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

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 ...

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 $K\gg S_t$,...

3

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 ...

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

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 ...

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$(-\infty,\infty). So ... 2 Assuming zero interest, the put option has the price \begin{align*} KN(-d_2)-S_0N(-d_1), \end{align*} and delta-N(-d_1)$. 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 ... 2 FX options are essentially the same mathematically as options on stocks that pay a continuous dividend. So the same arguments apply. If you are deeply in the money, it may be time to exercise a put. 2 The intuitive explanation is given in @Alex C's comment. You should stick to that if you understand it. Yet, if you are more comfortable with a mathematical approach: Payoff of being long a forward contrat with maturity$T$:$(S_T - X)$. Interpretation: at time$T$, you pay a certain price$X$in exchange for which you receive the underlying$S_T$Payoff ... 2 The short answer: Your observation is caused by some sort of central limit theorem. The long answer: The reason for the volatility smile/skew is the non-normality of the assumed return distribution. If the implied volatility of out-of-the-money put options is higher than of at-the-money put options, it implies that the market assumes that the risk-neutral ... 2 The PDE doesn't equal zero when I replace with the expression of$V(S)$you gave. Threre is an issue in your PDE's the boundary conditions and its solution. Let's start with the regular american put PDE, and deduce the perpetual american put ones, then solve for the price: Regular american put Black scholes PDE is satisfied by the price for$S(t) > S^*(...

2

I firstly fill in the gaps from Slade whose comments outline the answer and then I provide an alternative approach. Let $K_1\leq K_2$. You want to prove $P(S_t,K_1,T)\leq P(S_t,K_2,T)$. Recall firstly that $P(S_t,K,T)=e^{-r(T-t)} \mathbb{E}^\mathbb{Q}[\max\{K-S_T,0\}\mid\mathcal{F}_t]$ which is the result from risk-neutral pricing. Probably you know that ...

1

Please recheck, from what i can understand this question is trying to test pricing for currency options on currency forwards.Spot price is $0.6868$. If underlying is a currency forward, the underlying price $S0$ would be the forward price calculated using the interest rate parity. $$S0 = 0.6868*\exp((0.028 - 0.0145)*0.5)$$ ATM options means strike price is ...

1

No, your first assumption isn't quite right: you forgot a scaling factor, and in fact we have $P(S_0,S_0.B_T) = P(S_0/B_T,S_0) * B_T$. This is turn will also approach $S_0$ when $T \to \infty$.

1

I would look to these papers below by Dan Stefanica et al. Very easy to code and yields better results. An Explicit Implied Volatility Formula Tighter Bounds for Implied Volatility

1

While this question is basic, I am answering because putting models and/or formulae into practice is a part of quant finance that is not covered extensively on this (or any other) SE. (The below will be from the perspective of a listed equity) In practice, when/if you find options that are not trading in parity with their underlying the first thing you ...

1

It is as simple as just taking the max(). The problem is that you took the wrong one. You must consider the max between the intrinsic value of the option on the one hand and its discounted continuation value (which is an expectation in the risk-neutral world) on the other. In your final loop, you should therefore replace the line W(j+1,1) = max(K-W(j+2,1)...

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