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3

Since you have that proportionality between the stock price $S$ and the futures price $F = Se^{rT}$, you just have to slightly shift your graph but the shape is the same.


2

If the stock prices falls "gently" and the option remains in the money, your delta will converge to 1 and you will have to buy stocks: the gains on your long stock positions will be lower, but the payoff of the option you wrote will be lower too. If the stock prices falls more sharply and gets closer to the strike, there are higher chances that the option ...


0

I think it is simpler than looking at the distribution of returns. Since options can be perfectly hedged using synthetic positions, the distribution should not matter. Let's keep it in the put-call-parity universe. To make a synthetic call, you need to buy a put, buy a stock, and borrow the money at rate r. You're paying that back over time. To make a ...


1

There are two ways (or shall i say at least two ways) to look at this. 1) The option price does not depend on the promise of the pay-offs alone, but also on the probability of those payoffs. As you alluded to, if you look at the probability distribution, you will see the unlimited payoffs is so unlikely as to be irrelevant. 2) The put call parity gives you ...


0

Multiple "things" = Multi-Asset option. Based on your description it seems to me that what you need is a 90% basket put option. Call 3-4 banks and ask them to price it, pick the best price. It's far from trivial to price a basket option. Delta-hedging does not have the same effect as an option. Rather, it neutralizes moves in the spot prices, but leaves ...


3

A call struck at $100$ costs $2.97$, therefore a call with a strike higher than $100$ must cost less than $2.97$.


3

First, you'd rather simulate $\log(X)$ rather than $X$; thus, there is no level dependency in your discretisation scheme, making it more accurate. $$Z_t = \log(S_t)$$ $$dZ_t = \left(r - \frac{\sigma^2}{2}\right)dt + \sigma dW_t$$ You can even run one single time step, and the distribution of your final price will still be as accurate! Second, the different ...


5

Just to add to the answer by @KeSchn : There are at least two things going on here. First of all let $\{Q_i \}$ denote a set of equivalent probability measures, which includes your $P$ and $Q$ above. Any $F^i(t)$ defined as $F^i(t) = E_t^{Q_i} [P_T]$ will be a martingale by application of the tower law. With the definition above, it will not be the case ...


5

You probably wonder whether $\mathbb{E}^\mathbb{P}[P_T\mid\mathcal{F}_t]= \mathbb{E}^\mathbb{Q}[P_T\mid\mathcal{F}_t]$. Note the $T$ as index, i.e. the future unknown payoff and not the current price $P_t$. Now, why should $P_t$ be a martingale under both, $\mathbb{P}$ and $\mathbb{Q}$? Most likely, it is not. Indeed, the reason why you use $\mathbb{Q}$ in ...


3

You can compute the SV - LV price difference and see if it is substantial or not. This is easily done and will give you an indication of whether your product can be safely priced with LV only. start with a pure SV model: choose $\sigma(t,S)=1$ and do a rough calibration of the parameters that drive $u_t$, to historical data for instance Price your ...


4

Local volatility models capture skew today but not dynamics tomorrow. Stochastic vol captures dynamics tomorrow but not necessarily skew today (how well does your calibrated vol surface match observation?). To answer your question: if you're pricing exotic options that are path dependent, stochastic local vol is more accurate. If you're pricing vanilla ...


6

As ilovevolatility pointed out, the main application of the COS method is to price options. The initially proposed method simply approximates the probability density function by a cosine expansion using the characteristic function of the log spot price. So, if you know the characteristic function (say for exponential Levy models and many stochastic ...


5

With a long time to maturity, your options have a low theta because their time value decays quite slowly. If there are many months to go, the passage of one day does not change the exercise probabilities too much, whereas short life options with only a few days left have a much higher time value decay. Hence, the larger the time to maturity, the lower theta. ...


0

This is an old post, but here is a recent paper on a market model for pricing and hedging volatility derivatives: A multi-factor shifted lognormal model for forward starting variance swaps


2

I think the variance of the instantaneous shifts in the spread is meant: $V \left[ dX \right]=V \left[ dS_1-dS_2 \right]$ And the individual variances (in the conditional and local sense) are: $V \left[ dS_1 \right]= \sigma_1^2 S_1^2dt$ $V \left[ dS_2 \right]= \sigma_2^2 S_2^2dt$ And the covariance term is, assuming the two Brownians are correlated:...


1

Under the condition $r=\frac{\sigma^2}{2}$, it is true that $S_t = S_0e^{\sigma W_t}$. Since \begin{align*} E\Big( S_T - \min_{0 \le t \le T} S_t\Big) = E\big( S_T\big) - E\Big(\min_{0 \le t \le T} S_t\Big), \end{align*} what you need is the expectation $E\big(\min_{0 \le t \le T} S_t\big)$. Note that \begin{align*} \min_{0 \le t \le T} S_t = S_0e^{\sigma \...


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