Tag Info

New answers tagged

0

I think the delta-replicating of $\sigma_2$ call is just a fancy way of saying "hedging the call option bought at $\sigma_1$ volatility, with deltas based on $\sigma_2$ volatility". This is full arbitrage in case the hedging/replicating is optimal, and just a statistical arbitrage in real life. You probably do not need such sophisticated proof of why this is ...


0

Take a look here: here Option values don't always go up with maturity.


3

First, as far as I can tell, you are not taking into account dividends. Second, If you simply take the forward price of the SPX @ $5.5\%$ which is what you are using, you get $1411 \cdot \text{exp}(0.055 \cdot 2.99) = 1663$. Given a strike of $1300$, the call should have an intrinsic value of $1663-1300= 363$. You have a price of $272$. The price is less ...


0

This is pretty straight forward: The market prices vanilla options via implied volatility. You can like it or not like it but that is the way it is. So, the fair price of the option is the equivalent of the implied vol via BS. Now, if you believe the true price of an option should be different from the traded market price and you figure out that you have ...


1

i would guess it is the difference between $$\exp(-0.5\sigma^2 T + \sqrt{T}\sigma Z)$$ and $$\exp(\sqrt{T}\sigma Z)$$ The first is correct, the second is wrong.


0

It the value of the forward contract and not the forward price that has drift r under the risk-neutral measure. In fact, in the simplest case where the risk-free interest rate is a constant r, then the forward price process f(t,T) has zero drift under the risk-neutral measure: If the spot price process satisfies dS(t)=S(t)(rdt+bdW(t)), then ...


0

It depends what you want volatility for. Theory will tell you that: "Implied variance of short maturity ATM options is approximately equal to the expectation of the realised integrated variance of the underlying over the life of the option and under the risk neutral measure" In math: $\sigma^2_{ATM}\approx E^Q\left(\frac{1}{T}\int_0^T\sigma^2_t dt\right)$ ...


0

The main difference is that one approach assumes that a certain dynamical structure properly describes the underlying instrument, while the other approach is really only a re-writing of the price in terms of an implied volatility. Implied volatility Implied volatility really only needs two things: the underlying stock price and the call option price (apart ...


2

The short answer is: As long as a derivative can be perfectly replicated via hedging in the underlying asset then the price of the derivative should be independent of investors' risk aversion and hence the application of risk-neutral probabilities and discounting of the future expected payoff under risk neutral probability leads to the same price of the ...


2

The risk-neutral probability is used as a convenient mathematical tool but, strictly speaking, it is not a necessary ingredient for the BS formula. In fact, the formula can be derived by computing the expectation (under the physical probability) of the option payoff, properly discounted with the right stochastic discount factor: $$ p_t = E_t[\;m_{t,T}\; ...


-1

Because BS is about derivatives and not about the underlying. In a way if you priced derivatives with real world measures (all else being equal) you would double count risk preferences because these are already included in the underlying - think about it this way (beware, oversimplification ahead): You want to price a derivative on gold, a gold certificate. ...


0

If we are talking about BS world, one plausible explanation is every move in the risk-free world is "perfectly hedgeable". In other terms if you consider the drift term $\mu = r + \lambda\sigma$ BS formula allows you to reduce $\lambda$ to zero by constantly hedging with the underlying. That is why you can get a correct price with $r$ under BS world. ...


0

Risk-neutral probabilities and physical probabilities agree on what is possible and what is impossible. Also, a hedge under Risk-neutral probability works almost surely and so does the hedge under physical probability.



Top 50 recent answers are included