# Tag Info

6

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...

5

You only have one asset in your portfolio which means that you can only statically hedge. By the definition of self financing, $V_0=\phi_0 S_0$, $V_1=V_0+\phi_1 (S_1-S_0)$, and $V_1= \phi_1 S_1$. Putting these last two together, $V_0=\phi_1 S_0$. Hence $\phi_1=\phi_0$ and you have a static position. Intuitively, this is because you cannot trade in ...

5

By definition the fair value of an option is given by an expectation value of the payoff, $\mathbf{E}\left[\textrm{payoff}(\textit{paths})\right]$. The probability distribution of the paths is the risk neutral measure. This is just an integral expression of the form you wrote. This applies to all option prices. Many options are, of course, special in the ...

5

importance sampling is well known to be tricky. See the extensive discussion in Glasserman's book. I presume that you are simply meanshifting and multiply by the ratio of normal densities. For this sort of problem, I'd use a more stratified algorithm instead and force every path to end in the money. To do this I'd compute the uniform that goes to the ...

4

It doesn't imply $$\ln S_T=\ln S_0+rT+σW^Q_T,$$ it implies $$\ln S_T=\ln S_0+(r-0.5\sigma^2)T+σW^Q_T.$$ Look up Ito's lemma. This is covered in just about any book on financial maths including my own Concepts etc.

4

Your characterisation is correct but incomplete. 1) The most important part of Black-Scholes is not the model but the more general framework of dynamic hedging: you can replicate your payoff by continuously trading the underlying and the amount (delta) you should hold is the derivative of the current premium with respect to the current spot. This is a much ...

4

This is a bit of an old question, but I thought I'd contribute to add more weight to to what some people have been saying. A CSO (calendar spread option) is NOT a calendar spread of options. If you read it carefully, you can see the Hull quote Max Li posted is talking about a calendar spread, not a CSO. A CSO needs to be priced the same way as a spread ...

4

If $\mu$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ...

4

Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ...

4

Since American style options allow early exercise, put-call parity will not hold for American options (unless they are held to expiration). In practice, there is also a difference between calls and puts for European options as well. The full description is here: What causes the call and put volatility surface to differ?

4

This drift comes from making the discounted stock a martingale in the risk-neutral measure $\mathbb Q$ You start with a stock in $\mathbb P$ having this form: $$dS_t = \mu S_t dt + \sigma S_t dW_t$$ You also have a discount factor $e^{rt}$. The idea is to remove the drift of the discounted process in $\mathbb Q$ so you get (after applying Girsanov's ...

4

I know two papers explaining how to calibrate this kind of models, and one of them explain the impact of the quality of the fit on a pricing model: Aït-Sahalia, Y. (2002, January). Maximum likelihood estimation of discretely sampled diffusions: A closed-form approximation approach. Econometrica 70 (1), 223-262. Azencott, R., Y. Gadhyan, and R. Glowinski ...

4

Let $\{P_t \mid t \geq 0\}$ be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and $N_t$ is a Poisson process with intensity $\lambda$ and jump times $\tau_i$, $i = 1, \ldots, \infty$. Let $Y_i=\ln V_i$ and $f(x)$ be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ...

4

When a pay-off is piecewise linear plus jumps, it the same as the portfolio of calls and digital calls. Its price must agree with that of the portfolio by no arbitrage. Every time there is a jump we add in a digital call and every time there is a change in gradient we add in calls equal to the gradient change. Here we have a call struck at $K$. Just below ...

4

It is not the fact that volatility is time varying that creates the skew per se, but the fact that volatility is negatively correlated with the spot. That is to say, as the stock/index price declines volatility will tend on average to increase, and vice versa. Time varying volatility itself would create a more symmetric 'smile'. Edit: Suppose that you ...

3

I found and answer to my own question. So, I post it here for people who maybe have the same problem. The answer, however, is quite intuitive. The last observation used for the estimation of the physical density is also the time point where the investors know the most about the physical density because at this point the most possible historical observations ...

3

"Intuitively, everything else being equal, if a stock has higher drift, shouldn't it have higher probability of finishing in-the-money (and higher probability of having higher payoff), and the call option should be worth more?" All these other answers are focusing on the wrong aspect of the question - it is true that the maths makes the drift drop out from ...

3

you don't need $ud=1.$ In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have $ud=1.$ All you need is $$d < e^{r \Delta t} < u$$ The tree recombines provided $u$ and $d$ don't change from step to step. See my book More Mathematical Finance for a comprehensive review and ...

3

IV is one of the inputs for your option pricing model, vega measures the actual impact (e.g. in Dollars, Euros...) of any change in IV. Intuitively IV is the price of the option while vega is the sensitivity to IV. Bottom line: There is a clear distinction!

3

this is probably the most asked question in quantitative finance... There are many answers. One nice example to consider is what if the calls were struck at zero. The call then pays the stock price at time $T$ and so it's value today must the stock price today since we can replicate by holding one unit of stock. This will be true regardless of the drift of ...

3

I don't know the BS formula you are trying to use. The price is the expected value of the discounted payoff under the risk neutral probability measure (I.e. Under which S is a martingale) So the you need to compute the risk neutral probabilities for S to go up or down. The probabilities given in the problem have no impact. They are just there to trick the ...

3

The above equation is the price of a call option. It has nothing stochastic inside it. It only depends on the current price and the time. So no Ito is needed. You should just compute the derivatives of your solution v (like you do for any deterministic multivariable function), plug them into the PDE and verify that it's satisfied.

3

There is a good quick well-known approximation for at-the-money options: $$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}.$$ See further discussion at What are some useful approximations to the Black-Scholes formula?.

3

if put call parity seems to be violated there could be things you are ignoring like dividends or hard to borrow fees. Hard to borrow will make puts more expensive

3

The typical investor is long. To protect the portfolio, he buys puts, thus driving up the price. To generate income against his long position, he sells covered calls, thus driving down the price. This is the most basic explanation for the difference in put call prices that are equidistant from the money. Obviously other factors are there as pointed out by ...

3

Its a stylized fact in academia that put options are overpriced. E.g., the monthly average return on S&P500 put options is around -40% for ATM options. The most often quoted reason for this phenomenon are hedging costs: A put is more difficult to hedge from a market maker's perspective, hence the prices artificially go up. An important paper on this ...

3

Options have an asymmetric payoff profile: The payoffs are zero for almost all cases and positive else (as we well know). If the option is OTM, most of its payoffs are zero. A rise in volatility will hence increase the likelihood for instead positive payoffs from a change in the underlying price (i.e. delta increases). If the option is already ITM, ...

3

if you let $\delta t$ be small enough, this won't happen. So the solution is to take more steps. The CRR tree is very out dated in any case.

3

Fact 1: if you are not good at pricing options, of course you can create a lot of arbitrage opportunities for the rest of the market. It does not matter whether the reason is in dividends or anything else. Fact 2: if you are good in pricing options, you price the dividend effect in advance. Consider the situation of the European calls, and suppose that both ...

3

The No-Arbitrage bounds for a European put are: $$(Ke^{-rT}-S)^+ \leq P \leq K e^{-rT}$$ This is because the maximum payoff at maturity is $K$ (discounted) and the minimum value is the discounted intrinsic value (since $E(e^{-rT}S_T)=S_t$ by the martingale condition and the payoff being always semi-positive).

Only top voted, non community-wiki answers of a minimum length are eligible