New answers tagged

0

Applying Itô's lemma to the Black-Scholes SDE and integrating from $t$ to $t+\Delta t$ gives: $$ S_{t+\Delta t} = S_t e^{(r-\frac{1}{2}\sigma^2)\Delta t + \sigma \sqrt{\Delta t}Z} $$ with $Z \sim N(0,1)$, showing that $S_{t+\Delta t}$ given $S_t$ is log-normally distributed. It is then straightforward to write, for any compact $\mathcal{A} = [a_1,a_2]$ ...


0

Note that $${{f}_{W(t)\left| W(s) \right.}}\left(x\left| y \right. \right)=\frac{{{f}_{ W(s),W(t)}}\left( x,y \right)}{{{f}_{ W(s)}}\left( y \right)}=\frac{1}{\sqrt{2\pi(t-s)}}\exp \left[-\frac{{{(x-y)}^{2}}}{2(t-s)} \right]$$ By application of Ito's lemma we have $$ln\,S_{t+\Delta t}=ln\,S_t+\left((\mu-\frac{1}{2}\sigma^2)\Delta t+\sigma(W_{t+\Delta ...


3

You are trying to price an option through Monte Carlo simulations. Here is how it should work, assuming the Black-Scholes diffusion framework. Under the Black-Scholes model's assumptions, the value of a risky asset $S$ at the time $t=T$ is a random variable which reads $$ S_T = S_0 e^{\left(\mu-\frac{\sigma^2}{2}\right)T + \sigma \sqrt{T} Z}\tag{1}$$ with ...


5

You should see this as a comment to @Kiwiakos answer which already hit the bull's eye. In the SE question you're referring to and to which I have answered, the idea was simply to provide you with a sound way of simulating returns out of a NIG distribution. It so happens that, for whatever your reason was, you decided to calibrate your NIG parameters based ...


5

I believe that the confusion arises because of the wrong treatment of NIG. The answer to the question you link is misleading, as it simulates under P which is not appropriate for option pricing. None of the NIG parameters under P carries over to Q in general, but especially the drift is the problem here. First use the mom gen function of NIG to find the ...


3

The ADF test assumes the DGP $$ \Delta y_t = \alpha +\beta t +\gamma y_t +\delta_1 \Delta y_{t-1}+\cdots +\delta_k \Delta y_{t-k}+\epsilon_t $$ The parameters are estimated using OLS on a sample of length $T$. You might impose $\alpha=0$ and/or $\beta=0$, this will give you different null hypotheses to test. But your test is always $\gamma=0$, and the ...


1

Your valuation is NOT for the knock-out option that you have specified. Let \begin{align*} \tau = \inf\{t \mid 0 \le t \le T, S_t \ge L\}. \end{align*} Here, we set the infimum of an empty set to $\infty$. Then, the payoff of the knock-out option is of the form \begin{align*} (S_T-K)^+ 1_{\tau = \infty}. \end{align*} Under the Black-Scholes setting, this ...


6

By definition, the payoff of a log-contract of maturity $T$ writes $$ \phi(S_T) = \ln\left(\frac{S_T}{S_0}\right) $$ Let $\Pi_t$ denote the $t$-value of such a contingent claim. We are interested in the price at $t=0$, best known as the option premium. Theory tells us that the latter premium can be computed as $$ \Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ ...


0

Correlated (simulated) features or variables are used when the source data has correlated features. If the original features in the source data are not correlated and are orthogonal, then there is no reason to use correlation when simulating. Many assets are correlated, mostly through volatility clustering and sentiment. Secular bull and bear markets can ...


0

Thank you for your answer @MarkJoshi. I followed you advice and achieved in deriving the approximation formula. However, I can not fully understand why the fact that Black's formula is linear in $\sigma$ for ATM strikes causes the Rebonato approximation only to be accurate for ATM strikes and not OTM and ITM strikes. I would be grateful if somebody can ...


4

Surely what is meant is that the 100 components are pairwise correlated but the 1000 draws are independent.


1

Two comments: Normal returns should always be in $[-1,+\infty)$. I believe that the way you sample $R_i$ from Stable directly violates that. You might want to sample $\log (1+R_i)$ from Stable instead. The question is very poorly worded. For the sampling distribution of a percentile you can invoke order statistics. It will follow a transformation of Beta ...


0

Your approach is sensible for a single variate case if the cdf is available, but does (as your friend said) break down for more variates. One issue with multivariate case is the "curse of dimensionality" - as the number of variates increases your number of samples will get infeasibly large very quickly. To address this, one can use a low discrepancy ...


0

That all looks correct to me. It might be a bit more natural to convert the one day returns into prices and then compute the five day returns from those, but it's of course equivalent. For your final question, you are generating a sequence of random variables (the quantiles) and want to know how good your estimate of the mean is. A practical choice would ...


2

What you describe is a very simple quasi monte carlo, where the 'random' points are equally spaced in probability space. Like numerical integration. Sometimes you can use it, but in general you will need the cumulative distribution to do percentile mapping. This very frequently is not known in closed form, and can be very expensive to compute numerically. ...


1

In order to define option price we should follow Black Scholes construction to construct riskless portfolio at t then to state that instantaneous rate of return of this portfolio equal risk free rate r ( t ) where r is a random on [ t , t + dt ] interval. We actually then arrive at the problem which could not be embedded in BS pricing world.


3

it certainly works best at the money. Why? I think it comes from the fact that Black's formula is approximately linear at the money. The approximation $$ \frac{1}{\sqrt{2\pi}} \operatorname{SR} \sigma \sqrt{T} A, $$ with $A$ the annuity is remarkably good. One way of deducing these formulas is to do an asymptotic/Taylor expansion about $\sigma=0.$


0

Andersen--Broadie converts an exercise strategy into an upper bound. The better the exercise strategy the better the upper bound. You can get the exercise strategy by using regression to approximate the continuation value and this is pretty standard -- the LS Method is widely used but does have defects. Once you have an exercise strategy you need the value ...



Top 50 recent answers are included