# Tag Info

17

Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from $0$ and hitting $x$ before time $t$, or its first passage time $\tau_x$ being less than $t$. This can be derived through the reflection principle. The paths crossing $x$ are exactly paired up by the ...

10

If you use a risk-neutral pricing model and consider the probability there, then you get the probability with respect to a risk neutral measure, in addition that probability depends on the chosen numeraire. For example, in Black-Scholes model taking the risk-neutral measure with respect to the bank account $B$ gives $$P(S(T)<K) = Q^{B}(S(T)<K) = \Phi(... 10 In the derivatives context, "arbitrage free" means almost surely for the probability measure under consideration. This is in opposition with statistical arbitrage used at high frequencies for example. More precisely the assumption is that there is no T\geq 0 and self-financed portfolio V such that V_0 = 0, P(V_T < 0) = 0 and P(V_T > 0) > ... 10 I'll outline how you can estimate the (implied) real-world density function from (observed) option prices. Having found this real-world density, you can then compute all sorts of probabilities and quantify the market's expectation of future prices. Recall firstly that (European-style) options are priced as risk-neutral expectation of the discounted payoff. ... 9 If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list. Pricing. For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ... 9 The way that I understand your question is that you are looking to fit the market prices of European plain vanilla options of a single maturity and then back out the corresponding implied probability density function. There are multiple ways that you could approach your problem. 1) Modelling the Market Prices The market prices of European plain vanilla ... 8 In this scenario, the "joint dynamics" are trivially computed since the option value is a known deterministic function of stock price. For example, the mean of the option value for time \tau is$$ \mu_O = \int_0^\infty BSM( S_\tau ) p(S_\tau) dS_\tau $$which is best computed using quadrature as available in standard numerical libraries like scipy. The ... 8 \mathbb{P} is the true probability measure. Measure \mathbb{Q} is a measure of convenience that allows risk neutral pricing. Stochastic discount factor M takes you between the two. If you care about prices you can either: (1) work under \mathbb{Q} or (2) work under \mathbb{P} with a stochastic discount factor M. There's an isomorphic ... 8 In order to estimate a probability of an event with small probability, you might want to try to estimate the probability for a changed random variable that allocates a larger probability mass for the event happening. So, in your case, you might want to change the original N(0, 1) to N(100, 1) because for the second r.v. the probability of it being higher ... 7 What you refer to as the 99.5th percentile is known as the "Value-at-Risk." You are correct that you will need to make a distributional assumption, and there is a popular and well-researched approach to this problem, though I'm not certain it could be called "standard." I would recommend you use the "truncated Levy flight" distribution. James Xiong at ... 7 This leads to the same result as Alexeys answer. However, my reasoning is different.$$ E[F_X(X+a)]=\int_{-\infty}^{\infty} F_X(x+a) f_X(x)dx=\int_{-\infty}^{\infty} \int_{-\infty}^{x+a}f_X(y)dy f_X(x)dx=\\ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 1_{(-\infty,x+a]}(y) f_X(y) f_X(x)dydx= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 1_{_{\{y-x\le a\}...

7

In this case, the t-statistic is used to determine if the returns are statistically different from zero (the theoretical mean). A small t-statistic would imply that the null hypothesis (no significant excess return) cannot be rejected. Newey-West standard errors are used to correct for the correlations of error terms over time. I have written a Matlab ...

7

you should have a look at implied probability densities. They do exactly what you are asking - extracting the pricing density from option prices. This is done by differentiating the option price with respect to the call. Here are two links. The first one explains the procedure the second one deals with where such densities can be applied Implied state ...

7

I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$\int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2}$$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\... 7 Let \tau = T-t. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where Z is a standard normal random variable, independent of \mathcal{F}_t. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, ... 7 Firstly, m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T. So, you can either consider the running maximum or minimum. Let \tau be a stopping time and (B_t) a Brownian motion. Then, \begin{align*} W_t =\begin{cases} B_t & t\leq \tau, \\ 2B_\tau - B_t & t\geq \tau, \end{cases} ... 6 You can point out to your friend that, statistically speaking, having more observations reduces uncertainty in estimators. Mathematically, SE_\bar{x}\ = \frac{s}{\sqrt{n}}, showing that the standard error of a statistical estimator decreases with increased observations. This argument is concise and consistent with the Taleb quote. From wikipedia on ... 6 Its a simple expected value question: Probability of throwing a 6 is 1/6 Probability of not throwing a 6 is 5/6 thus expected pay off per roll: 10 dollars * 1/6 + (-1 dollar) * 5/6 = 5/6 dollars Edit: And several of your above assumptions are plain wrong: "Probability tells me that every 6 throws I get one 6 and 5 different numbers." -> That is NOT what "... 6 Not sure about all of the complicated math and programming above, but I can tell you that, if you want to calculate for 1 Standard Deviation from the current stock price X days away, the following calculation will give you a +/- value from the current stock price. 1 StdDev Move = (Stock Price X Implied Volatility X the Square Root of 'how many days') all ... 6 One approach is to take the entire option chain, and calculate the prices for adjacent butterflies along the chain. The risk / reward of each of the butterflies represent the empirical probability that the market is pricing for the underlying to move between the strikes of the butterfly. To make sure it is a proper probability distribution, you will want ... 6 Since W_{2t}-W_{t} is independent of W_t and has the same law as W_{2t-t}=W_t we only have to computeP(X(X+Y)<0)where (X,Y) follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either X<0 and X+Y>0 or the opposite. The two events have the same probability since (-... 6 Consider two jointly normal random variables X_1 \sim N(u_1, \sigma_1^2) and X_2 \sim N(u_2, \sigma_2^2). Note that, \begin{align*} \max(X_1, X_2) = X_2 + \max(X_1-X_2, \ 0). \end{align*} Moreover, X_1-X_2 is a normal random variable with mean \mu=\mu_1-\mu_2 and variance \begin{align*} \sigma^2 = \sigma_1^2+\sigma_2^2 - 2 \rho\sigma_1\sigma_2, \end{... 6 IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure \mathbb{Q} equivalent to \mathbb{P} and under which X_t is a martingale. Just take\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$for any a \in \... 6 Quantiles are preserved under monotonic transformations, hence the quantile for Y is simply the exponential of the quantile of X, no need for corrections whatsoever (see here for instance). Put otherwise, let q denote the quantile \alpha of X i.e.$$\Bbb{P}(X \leq q) = \alpha then \begin{align} \Bbb{P}( X \leq q ) &= \Bbb{P}( \underbrace{\...

6

The question requires you to provide a method which uses uniform random variables and transforms them to generate realizations of the described asset values. To give a bit more general answer: this is solved by the inverse transform sampling method. The main idea is to obtain realizations of a random variable $x$ with any given distribution function $F(x)$, ...

5

I'll throw this in as an "application of RMT" ... EDHEC and FTSE use RMT to decide the optimal number of principal components in their covariance estimation procedure for which they use PCA (Principal Component Analysis). For details look here or here in Appendix C section 4 for details.

5

If you are happy with the OptionStrategist probability values, why don't just do it exactly how they do it: function Covered() { form=document.callreturn; p=form.price.value; q=form.strike.value; t=form.days.value/365; v=form.volatility.value/100; vt=v*Math.sqrt(t); lnpq=Math.log(q/p); d1=lnpq / vt; y=Math.floor(1/(1+.2316419*Math.abs(d1))*100000)/...

5

This might be a surprise to you, you can evaluate the option using Black Scholes. The key concept is change your numéraire from dollar to the asset associated with $V$. The $V$ in your payout $\max(U_t-V_t,0)$ will effectively get replaced by a constant, the par forward of asset $V$ at maturity $t$. Since $U_t$ and $V_t$ are independent, you can ...

5

When I run this simulation I see the same results, and it makes sense. For the straight 50%/50%, I found that my win ration was about 38% and my loss ratio 61%. The reason it wasn't 50/50 was that if I had consecutive up flips my value could keep going up, but if I had consecutive down flips I would 0 out and the sequence would have to end as I had lost ...

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