New answers tagged

6

I think you're confusing the Lebesgue integral with Itô calculus. They are related concepts. I'll explain. Lebesgue vs Riemann The simplest explanation of the difference between Lebesgue and Riemann integration - that I know of - follows. Imagine a bunch of bank notes tossed on a carpet. Riemann would count the money by first drawing a rectangular grid on a ...


1

Because a bootstrap by definition involves drawing a random sample (with replacement) which has the same sample size as the original dataset, i.e., $n$ objects. On average, when randomly drawing $n$ objects with replacement from a sample of $n$ objects, the probability that an object will not be selected is $(1-1/n)^n=\exp(-1)=0.368$. Thus, the probability ...


1

A GARCH process with constant conditional mean would be an example of a martingale process that is not a random walk*: \begin{aligned} y_t &= \mu+\varepsilon_t, \\ \varepsilon_t &= \sigma_t z_t, \\ \sigma_t^2 &= \omega+\alpha_1\varepsilon_{t-1}^2+\beta_1\sigma_{t-1}^2, \\ z_t &\sim i.i.D(0,1) \end{aligned} where $D$ is some distribution with ...


1

A stopping time $T$ for a sequence $\{X_n\}$ is a measurable random variable for which the event $$ \{ \omega : T(\omega) \leq n \} $$ is $\mathcal F_n$-measurable where $$ \mathcal F_n = \sigma \left(X_1,\dots,X_n \right) $$ If $M$ and $N$ are stopping times for $\{X_n \}$ then for $n \geq 0$, ($\wedge$ = min, $\vee$ = max): \begin{align*} \left \{ \omega : ...


0

The example is correct. There are two sets to consider: The set of values that each $X_n$ can take The index set (i.e. the values that $n$ can take) The state-space is the first one. In this example, this is the positive real line, so it is a "continuous-state" process. The fact that $X_n$ models a "time" concept (waiting time) is ...


Top 50 recent answers are included