Table of Contents
- 1 Is a martingale a local martingale?
- 2 Is a local martingale integrable?
- 3 What is martingale property?
- 4 What is exponential martingale?
- 5 What is stopping time in stochastic process?
- 6 How do you show a stochastic process is a martingale?
- 7 Are martingales independent?
- 8 What is the difference between a local martingale and a submartingale?
- 9 Is Brownian motion a martingale or a constant?
Is a martingale a local martingale?
Every martingale is a local martingale; every bounded local martingale is a martingale; in particular, every local martingale that is bounded from below is a supermartingale, and every local martingale that is bounded from above is a submartingale; however, in general a local martingale is not a martingale, because its …
Is a local martingale integrable?
There do exist integrals, with respect to local martingales, which are not locally integrable and, hence, cannot themselves be local martingales.
Is Brownian motion local martingale?
In particular, a (continuous) local martingale is a deterministic time change of Brownian motion if and only if its quadratic variation is a deterministic function (and absolutely continuous with respect to Lebesgue measure).
What is martingale property?
The Martingale property states that the future expectation of a stochastic process is equal to the current value, given all known information about the prior events. Both of these properties are extremely important in modeling asset price movements.
What is exponential martingale?
Exponential Martingales. In what follows, (Ω,F,P) is the canonical sample space of the Brownian motion (Bt)t≥0 with B0 = 0; other notation is that used in class. Given H ∈ L2. loc let M denote the associated local martingale: (1)
How do I show a process in martingale?
Note that to check a process is a martingale, it suffices to check property (iii) (which is usually called “the martingale property”) since if it holds, then the condi- tional expectation makes sense, so (ii) holds, and since the conditional expectation is measurable with respect to the σ-field being conditioned on, it …
What is stopping time in stochastic process?
In probability theory, in particular in the study of stochastic processes, a stopping time (also Markov time, Markov moment, optional stopping time or optional time) is a specific type of “random time”: a random variable whose value is interpreted as the time at which a given stochastic process exhibits a certain …
How do you show a stochastic process is a martingale?
Formally, a stochastic process as above is a martingale if E[Xt+1|ℱt] = Xt. Often we replace ℱt with the σ-algebra generated by X0… Xt and write this as E[Xt+1|X0… Xt] = Xt.
Is geometric Brownian motion a martingale?
When the drift parameter is 0, geometric Brownian motion is a martingale. If , geometric Brownian motion is a martingale with respect to the underlying Brownian motion . This is the simplest proof.
Are martingales independent?
Martingales as sums of uncorrelated random variables. , where E[Δi|Δ1…Δi-1] = E[Xi-Xi-1|ℱi] = 0. In other words, Δi is uncorrelated with Δ1…Δi-1. This is not quite as good a condition as independence, but is good enough that martingales often act very much like sums of independent random variables.
What is the difference between a local martingale and a submartingale?
Every martingale is a local martingale; every bounded local martingale is a martingale; in particular, every local martingale that is bounded from below is a supermartingale, and every local martingale that is bounded from above is a submartingale; however, in general a local martingale is not a martingale,…
Is driftless diffusion a martingale?
In particular, a driftless diffusion process is a local martingale, but not necessarily a martingale. Local martingales are essential in stochastic analysis (see Itō calculus, semimartingale, and Girsanov theorem ).
Is Brownian motion a martingale or a constant?
For the times before 1 it is a martingale since a stopped Brownian motion is. After the instant 1 it is constant. It remains to check it at the instant 1. By the bounded convergence theorem the expectation at 1 is the limit of the expectation at (n-1)/n (as n tends to infinity), and the latter does not depend on n.