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Cardinality of Sample Space for Brownian Motion

  1. Jan 29, 2012 #1
    Hello,

    Given a Brownian Motion process B(t) for 0≤t≤T,
    we can write it more explicitly as B(t,ω) where ω[itex]\in[/itex]Ω,
    where Ω is the underlying sample space.

    My question is: what is the cardinality of Ω. I.e. what is |Ω|?

    My thoughts are that it is an uncountable set, based on the observation
    that B(t) ~ N(0,t), and thus takes values in the real numbers.

    Am I correct, is this simple observation enough, or is a more rigorous
    proof needed?
     
  2. jcsd
  3. Jan 29, 2012 #2
    It depends on whether you want to consider the space in terms of points or trajectories. In a Euclidean space, you need only consider the points in what could be considered a vector space with every point being described by a tuple. You would consider the particle as "jumping" from point to point. The history of the particle's motion would be the ordered progression of tuples. In this case you are dealing with a countable set.

    If you want to consider it terms of a diffusion model based on a normal distribution around the point of origin, the space could be considered as an uncountable set or continuum.
     
    Last edited: Jan 29, 2012
  4. Jan 29, 2012 #3
    Thanks for the response SW VandeCarr.
    I'm not sure if I understand what "countable set" you mean. Certainly the ordered list of tuples would be countable. But I'm interested in Ω.

    Even if we are dealing with an ordered list of tuples (let's assume we are dealing with one dimension), wouldn't each element of this list be chosen out of ℝ? I.e. if we choose the n-th tuple with the random variable X:Ω→ℝ then we would need an ω[itex]\in[/itex]Ω for every possible x[itex]\in[/itex]ℝ, and thus |Ω|=|ℝ|.

    Thanks again.
     
  5. Jan 29, 2012 #4

    lavinia

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    The set of sample paths in a Brownian motion is uncountable. B(t) is the height of the path at time t. It is normally distributed. So there must be uncountably many different paths at time,t
     
  6. Jan 29, 2012 #5
    Ok, thanks lavinia. Makes sense.

    The reason I ask is because I have only dealt with stochastic processes in discrete time, in the case of stock prices. For example, we may have a stock St where t=0,1,2, that can go up or down between time 0 and 1, and again up or down between time 1 and 2. Thus there would be four possible outcomes (up up, up down, down up, down down) corresponding to ω1, ω2, ω3 and ω4. So it is easy to visualize how the "state of the world" (ω) corresponds to the value of S2.

    I have been trying to get an analogous understanding in the continuous case for Brownian motion. So I guess it works exactly the same, except there are now an uncountable number of different "states of the world" corresponding to the terminal value B(T) . But even further, for a given B(T) there would be an uncountable number of "states of the world" corresponding to the different paths that could potentially get there.
     
  7. Jan 30, 2012 #6

    lavinia

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    That is correct.

    BTW: One might be able to think of actual stock prices, prices where the stock actually trades, as samples from an underlying continuous process. The continuous Brownian motion is always taking place but only observed when it materializes in a trade price. The advantage of this point of view is that you do not need to have a fixed time increment to describe the successive prices. In a discrete model it seems that you would need to assume that the stock trades at specific points in time.

    Such a model could also allow for sudden prices changes e.g. reactions to unexpected economic reports, as a sample from an outlier continuous path in the tail of the normal distribution.

    So what would the stochastic process be? One starts with a Brownian motion then samples it at random times?

    The question I would ask is whether the evolution of the underlying sigma algebra of the stock price process is adequately modeled by the Brownian motion sigma algebra which is generated entirely by paths of prices.
     
    Last edited: Jan 30, 2012
  8. Feb 1, 2012 #7
    Sorry I didn't back to you sooner. Yes, I agree with you and Lavinia that [itex]\Omega[/itex] is an uncountable set and I referred to the diffusion model myself in my previous post. My only point was that in two or more dimensions, the physical model of Brownian motion is imagined in terms of directed line segments joined at (countable) singular points. I guess one could argue that the space itself is continuous, if that's what you mean. However, wouldn't that argument imply that discrete probability distributions, such as the binomial distribution, are in some sense continuous since each of n points maps to R in terms of its probability?
     
    Last edited: Feb 1, 2012
  9. Feb 2, 2012 #8
    Well I understand that the standard stock price model is given by the stochastic differential equation dSt = μStdt + σStdWt with solution St = S0exp(Wt + (-0.5σ2)t). So this would be a stochastic processes, depending on Brownian motion, and I suppose the times that investors needed access to prices could be considered random. Is this what you had in mind? However this does not capture the jumps due to sudden events etc. I don't know how to incorporate that into the above.

    I'm going to guess that the information revealed over time by the Brownian motion is the same information needed to model the stock price (since the Brownian motion is the only random part of the stock price), but I don't know enough to be sure I understand what you're asking.

    Ok, such as if you were simulating a Brownian path numerically on a computer.

    I don't think so. It is part of the definition of Brownian Motion that it depends continuously on its parameter.
     
  10. Feb 2, 2012 #9

    lavinia

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    What i meant was that a continuous Brownian motion can never be observed in actual prices. So if you believed that such a process was correct then actual stock prices are a random time sampling of this process. So the observed process is not Brownian motion. I am sure what it is.

    Outlier events have some probability in Brownian motion. So what appears to be a jump might actually be the discrete time observation of a continuous path that happens to moved rapidly away from the current price.

    Actual stock returns do not appear to follow a Brownian motion. Much statistical evidence suggests that outlier events occur too frequently. the distribution has fat tails.

    Also it is possible that the volatility is not constant but actually stochastic. There is a large body of statistical research on that question as well.

    The usual model that you wrote down is useful for the pricing of options and is not at all considered to be a true model especially over medium to longer time frames. First of all, the riskless rate will change significantly over long time periods and secondly it is not at all clear that the drift rate should be constant except over short time periods.

    In an efficient market,all of the information relevant to the stock price is incorporated into the price. In a Brownian motion the only information is the history of the stock price. I just said that this may not actually model the information set.
     
  11. Feb 4, 2012 #10
    Thanks lavinia, you've given me a bit to think/read about.
     
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