Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Stochastic Calculus

  1. Jan 27, 2005 #1
    Hello all

    If you throw a head I give you $1. If you throw a tail you give me $1. If [tex] R_i [/tex] is the random amount ($1 or -$1) you make on the [tex] ith [/tex] toss then why is: [tex] E[R_i] = 0, E[R^2_i]=1, E[R_iR_j] = 0 [/tex]? If [tex] S_i = \sum^i_{j=1} R_j [/tex] which represents the total amount of money you have won up to and including the ith toss, then why does [tex] E[S_i] = 0, E[S_i^2] = E[R_1^2 + 2R_1R_2 + ...] = i [/tex]? I know that if there had been five tosses already then [tex] E[S_6|R_1,...,R_5] = S_5 [/tex]

    Any help is appreciated

    Thanks
     
  2. jcsd
  3. Jan 27, 2005 #2
    I know that Expected Value is defined as: [tex] E[X] = \sum_x f(x)P(x) [/tex] for a discrete variable.
     
  4. Jan 27, 2005 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right, and Ri is what is called (in my probability theory class I'm taking anyways) a Bernoulli Random Variable. There are only two possible outcomes: success, or failure, usually denoted by 1 or 0 respectively. However, since Ri represents the winnings, and they are -1 for tails, the expectation, given your definition is (letting H denote a heads and T a tails):

    [tex] E[X] = \sum_x xp(x) = 1\cdot P(H) -1 \cdot P(T) = 1/2 - 1/2 = 0 [/tex]

    The expectation can be interpreted as the average value (ie the value expected over time) that the random variable takes on. For a random variable that can only take on the values 1 and -1, what is the average? Since both outcomes have equal weighting (p = 1-p = 1/2), on average you'd expect no net winnings, since the chances of getting a dollar are about equal to that of having to pay up. I hope this gets you started. Now consider the other cases: what is Ri squared? Well, what is 1^2? What is (-1)^2?
    ETC...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook