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Calculate expected value and variance of d, d = sqrt(x^2+y^2)

  1. Jul 14, 2011 #1
    I'm bad at stochastics so really glad for any help

    1. The problem statement, all variables and given/known data

    I have two normally distributed NON INDEPENDENT stochastic variables X~N(muX,sigX^2) and Y~N(muY,sigY^2)
    A third variable D is defined as D = sqrt(X^2 + Y^2).
    Since Y and X are stochastic D will also be stochastic.

    2. Relevant equations

    But how to calculate expected value muD and variance sigD^2 properly?

    3. The attempt at a solution

    Calculate the variance sigD^2
    D = sqrt(X^2 + Y^2) (1)
    D^2 = X^2 + Y^2 (2)
    E[D^2] = E[X^2 + Y^2] = E[X^2] + E[Y^2] (3)
    sigD^2 = E[(D - muD)^2] = E[D^2] - muD^2 (4)
    Using (3) in (4)
    sigD^2 = E[X^2] + E[Y^2] - muD^2 (5)
    sigD^2 = E[X]^2 + Var[X] + E[Y]^2 + Var[Y] - muD^2 (5)
    sigD^2 = muX^2 + sigX^2 + muY^2 +sigY^2 - muD^2 (6)
    So far so good.
    Calculate muD
    At this point I thought I'm done.
    But what about muD? I dont have it!
    At first tempting to assume
    muD = sqrt(muX^2+muY^2)
    But I dont think it's true, since X and Y are not independent. And even if it is true, how to show it.
    If I start out
    muD = E[D] = E[sqrt(X^2 + Y^2)]
    I dont manage to come to a solution.

    Really appreciate any help
     
  2. jcsd
  3. Jul 15, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    I think that finding Ed (d = sqrt(X^2 + Y^2) ) is possible only numerically. Your hoped-for result Ed = sqrt(muX^2 + muY^2) is false: it is an example of the so-called "fallacy of averages". In (x,y) space the function f(x,y) = sqrt(x^2 + y^2) is *convex*, so E f(X,Y) >= f(EX,EY), and for a spread out distribution like yours (which straddles the origin) the inequality will be strict. Google "Jensens inequality" for more on this.

    RGV
     
    Last edited: Jul 15, 2011
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