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Moments and Moment-gen. function.

  1. Oct 23, 2011 #1
    Hi people!

    Im having problems with my Prob.Theor. assignment=(
    I was hoping that u might be able to help me...

    I have 2 problems that i ve no idea how to solve!Oo

    Heres 1st one
    We re given rand. var. X, its mean value U, the stand deviation S (sigma).
    We need to show that E(z)=0 and var(z)=1
    if the relation between X and Z is this eq. Z=X-U/S

    Show if a rand. var. has the prob. density
    f(x)=1/2*Exp[-lxl] -inf<X<inf
    lxl-abs value
    then its moment gen func. is

    im not sure about this one bu heres what i got
    we re using the formula from the definition
    and gettin this
    1/2(Int[Exp[tx]*Exp[-lxl]) -inf<X<inf

    but lxl=+-x

    then we get 2 integrals
    both in -inf<X<inf

    now if we integrate it we get
    both in -inf<X<inf

    whats next?=(

    Would appreciate any help!=(
  2. jcsd
  3. Oct 23, 2011 #2


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    Science Advisor

    For your first problem z=(x-U)/S. Direct calculation will give you the results.

    For your second problem, the ranges of integration of the split integrals are (-∞,0) for the +x integral and (0,∞) for the -x integral. Plug in x=0 and the resulting terms (be careful of signs) add to get the answer.
  4. Oct 23, 2011 #3
    Ok, i think i got the second one.., but what do i do with 1/2 in front of 1/1+t and 1/1-t when i multiply them?

    Now the to first one. You mean i need to integrate it? Like this: Int[Z*(X-U)/S], but what range do i pick???
    And then var(Z) i can find from S^2=U'2-U^2
    which is S=Sqrt[U'2-U^2]
    where U^2 is E(Z)
    and U'2=Int[(Z^2)*(X-U)/S]
    but again what range???
    Thank you!!!
  5. Oct 24, 2011 #4


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    Science Advisor

    1/(1-t) + 1/(1+t) = 2/(1-t2). So multiply by 1/2 to get your answer.

    The range for the integration is the entire real line.
    A simpler approach is as follows:
    E[(X-U)/S] = {E(X) - U}/S = 0
    E[({X-U)/S}2] = {E(X2) -2UE(X) +U2}/S2 = 1
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