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Expectation of Normal Distribution

  1. Dec 2, 2007 #1
    Let Y = a + bZ + cZ2 where Z (0,1) is a standard normal random variable.
    (i) Compute E[Y], E[Z], E[YZ], E[Y^2] and E[Z^2].
    HINT: You will need to determine E[Z^r], r = 1, 2, 3, 4. When r = 1, 2 you should
    use known results. Integration by parts will help when r = 3, 4.



    I am struggling with the part of the question involving E[Z^3] and E[Z^4]. Clearly E[Z]=0 and E[Z^2]=1 but I do not no where to proceed when computing higher powers of Z. Any help would be greatly appreciated.

    Thanks
     
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2

    EnumaElish

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    Expected value of a function g of z is E[g(z)] = [itex]\int_{-\infty}^{+\infty}g(z)f(z)dz[/itex] where f is the pdf of z.
     
  4. Dec 2, 2007 #3
    So am I right in thinking I let g(z) = z^3 and f(z)= 1/(sigma * sqrt2pi) .... and then use integration by parts?
     
  5. Dec 2, 2007 #4

    EnumaElish

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    That should be it.
     
  6. Dec 2, 2007 #5
    I'm not really getting anywhere with the integration by parts. Is there any hint you could offer or show me a step I could be missing out?
     
  7. Dec 2, 2007 #6

    EnumaElish

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    Can you post what you have so far?
     
  8. Dec 2, 2007 #7
    Ok so,

    INT u dv = uv - INT v du

    Then i let u = Z^3 and dv = the pdf of the normal dist.

    Im lost when it comes to integrating the pdf and i'm not sure how it will condense down to a manageable solution. Sorry about the lack of symbols, i'm not sure how to properly upload images yet.
     

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  9. Dec 2, 2007 #8

    EnumaElish

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    If dv = the pdf of the normal dist., then what is v?

    (If dv = f(z) and v = F(z), what is F?)
     
  10. Dec 2, 2007 #9
    F(z) is the cdf of the normal distribution.

    So E(Z^3) = Z^3 * F(Z) - INT 3Z^2 * F(Z).

    Do i then need to apply integration by parts again? I'm sorry if i'm asking too many questions I just really don't understand what i'm trying to achieve as a solution.
     
  11. Dec 2, 2007 #10
    Or more accurately, what form the solution will take i.e an integer, function of Z...
     
  12. Dec 3, 2007 #11

    EnumaElish

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    u = Z^3 and dv = f(z)dz ===> v = F(z)

    E(Z^3) = INT u dv = [uv] - INT v du = [Z^3 * F(Z)] - INT F(Z) 3Z^2 dZ, where each of the [uv] expression and the last INT are evaluated over (-infinity, +infinity).

    E.g., [uv] = u(+infinity)v(+infinity) - u(-infinity)v(-infinity) ... but (infinity)^3 is not a real number, so something's amiss. See http://en.wikipedia.org/wiki/Integration_by_parts
     
  13. Dec 3, 2007 #12

    EnumaElish

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