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Varaiance of Normal Distribution

  1. Nov 1, 2006 #1
    When solving for the variance of the normal distribution one needs to evaluate the following integral:

    INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

    I proceed using integration by parts:

    [-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) 2*e(-x^2/2)dx]

    However apparently correct answer is:

    [-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) e(-x^2/2)dx]

    but I don't see how the 2 cancels?
     
    Last edited: Nov 1, 2006
  2. jcsd
  3. Nov 1, 2006 #2

    mathman

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    The integral of xe^(-x^2)/2) is xe^(-x^2)/2), while the derivative of x is 1. I don't see where you got the 2 from.
     
  4. Nov 1, 2006 #3
    Hi, thanks:

    First I'll take you through what I did before I follow your solution through fully:

    When I tried to solve the following integral:

    INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

    To solve using integration by parts: INT uv` = uv - INT u'v
    I take u=x^2
    u`=2*x

    v`=e^(-x^2/2)
    v =-(e^(-x^2/2))/x

    hence

    INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

    = x^2*-(e^(-x^2/2))/x|(inf -inf) - INT(inf - inf) (2*x)*-(e^(-x^2/2))/x

    = x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (2)*(e^(-x^2/2))

    clearly first term equates to 0

    INT|(inf -inf) (2)*(e^(-x^2/2))
    =2*INT|(inf -inf) (e^(-x^2/2))

    I know that INT|(inf -inf) (e^(-x^2/2)) = SQRT(2pi).
    so I get final answer of 2*SQRT(2pi) but the correct answer is SQRT(2pi)






    It seems from the above you wish to take:

    u=x
    u`=1

    v`=x.e^(-x^2/2)
    v =-e^(-x^2/2)

    correct?

    In which case:

    INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

    = x*-(e^(-x^2/2))|(inf -inf) - INT(inf - inf) (1)*-(e^(-x^2/2))

    = x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (1)*(e^(-x^2/2))

    Which seems to give the correct answer of SQRT(2pi)

    However I cannot find any error in my working above if you follow it through the answer still seems to be 2*SQRT(2pi)?
     
    Last edited: Nov 2, 2006
  5. Nov 2, 2006 #4

    quasar987

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  6. Nov 2, 2006 #5
    Hi Quasar,

    Thanks for that it confirms that using u=x and dv/dx= x*exp(-x^2/2)dx in the integration by parts it gives the correct answer. Which I agree with.

    However my remaining question is that how come when I work through using u=x^2 and dv/dx= exp(-x^2/2)dx in the integration by parts the calculation out by a factor of 2?

    I just would like to know in case I face similar problems in future. To me it seems more "natural" to choose u=x^2 and I guess it must yield the same answer if worked through correctly. My question is where am I going worng at present?

    I take u=x^2
    u`=2*x

    v`=e^(-x^2/2)
    v =-(e^(-x^2/2))/x

    hence

    INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

    = x^2*-(e^(-x^2/2))/x|(inf -inf) - INT(inf - inf) (2*x)*-(e^(-x^2/2))/x

    = x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (2)*(e^(-x^2/2))

    clearly first term equates to 0

    INT|(inf -inf) (2)*(e^(-x^2/2))
    =2*INT|(inf -inf) (e^(-x^2/2))

    I know that INT|(inf -inf) (e^(-x^2/2)) = SQRT(2pi).
    so I get final answer of 2*SQRT(2pi) but the correct answer is SQRT(2pi)
     
  7. Nov 2, 2006 #6

    HallsofIvy

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    This is incorrect as you can see by differentiating your v. [tex]e^{-x^2/2}[/itex] does not have an elementary anti-derivative.

     
  8. Nov 2, 2006 #7

    mathman

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    wrong (take the derivative of your v and you won't get your v')!
    you should use
    v'=xe^(-x^2/2)
    v=-e^(-x^2/2)
     
    Last edited: Nov 2, 2006
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