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A limit in probability (possibly dominated or monotone convergence theorem)

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data

    ## \sqrt{2\pi} a \exp({a^2 \over 2} ) P( \xi \geq a) \to 1 ##

    ## \xi \text{ ~ } N(0,1) ##

    2. Relevant equations


    This implies

    ##\sqrt{2\pi} a \exp({a^2 \over 2} ) \int_a^\infty a \exp({a^2\over 2}) \exp (-{x^2 \over2}) dx \to 1 ##

    3. The attempt at a solution

    The integral doesn't seem to converge to anything obvious to me. I think that as a grows,the integrand grows large but the interval in which we integrate over is getting smaller.

    Try and use a substitution.

    ##x^2 = a^2 + 2y ## and do some rearranging.

    ## \int_0^\infty y \frac{\exp({-y\over 2})}{\sqrt{1+2y/a^2}} dy ##

    Monotone convergence theorem

    we have that ## y \frac{\exp({-y\over 2})}{\sqrt{1+2y/a^2}} \nearrow y \exp({-y\over 2}) ##

    So

    ## \mu (f_a) \nearrow \mu(f) < \infty ##

    ##\mu(f) = 1 ##

    Is this right? Do I need to show that ##\mu(f_n) \nearrow \mu(f) ## ? also the subscript n doesn't appear in the integrand, I couldn't use a could I, and a seems to be continuous and so it isn't a sequence of functions?
     
    Last edited: May 11, 2012
  2. jcsd
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