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

  • Thread starter Gregg
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Homework Statement



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

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

Homework 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 ##

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?
 
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