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

[Gregg]

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?

 

[mighty2000]

Your approach is correct. You can use the Monotone Convergence Theorem to show that the integral converges to a finite value as $a \to \infty$. You can also use the Dominated Convergence Theorem, since the integrand is bounded by $y e^{-y/2}$, which is integrable on the given interval.

You are correct that the subscript $n$ is not needed in the integrand, since it is a continuous function of $a$.