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The Q-function is defined by : $$Q(x) =\frac{1}{\sqrt{2\pi}} \int_{x}^{\infty}\exp(-\frac{u^2}{2}) \ \mathrm{d}u \ \ (1).$$

According to the wiki page there is an alternative form of the Q-function based on John W. Craig's work that is more useful is expressed as: $$Q(x) =\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\exp\left(-\frac{x^2}{2\sin^2(\theta)}\right) \ \mathrm{d}\theta \ \ (2).$$

Craig's proove is based on probabilistic approach, there for I look for an analytic one.
any help will be appreciated.


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Thanks @DavideGiraudo for fixing it. – jack May 27 '12 at 20:50
here a way to solve this – user33125 Jun 7 '12 at 0:46

1 Answer 1

up vote 4 down vote accepted

Since both expressions coincide at $x=0$, it suffices to show that their derivatives coincide on $x\geqslant0$. Since the derivative of the first expression of $Q(x)$ is proportional to $\mathrm e^{-x^2/2}$, this happens if $R(x)$ is constant on $x\geqslant0$, where $$ R(x)=\mathrm e^{x^2/2}\int_0^{\pi/2}\frac{x}{\sin^2\theta}\mathrm e^{-x^2/(2\sin^2\theta)}\mathrm d\theta=\int_0^{\pi/2}\frac{x}{\sin^2\theta}\mathrm e^{-x^2\cot^2\theta/2}\mathrm d\theta. $$ The change of variables $v=x\cot\theta$ yields the range $v\gt0$ and the Jacobian $\mathrm dv=x\mathrm d\theta/\sin^2\theta$, hence $$ R(x)=\int_0^{+\infty}\mathrm e^{-v^2/2}\mathrm dv, $$ which does not depend on $x$. QED.

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I am not sure that Craig's proof is based on a probabilistic approach. – Did May 27 '12 at 21:19
Well Done @Didier, About the Craig's proof I refered to this [pdf document][1]. When I searching I found a book called : "Probability Distributions Involving Gaussian Random Variables" state that : <<The form in (l) is not readily obtainable by a change of variables directly in (2). However, by first extending (1) to two dimensions (x and y) where one of the dimensions (y) is integrated over the half plane, a change of variables from rectangular to polar coordinates readily produces (2).>> pg.123. If this is clear for you, how can be done? [1] – jack May 27 '12 at 22:29
Reference [1] is already in your post. As regards the other citation, its meaning is not clear to me at the moment. – Did May 28 '12 at 0:25
Alright then, thanks anyway :) – jack May 28 '12 at 0:41

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