# Half-normal distribution

1. Sep 4, 2011

### Zoran

Hi: I am reading an article that deals with the distribution function associated with the half-normal distribution. The author presents a formula for the c.d.f. as:

$$\left [ G\left ( x \right ) \right ]^{2r-2}=\left ( \frac{2}{\pi } \right )^{r-1}\left \{ \int_{0}^{x}\exp \left [ -\frac{1}{2} \left (w ^{2}\right )\right ]dw \right \}^{2r-2}$$

Note that r=2,3,...

The expression above is also equal to:

$$\left [ G\left ( x \right ) \right ]^{2r-2}=\left ( \frac{2}{\pi } \right )^{r-1}\left \{ \int_{0}^{x}\int_{0}^{x} \exp \left [- \frac{1}{2}\left ( x_{1}^{2}+x_{2}^{2} \right ) \right ]dx_{1}dx_{2}\right \}^{r-1}$$

which I have no problem with. The author then states that the second equation above is also equal to:

$$\left [ G\left ( x \right ) \right ]^{2r-2}=\left ( \frac{4}{\pi } \right )^{r-1}\left \{ \int_{0}^{\frac{\pi }{4}}\left [ 1-\exp \left ( -\frac{1}{2}x^{2}\sec ^{2}\theta \right ) \right ]d\theta \right \}^{r-1}$$

Can someone explain to me how the author gets from the second equation to the third (or last) equation above.

2. Sep 5, 2011

### torquil

Since an angle has been introduced in the final result, I would try to express the integral using polar coordinates in the x1,x2 plane.

3. Sep 5, 2011

### susskind_leon

You basically want to show this:

$$\int_{0}^{x}\int_{0}^{x} \exp \left [- \frac{1}{2}\left ( x_{1}^{2}+x_{2}^{2} \right ) \right ]dx_{1}dx_{2} = 2 \int_{0}^{\frac{\pi }{4}}\left [ 1-\exp \left ( -\frac{1}{2}x^{2}\sec ^{2}\theta \right ) \right ]d\theta$$

Since you are integrating over a rectangle from (0,0) to (x,x) here, you can't really use spherical coordinates.

You can, however, use an angle theta=0..pi/4 and one cartesian coordinate, i.e. x_1 to describe half of your rectangle. The transformation should look kinda like

$$x_1=x_1$$
$$x_2 = x_1 \tan \theta$$

For the argument inside the exponential function, you will get
$$x_1^2 + x_2^2 = x_1 (\tan^2\theta + 1) = x_1^2 \sec^2 \theta$$
For your Jacobian, you will get:
$$x_1 (\tan\theta)' = x_1 \sec^2\theta$$
Once you do the x_1-Integration, the theta-Part of the Jacobian will go away.

Last edited: Sep 5, 2011
4. Sep 5, 2011

### Zoran

Okay, thanks.

Note that I think you meant to write:

$$x_1^2 + x_2^2 = x_1^2 (\tan^2\theta + 1) = x_1^2 \sec^2 \theta$$

Last edited: Sep 5, 2011
5. Sep 5, 2011

### susskind_leon

Yup, of course, sry ;)