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B–7 Examples of Important Distributions 699<br />

statistics, and many theorems have been developed by statisticians that are based on<br />

Gaussian assumptions. It cannot be overemphasized that the Gaussian distribution is very<br />

important in analyzing both communication problems and problems in statistics. It can also<br />

be shown that the Gaussian distribution can be obtained as the limiting form of the binomial<br />

distribution when n becomes large, while holding the mean m = np finite, and letting the<br />

variance s 2 = np(1 - p) be much larger than unity [Feller, 1957; Papoulis, 1984].<br />

DEFINITION. The Gaussian distribution is<br />

1<br />

>(2s<br />

f(x) =<br />

2 )<br />

(B–45)<br />

12ps e-(x-m)2<br />

where m is the mean and s 2 is the variance.<br />

A sketch of Eq. (B–45) is given in Fig. B–5, together with the CDF for the Gaussian<br />

random variable. The Gaussian PDF is symmetrical about x = m, with the area under the PDF<br />

1<br />

1<br />

being<br />

2<br />

for (-q … x … m) and<br />

2<br />

for (m … x … q). The peak value of the PDF is 1>112ps2,<br />

so that as s S 0, the Gaussian PDF goes into a δ function located at x = m (since the area<br />

under the PDF is always unity).<br />

We will now show that Eq. (B–45) is properly normalized, (i.e., that the area under f(x)<br />

is unity). This can be accomplished by letting I represent the integral of the PDF:<br />

q<br />

q<br />

1<br />

>(2s<br />

I ! f(x)dx = 2 ) dx<br />

(B–46)<br />

L -q L -q 12p s e-(x-m)2<br />

Making a change in variable, we let y = (x - m)/s. Then,<br />

q<br />

q<br />

1<br />

I =<br />

e -y2 >2 1<br />

(sdy) = e -y2 >2 dy<br />

(B–47)<br />

12p s L-q<br />

12p L-q<br />

The integral I can be shown to be unity by showing that I 2 is unity:<br />

q<br />

q<br />

I 2 1<br />

= c e -x2 >2 1<br />

dx dc e -y2 >2<br />

dy d<br />

12p L-q<br />

12p L-q<br />

= 1 q q<br />

e -(x2 +y 2 )>2<br />

dx dy<br />

(B–48)<br />

2p L-q L-q<br />

Make a change in the variables to polar coordinates from Cartesian coordinates. Let r 2 = x 2 + y 2 ,<br />

and let u = tan -1 (y>x); then,<br />

I 2 = 1 2p q<br />

c e -r2 >2 r dr d du = 1 2p<br />

du = 1 (B–49)<br />

2p L0<br />

L 0<br />

2p L0<br />

Thus, I 2 = 1, and, consequently, I = 1.<br />

Up to now, we have assumed that the parameters m and s<br />

2 of Eq. (B–45) were the mean<br />

and the variance of the distribution. We need to show that indeed they are! This may be<br />

accomplished by writing the Gaussian form in terms of some arbitrary parameters a and b:<br />

f(x) =<br />

1<br />

12p b e-(x-a)2 >(2b 2 )<br />

(B–50)

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