Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 3.3 Moments of Continuous Distributions
In this example we will derive relationships for the expected values for three distributions, the
uniform, exponential and gaussian and give examples of their use.
UNIFORM - A train is expected to arrive between the hours of 1 and 4 PM. We assume a uniform
distribution of arrival time in the 1 - 4 interval. What is the expected arrival time of the train and what is
the variance?
SOLUTION: The problem can be solved by computing the expected value of a uniform distribution with
f(x) = 1 / (4 - 1) for 1 < x < 4 or
4
1
E  X    u   du
3
1
syms x u
Ex=int(u/3,u,1,4)
Ex =
5/2
We derive a general formula for a rectangular distribution
syms a b
Ex=int(u/(b-a),u,a,b)
Ex =
-1/2*b^2/(-b+a)+1/2*a^2/(-b+a)
Ex= simplify(Ex);
pretty(Ex)
1/2 a + 1/2 b
The second moment may be similarly computed
Ex2=int(u^2/(b-a),u,a,b)
Ex2= simplify(Ex2)
Ex2 =
-1/3*b^3/(-b+a)+1/3*a^3/(-b+a)
Ex2 =
1/3*a^2+1/3*b*a+1/3*b^2
The variance can be computed using the formula for moments Equation 3.2-5
VAR[X] = E[X2] - E[X]2 =
VARx=Ex2-Ex^2
VARx=simplify(VARx)
VARx =
1/3*a^2+1/3*b*a+1/3*b^2-(1/2*a+1/2*b)^2
VARx =
1/12*a^2-1/6*b*a+1/12*b^2
factoring the expression we obtain
VARx=factor(VARx)
VARx =
1/12*(-b+a)^2
The last simplification was performed by Matlab using simplify and factor. The solution to the train arrival
problem is therefore
subs(Ex,{a,b},{1,4})
ans =
2.5000
subs(VARx,{a,b},{1,4})
ans =
0.7500
sqrt(0.75)*60
ans =
51.9615
minutes
The expected hour of arrival is 2.5 or 2:30 PM with a variance of 0.75 hours 2. It is customary to use the
root of the variance since the units are the same as the expected values. We restate the results as the train is
expected to arrive at 2:30 PM with a standard deviation of sqrt((0.75)60 = 52 minutes.
EXPONENTIAL - A lifetime of a part in hours follows an exponential distribution with a parameter =
0.1. What is the expected lifetime of the part what is the variance of the lifetime?
SOLUTION: The problem can be solved by computing the expected value of a exponential distribution
with
f t     e  t for 0 < t < or
syms lam
gt = sym('lam>0');
maple('assume',gt);
ET=int(u*lam*exp(-lam*u),u,0,inf)
ET =
1/lam
In the solution of the integral, Maple evaluated the integral with n as the upper limit. Then the limit was
taken when n  and we obtained E[T] = 1 / . The parameter  has been shown to be the reciprocal of
the expected lifetime. The variance can be similarly computed by computing the second moment and using
Equation 3.4-5
ET2=int(u^2*lam*exp(-lam*u),u,0,inf)
ET2 =
2/lam^2
In the solution of the integral, Maple evaluated the integral with n as the upper limit. Then the limit was
taken when n  and we obtained E[T2]= 2 / 2. The variance is VAR[T] = E[T 2] - E[T]2 =
VART=ET2-ET^2;
VART=simplify(VART)
VART =
1/lam^2
The solution to the lifetime problem is
subs(ET,lam,0.1)
ans =
10
hours
VAR[T] =
subs(VART,lam,0.1)
ans =
100.0000
hours2
GAUSSIAN - For the gaussian distribution we will demonstrate the normalization factor = 1 and identify
the two parameters, a and 2, of the distribution as the mean and variance of the distribution, respectively.
Equation 2.5-8 shows the normalization factor
of the integral, I,
2   for the gaussian. We begin by setting up the kernel
syms sig u v
f=exp(-1/2*u^2)
f =
exp(-1/2*u^2)
g=subs(f,u,v);
pretty(f)
2
exp(- 1/2 u )
The integral I cannot be directly evaluated. Let us instead find an expression for I 2. We must changes
variables to avoid confusion in the dummy variables of the two integrals
I2= int(int(f*g,u,-inf,inf),v,-inf,inf);
double(1/(2*pi)*I2)
ans =
1
Using Matlab we find that the integral for I2 is indeed equal to unity. We can show the details of the
computation by expanding the integral and changing variable.
syms u v
syms r alp
fg=f*g;
pretty(fg)
2
2
exp(- 1/2 u ) exp(- 1/2 v )
Let u = r cos() and v = r sin() and dv du = r dr d to maintain the same elemental area of integration we
have u2 + v2 = r2. The limits of the area integral have to be change to encompass the same area and
0    2   and 0 < r < . Rewriting the integral we have
er=subs(fg,{u,v},{r*cos(sig) r*sin(sig)});
er=simple(er)
er =
exp(-1/2*r^2)
u=r*cos(alp);
v=r*sin(alp);
J= det(jacobian([u ; v],[r alp]));
J=simple(J)
J =
r
1
I 
2 
2
2 
 r e
0
1
 r 2
2
 dr  d
0
This two-dimensional integral is separable and each integral can be evaluated separately, The integral for r
is found by Matlab and is equal to unity.
int(er*J,r,0,inf)
ans =
1
The other integral for  evaluates to 2 , which is exactly equal to the square of the normalizing factor.
maple('int','1,x=0..2*pi')
ans =
2*pi
The expected value is computed by the integral after we clear the symbols u and v so that they can reused
in another context.
syms u unreal
syms v unreal
clear u v
syms a u v
f=u*exp(-1/2*((u-a)/sig)^2)
f =
u*exp(-1/2*(u-a)^2/sig^2)
syms sig
gt= sym('sig>0');
maple('assume',gt);
EXG=1/sig*int(f,u,-inf,inf)
EXG =
a*2^(1/2)*pi^(1/2)
The result must be divided by 2   to obtain E[X] for a gaussian = a. The integral may also be
evaluated by first substituting for v = (u - a)/  and dv = du /  . The integral becomes
EXG=int(exp(-1/2*v^2)*(sig*v+a),v,-inf,inf)
EXG =
a*2^(1/2)*pi^(1/2)
for the same result. The second moment and the variance is similarly computed
E[X2] =
EX2G=int(exp(-1/2*v^2)*(sig*v+a)^2,v,-inf,inf);
EX2G=simple(EX2G)
EX2G =
2^(1/2)*pi^(1/2)*(a^2+sig^2)
Again we must normalize by dividing by 2   to obtain E[X2] = a2 +2. The variance is therefore, using
Equation 3.4- 5, VAR[X] = 2. We have shown that the two parameters of the gaussian are the mean = a
and the variance =  and  is called the standard deviation.
