Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 3.1 Histogram and Averages
 In this example we illustrate the different way averages may be computed for sequence of random
variables. We must first generate a sequence of random variables and set the values to five discrete values
of xi. We use the Matlab built in random number generator rand(1,N). Multiplying by 5.0 we get a 0 - 5
uniform number sequence. We discretize the result by the use of the integer (floor) and obtain a sequence
that has values from a set of possible outcomes, { 0 1 2 3 4}. We will work with a 15 number sequence to
illustrate the process.
y=floor(5*rand(1,15));
A companion array cj is generated to indicate the index on yi. In this example we used the Matlab built in
random number generator since it behaves in a manner similar to the generator of Example 2.9.
c=1:15
c =
Columns 1 through 12
1
2
3
4
12
Columns 13 through 15
13
14
15
5
6
7
8
9
10
11
4
3
2
0
4
2
3
y
y =
Columns 1 through 12
4
1
3
2
3
Columns 13 through 15
4
3
0
A histogram function can be computed for the sequence yi and tabulated to indicate the number of terms in
the sequence that correspond to each item, xi , of the set.
n=hist(y,5);
Comparing the items in the sequence, y, we find that there are
n(1)
ans =
2
items with value 0 in the yj sequence. The next quantity we would like to compute is the index of the items
in the sequence that are valued 0, 1 etc.
i=find(y==0)
y(i)
i =
8
15
0
0
ans =
The indices that correspond to x1 = 0 are recovered. The entire set of operation can be repeated for the other
values of u, such as u = 1
i=find(y==1)
y(i)
i =
2
ans =
1
and for other values of u, such as u =2
i=find(y==2)
y(i)
i =
4
7
10
2
2
2
ans =
and for other values of u, such as u =3
i=find(y==3)
y(i)
i =
3
6
11
12
14
3
3
3
3
3
ans =
Since the histogram counts the number of terms in the sequence that corresponds to each item of the set x i,
we may sum the elements in the histogram to recover N as
sum(n)/15
ans =
1
Averaging can also be performed, we may verify Equation 3.1-9 by first summing the relative frequency or
probabilities in the limit.
x=0:4
sum(x.*n)/15
x =
0
ans =
1
2
3
4
2.5333
The alternate way of computing the averages is by the sequence directly or
sum(y)/15
ans =
2.5333
The same averaging process can be used for
sum(y.^2)/15
ans =
8.1333
By the use of histograms
sum(x.^2.*n)/15
ans =
8.1333
For this sequence we can plot the histogram approximation to the probability density function. Remember
that in this example we have used a small number for N and we do not expect a good probability density
representation.
bar(n/15,.2)
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1
2
3
4
5
