Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 4.5 Joint Continuous Expectation
 In this example we will compute the expected values for random variables that have continuous
distribution functions. For this example we assume that the density functgion is uniform over a two
dimensional region that is triangular in shape. This form of dependence of the density function upon
a region that is described by a functional relation between the two variables, x and y, causes the
resultant distribution to be correlated. We will show this result. We first descibe this density
function. Let f(x,y) = C for 0 <y < x <1. Normalizing this density function over the region we
express the normalization intergral. The limist on the v (y) integration is obtained from the domain
of x, y expression, 0 < y < x, and therefore the upper limit for the inner integration is x (u).
syms u v C
F1=int(int(C,v,0,u),u,0,1);
C=solve(F1-1,C)
C =
2
When we integrate we find that C = 2. Computing the expected values we have
E[X] =
EX=int(int(2*u,v,0,u),u,0,1)
EX =
2/3
This expression can be evaluated numerically or symbolically. We have evaluated this expression
symbolically to obtain a rational arithmetic expression for the result. Maple performs numerical
evaluation using rational arithmetic. A similar expression can be obtained for E[Y]
E[Y] =
EY=int(int(2*v,v,0,u),u,0,1)
EY =
1/3
The Correlation can be computed using
E[XY] =
EXY=int(int(2*v*u,v,0,u),u,0,1)
EXY =
1/4
The marginal distributions can be computed and then the expectation operations performed
syms x y
fx=int(2,v,0,x)
fx =
2*x
fy=int(2,u,y,1)
fy =
2-2*y
and the expected values
E[X] =
EX=int(2*u*u,u,0,1)
EX =
2/3
E[Y] =
EY=int(2*v*(1-v),v,0,1)
EY =
1/3
Both procedures result in the same values for the expectation.
The variances and covariances can be computed
E[X2] =
EX2=int(int(2*u^2,v,0,u),u,0,1)
EX2 =
1/2
VAR[X] = E[X2] - E[X]2 =
VARX=EX2-EX^2
VARX =
1/18
E[Y2] =
EY2=int(int(2*v^2,v,0,u),u,0,1)
EY2 =
1/6
VAR[Y] = E[Y2] - E[Y]2 =
VARY=EY2-EY^2
VARY =
1/18
The Covariance can be computed COV[XY] = E[XY] - E[X] E[Y] =
COVXY=EXY-EX*EY
COVXY =
1/36
Two other computations can be performed
E[X + Y] =
EXPY=int(int(2*(u+v),v,0,u),u,0,1)
EXPY =
1
and it is easy to show that E[X + Y] = E[X] + E[Y] =
EX+EY
ans =
1
And similarly
E[(X + Y)2] =
EXPY2=int(int(2*(u+v)^2,v,0,u),u,0,1)
EXPY2 =
7/6
and it can be shown that
E[(X + Y)2] = E[X2] + 2 E[X Y] + E[Y2]=
EX2+2*EXY+EY2
ans =
7/6
The covariance matrix P can be written as
 VARX
K
COVXY
COVXY 
VARY 
K=[VARX COVXY;COVXY VARY]
K =
[ 1/18, 1/36]
[ 1/36, 1/18]
