Solution to ECE 504 Test #2 F02
1.
A random variable, X, has a pdf that is uniform between x1 and x 2 . What is the
numerical probability that X < x 0 , given that X > x 3 ?
x0
Pr ( x 3 < X < x 0 ) FX ( x 0 ) − FX ( x 3 )
Pr ( X < x 0 | X > x 3 ) =
=
=
1 − FX ( x 3 )
Pr ( X > x 3 )
∫ p ( x )dx
X
x3
∞
∫ p ( x )dx
X
x3
1
( x − x3 ) x − x
x 2 − x1 0
3
Pr ( X < x 0 | X > x 3 ) =
= 0
1
x
−
x
( x − x3 ) 2 3
x 2 − x1 2
2.
A random variable, X, has an expected value of E ( X ) and a variance, σ X2 . If a
number is computed by averaging N randomly-chosen values of X, what is the probability
that the number will be between E ( X ) − x 0 and E ( X ) + x 0 ?
X=
1
N
N
∑X
i =1
i
If N is large, the central limit theorem applies and the pdf of X is Guassian with
σ X2
2
. The probability is
E ( X ) = E ( X ) and σ X =
N
Pr (E ( X ) − x 0 < X < E ( X ) + x 0 ) =
Let u =
x − E( X )
⇒ du =
2σ X
E( X ) + x 0
∫
( )
E X − x0
−
1
e
2πσ X
( x − E ( X )) 2
2σ X2
dx =
2
2πσ X
E( X ) + x 0
∫
( )
−
( x − E ( X )) 2
e
E X
dx
.
2σ X
Then
Pr (E ( X ) − x 0 < X < E ( X ) + x 0 ) =
2 2σ X
2πσ X
x0
x0
2σ X
2σ X
∫
0
e − u du =
2
2
π
∫
0
e − u du .
2
2σ X2
dx
 x 
Pr (E ( X ) − x 0 < X < E ( X ) + x 0 ) = erf  0 
 2σ X 
 x − E ( X )
rect 

 x 2 − x1 
3.
The pdf of X is of the form, p X ( x ) =
and the pdf of Y is of the
x 2 − x1
form, pY ( y ) = aδ ( y − y1 ) + bδ ( y − y 2 ) , a + b = 1. The random variable, Z, is Z = X − Y .
So the pdf of Z is the convolution of the pdf’s of X and -Y . The pdf of -Y is
aδ (− y − y1 ) + bδ (− y − y 2 ) = aδ ( y + y1 ) + bδ ( y + y 2 ) . Therefore the pdf of Z is
 z − E ( X )
rect 

 x 2 − x1 
p Z ( z) =
∗ aδ ( z + y1 ) + bδ ( z + y 2 )
x 2 − x1
[
]
 z + y1 − E ( X ) 
 z + y2 − E ( X )
rect 
rect 


x 2 − x1 
x 2 − x1 


p Z ( z) = a
+b
.
x 2 − x1
x 2 − x1
The probability is the area under this function for z > z0 . This is the sum of two areas,
one for each of the two terms above. The area under the first term is
a
(z − z )
x 2 − x1 2 1
where z2 is the greater of E ( X ) − y1 +
x 2 − x1
and z0 and z1 is the greater of
2
x 2 − x1
and z0 . this function for z > z0 . This is the sum of two areas, one for
2
each of the two terms above. The area under the second term is
E ( X ) − y1 −
b
(z − z )
x 2 − x1 4 3
where z4 is the greater of E ( X ) − y 2 +
E( X ) − y2 −
of Z.)
x 2 − x1
and z0 and z3 is the greater of
2
x 2 − x1
and z0 . (This last result is best seen by drawing a sketch of the pdf
2
4.
Two random variables, X and Y, have a joint pdf, which is one-half in the regions,
0 < x <1 , 0 < y <1
and
−1 < x < 0 , − 1 < y < 0
and zero elsewhere. That is, the joint pdf is
p XY ( x, y ) =
1
1
1
1
1 




rect  x −  rect  y −  + rect  x +  rect  y +   .








2
2
2
2
2 
What is the numerical value of the correlation coefficient, ρXY ?
ρXY =
E( X ) =
∞ ∞
x

1
E ( XY ) − E ( X ) E (Y )
σ Xσ Y

1

1

1 
∫ ∫ 2 rect  x − 2 rect  y − 2 + rect  x + 2 rect  y + 2 dxdy
−∞ −∞
1 1
0 0
x
x
1 1
E ( X ) = ∫ ∫ dxdy + ∫ ∫ dxdy = − = 0
2
2
4 4
0 0
−1 −1
Similarly,
E (Y ) = 0
E ( XY ) =
1 1
E ( XY ) = ∫ ∫
0 0
∞ ∞
xy 
1
1
1
1 




rect  x −  rect  y −  + rect  x +  rect  y +  dxdy





2 
2
2
2
2 
−∞ −∞
∫∫
0 0
1
1
0
0
 1 1 1  1
xy
xy
1
dxdy + ∫ ∫
dxdy = ∫ y ∫ xdxdy + ∫ y ∫ xdxdy  =  +  =
2
2
2 0 0
 2 4 4  4
−1 −1
−1 −1
E( X 2 ) =
∞ ∞
x2 
1
1
1
1 




∫−∞ −∞∫ 2 rect  x − 2 rect  y − 2 + rect  x + 2 rect  y + 2 dxdy
1 1
E( X 2 ) = ∫ ∫
0 0
0 0
x2
x2
1 1 1
dxdy + ∫ ∫ dxdy = + =
2
2
6 6 3
−1 −1
Similarly,
E (Y 2 ) =
1
.
3
Therefore σ X2 =
Finally, ρXY
1
1
1
and σ Y2 = and σ X σ Y = .
3
3
3
1
3
= 4 =
1 4
3
5.
A random variable, X, has an expected value, E ( X ) , and a standard deviation,
σ X . Another random variable, Y, has an expected value, E (Y ) , and a standard deviation,
σ Y . X and Y are independent. Two other random variables are formed by Z1 = X + Y and
Z2 = X − Y . Find the numerical value of the covariance, σ 12 , between Z1 and Z2 .
σ 12 = E ( Z1Z2 ) − E ( Z1 ) E ( Z2 ) .
E ( Z1 ) = E ( X ) + E (Y ) and
E ( Z2 ) = E ( X ) − E (Y )
E ( Z1Z2 ) = E (( X + Y )( X − Y )) = E ( X 2 − Y 2 ) = E ( X 2 ) − E (Y 2 )
σ 12 = E ( X 2 ) − E (Y 2 ) − [E ( X ) + E (Y )][E ( X ) − E (Y )]
σ 12 = E ( X 2 ) − E (Y 2 ) − [E ( X )] + [E (Y )] = σ X2 − σ Y2
2
2