Solution to EE 504 Test #2 F03
1.
Three random variables, X, Y and Z have the following characteristics:
E ( X ) = 1 , E (Y ) = 0 , E ( Z ) = −3
σ X2 = 4 , σ Y2 = 4 , σ Z2 = 1
σ XY = 0 , σ XZ = 2 , σ YZ = 1
Find the numerical value of the correlation coefficient, ρXW , between X and
W = X − 2Y + Z .
σ
ρXW = XW
σ Xσ W
σ X = σ X2 = 2
Using the formula for the variance of a linear combination, Z, of N random variables, X i ,
N
σ = ∑a σ
2
Z
2
i
i= 1
we get
N
2
Xi
N
+ ∑ ∑ ai a jσ X i X j ,
i= 1 j = 1
i≠ j
σ W2 = (1) σ X2 + (−2) σ Y2 + (1) σ Z2 + 2(1)(−2)σ XY + 2(1)(1)σ XZ + 2(−2)(1)σ YZ
σ W2 = 4 + 4 × 4 + 1 − 4 × 0 + 2 × 2 − 4 × 1 = 21 ⇒ σ W = 21
2
2
2
The variance of W could also be found using
σ W2 = E (W 2 ) − [E (W )]
E (W ) = E ( X − 2Y + Z ) = E ( X ) − 2 E (Y ) + E ( Z ) = 1 − 2(0) − 3 = −2
2
2
E (W ) = E ( X − 2Y + Z ) = E ( X 2 ) + 4 E (Y 2 ) + E ( Z 2 ) − 4 E ( XY ) + 2 E ( XZ ) − 4 E (YZ )
2
(
)
E ( X 2 ) = σ X2 + [E ( X )] = 4 + 12 = 5
2
E (Y 2 ) = σ Y2 + [E (Y )] = 4 + 0 2 = 4
2
E ( Z 2 ) = σ Z2 + [E ( Z )] = 1 + (−3) = 10
2
2
E ( XY ) = σ XY + E ( X ) E (Y ) = 0 + 1(0) = 0
E ( XZ ) = σ XZ + E ( X ) E ( Z ) = 2 + 1(−3) = −1
E (W
E (YZ ) = σ YZ + E (Y ) E ( Z ) = 1 + 0(−3) = 1
2
) = 5 + 4(4) + 10 − 4(1)(0) + 2(1)(−3) − 4(0)(−3) = 5 + 16 + 10 − 6 = 25
σ W2 = 25 − (−2) = 21 ⇒ σ W = 21 Check.
2
The covariance of X with W is
σ XW
σ XW = E ( XW ) − E ( X ) E (W )
= E ( X ( X − 2Y + Z )) − E ( X ) E ( X − 2Y + Z )
σ XW = E ( X 2 − 2 XY + XZ ) − 1 × (1 − 0 − 3)
σ XW = E ( X 2 ) − 2E ( XY ) + E ( XZ ) + 2
123 123 123
−1
0
5
σ XW = 5 − 1 + 2 = 6
6
ρXW =
= 0.6547
2 × 21
2.
Two independent random variables, X and Y, have pdf’s given by
p X ( x ) = 0.3δ ( x − 2) + 0.7δ ( x + 1)
and pY ( y ) is uniform over the range, 2 < y < 7 .
(a)
Sketch the pdf of Z = X − Y .
1
 y − 4.5 
pY ( y ) = rect 

 5 
5
1
 y + 4.5 
 − y − 4.5  1
p( −Y ) ( y ) = pY (− y ) = rect 

 = rect 
 5 

5
5  5
1
 z + 4.5 
p Z ( z) = p X ( z) ∗ p( −Y ) ( z) = [0.3δ ( z − 2) + 0.7δ ( z + 1)] ∗ rect 

 5 
5
p Z ( z) =
1
 z + 4.5 
 z + 4.5  
0.3δ ( z − 2) ∗ rect 
 + 0.7δ ( z + 1) ∗ rect 




 5  
5
5
p Z ( z) =
1
 z + 5.5  
 z + 2.5 
0.3 rect 

 + 0.7 rect 

 5  
 5 
5
p Z (z)
0.20
0.16
0.12
0.08
0.04
-9 -8
(b)
-7 -6
-5 -4
-3 -2
z
-1
What is the numerical probability that Z < 4?
Pr ( Z < 4 ) = Pr (−4 < Z < 4 ) =
Pr ( Z < 4 ) =
4
∫ p (z)dz
Z
−4
1 
 z + 5.5  
 z + 2.5 
0.3 rect 
 dz
 + 0.7 rect 

∫
 5  
 5 
5 −4 
4
1
1
 z + 5.5 
 z + 2.5 
Pr ( Z < 4 ) = ∫ 0.3 rect 
 dz
 dz + ∫ 0.7 rect 
 5 
 5 
5 −4
5 −4
4
4
−3
1.2 + 0.7
0.7
0.3
0.7
0.3
Pr ( Z < 4 ) =
dz +
dz =
(4) + (1) =
= 0.38
∫
∫
5
5
5 −4
5 −4
5
0
Referring to the pdf graph, this is the area under the pdf from -4 to 0, which is 0.2(1) +
0.06(3) = 0.38. Check.
3.
An experiment is performed 200 times. The sample mean, x , of the experimental
outcomes is 25 and the sample standard deviation, SX , is 4. A sample size of 200 is large
enough to reasonably use the sample standard deviation as the population standard
deviation, σ X , and to reasonably assume that the pdf of the sample mean, X , is Gaussian.
A confidence interval on the sample mean is to be reported in the form, 25 ± k, for a
confidence level of 80%. Find the numerical value of k.
σ X2
16
=
= 0.08 ⇒ σ X = 0.08 = 0.2828 .
The variance of the mean is σ =
N 200
2
X
From the normal distribution tables, for a confidence level of 80%, the confidence interval
should be approximately 25 ± 1.3σ X = 25 ± 0.3676 . Therefore k is 0.3676.
4.
What characteristic of its probability density function indicates that a random
variable is continuous?
The pdf has no impulses in it.
OR
What characteristic of its probability density function indicates that a random variable is
discrete?
The pdf has only impulses in it.
5.
A deterministic random process has sample functions of the form,
X( t) = A cos(2πt + θ ) ,
where A and θ are random over the ensemble but constant for any single sample function.
Let θ be uniformly distributed over the range, −π < θ < π and let A be Gaussian
distributed with an expected value of 1 and a variance of 4. Let A and θ be independent,
implying that A and cos(2πt + θ ) are also independent. Find the mean-squared value of the
random process, E ( X 2 ( t)) .
 A2

E ( X ) = E ( A cos (2πt + θ )) = E  (1 + cos( 4πt + 2θ ))
 2

2
2
2
 A2
 A2 

E ( X ) = E   + E  cos( 4πt + 2θ )
 2
 2

2
Since A and cos(2πt + θ ) (and, by implication, cos( 4πt + 2θ ) ) are independent,
 A2 
 A2 
E ( X 2 ) = E   + E   E (cos( 4πt + 2θ ))
 2
 2
∞
1
E (cos( 4πt + 2θ )) = ∫ cos( 4πt + 2θ ) pΘ (θ ) dθ =
2π
−∞
π
∫ cos(4πt + 2θ )dθ
−π
π
1  sin( 4πt + 2θ ) 
1  sin( 4πt + 2π ) sin( 4πt − 2π ) 
=
E (cos( 4πt + 2θ )) =
−




2π 
2
2
2
 −π 2π 

E (cos( 4πt + 2θ )) =
1
4π


sin( 4πt + 2π ) − sin( 4πt − 2π )  = 0
4244
3 14
4244
3
14
= sin ( 4 πt )

 = sin ( 4 πt )
(This result, E (cos( 4πt + 2θ )) = 0 , should be obvious but the preceding three lines prove it
in case it is not.) Then
 A2  1
5
1
1
2
E ( X 2 ) = E   = E ( A 2 ) = σ A2 + [E ( A)] = ( 4 + 1) = = 2.5
2
2
2
 2 2
(
)