Download Solutions to ECE 863 Homework Set #4: Analysis of Stochastic Systems and more Exercises Probability and Statistics in PDF only on Docsity!
1
ECE 863 – Analysis of Stochastic Systems
Fall 2001
Solutions for Homework Set
Problem 1
4.3 (a)
P ^ X < 5, Y > 2, Z^2 ≥ 2
= P [ − 5 < X < 5 ] ( 1 − P [ Y ≤ 2 ] ) ( 1 − P ^ − 2 < Z < 2 )
( FX^ (^5 )^ FX^ (^5 )) (^1 FY^ (^2 )^ )^1 ( FZ^ ( 2 ) FZ^ ( 2 ))
4.3 (b)
P X > 5, Y < 0, Z = 1
= c 1 - FX b g 5 h F Y d i 0 -^ eF Z b g 1 - F Zd i 1 - j
4.3 (c)
P min bX, Y, Z g > 2 = P X > 2, Y > 2, Z > 2
= c 1 - FX b g 2 h c 1 - F Y b g 2 h c 1 - FZ b g 2 h
4.3 (d)
P max bX, Y, Z g < 6 = P X < 6, Y < 6, Z < 6
= F X d i 6 -^ F Y d i 6 - F Zd i 6 -
2
4.9 (a)
F (^) XY x, y ase as bte dsdt
(^2) /2 bt (^2) / 0
y 0
x b g =^ z zFH -^ IK FH - IK
= FH 1 - e- ax IK FH 1 - e- IK
(^2) /2 by (^2) /
4.9 (b)
P X Y axe ax bte dtdx
(^2) /2 bt (^2) / 0
x 0
= FH -^ IK FH - IK
= - FH - IK
(^2) /2 bx (^2) / 0
1 a a b
4.9 (c)
There are a couple of ways to solve this. We will use the joint cdf F (^) XY bx, y g.
F x
Lim X (^) y FXY^ x, y^1 e b g =^ b g ax^2 / Æ • =^ -^
fi f x = d = - dx x FX^ x^ a x e ax
(^2) / b g b g
Similarly, fY y byeby
(^2) / b g =^ -
4
( ) ( )
( )
=> − ρ + − ρ (^) − (^) − = (^) − (^) σ ρ (^) + ^ σ^ σ^ − ρ σ
1 2 2
(^2 ) 2 (^2 ) (^1 2 ) 2 1 2
a b (^) b 2 1 2
x m y^ m^1 y^ m 2 1 2
Therefore we can express the joint pdf as follows:
( )
( )
2 2 2 (^2 21 ) (^2 12 )
y m x^ m y^ m (^2 2 ) xy (^2) (^2 )
f x, y e^. 1 e (^2 2 )
− − ^ −^ ^ +σ ρ− −^ ^ σ σ (^) σ −ρ = πσ (^) πσ − ρ
This is the product of two pdf’s of two Gaussian RV’s.
The second pdf has a mean: 1 1 2 2
m ′ = m +σ ρ y^ −m σ
and a variance: ( σ′ )^2 = σ 1 2 ( 1 − ρ^2 )
Now, if we integrate the joint pdf with respect to x, we get the marginal pdf for Y:
( ) ( )
( )
( ) (^) ( ) ( )
2 2 2 (^2 21 ) (^2 12 )
(^2 ) (^22 )
y m x^ m y^ m (^2 2 ) Y XY 2 (^2 ) y m (^) x m (^22)
2
f y f x, y dx e^1 e dx (^2 2 )
e (^1) e dx 2 2
− − ^ −^ ^ +σ ρ− ∞ σ ∞ −^ ^ σ σ −ρ −∞ −∞ − − − ′ σ ∞ − (^) σ′
−∞
πσ (^) πσ − ρ
πσ πσ′
∫ ∫
∫
The second pdf integrates to 1.
( )
( 22 )^2 2
y m Y^2 2
f y 1 e 2
− − ⇒ = σ πσ
A similar method can be used to derive fX(x)
5
4.17 (a)
P X [ = 1, Y ≤ y (^) ] = P Y[ ≤ y /X = 1 P X] [ = (^1) ]
P N 1 y (^) z
e z dz
y (^1) a (^) a
R S
||
T
| |
e y 1 1 4
2 e y 1
y 1
y 1
a
a
b g
e b^ gj
Similarly, for X = -
P X 1, Y y
e y 1 1 4
2 e y 1
y 1
y 1
R S
||
T
||
a
a
b g
e b^ gj
4.17 (b)
F (^) Y b gy = P Y £ y, X = 1 + P Y £ y, X = - 1
Then we can use the above results from (a)
F y 1 2
P N y 1 1 2 Y b g^ =^ £^ -^ +^ P N^ £^ y^ +^1
f y d dy
F y 1 2
f y 1 1 2 Y b g^ =^ Y b g^ =^ N b^ -^ g^ +^ fN^ by^ +^1 g
4.17 (c)
We need to compare the two probabilities:
P X = 1 / Y > 0 and P x = - 1 / Y > 0
7
( )
( ) (^) ( )
( ) ( ) (^ )^ ( )
− −
∞
− ∞ −
∞ − −
− −
∫
∫
∫
2 2
2 2
2 /
2 2
xy ax /2^ bY /
x xy 0 ax /2 bY / 0 (^2) / /
x ax /2^ bY 0 x ax / y by / XY X^ Y
f x, y axe bye x, y, a,b 0
f x f x, y dy
axe bye dy
Let y y dy ydy 2 f x axe be dy
f x axe Similarly, f y bye f x, y f x f y
=> X and Y are independent.
8
4.25 (a)
We have shown this in class. Please see lecture notes.
4.25 (b)
P XY > 0 = P X > 0, Y > 0 + P X < 0, Y < 0
Since X & Y are independent,
P XY > 0 = P X > 0 P Y > 0 + P X < 0 P Y < 0
P X 0 1
e dx 1
x m1^2 /2 12 0
> = -^ -
= = F - HG^
I KJ
z
e 2
dt Q m
t^2 /2 1 m 1 1 1
p s s
Similar expressions can be derived for the other terms
fi > = F - HG^
I KJ^
F - HG^
I KJ^
I KJ
F HG^
I KJ^
I KJ
F HG^
I KJ
P XY 0 Q m^1 Q m^1 Q m^1 Q m 1
2 2
1 1
2 s s s s 2
10
4.38 (a)
Given X = x then Y = N + x is a Gaussian random variable with a mean M (^) Y = x and standard deviation s (^) Y = sN.
( ) (^ ) y x 2 /2^2 N Y/X N
f y/x 1 e 2
⇒ = −^ −^ σ π σ
4.38 (b)
fXY (^) bx, y g = fY/X (^) by / x g fX (^) b gx
( y x) 2 / (^2 2) N^ x /2^2 N
(^1) e 1 e 2 2
= −^ −^ σ^ − π σ π
( ) ((^ ) )^
y x (^2 2) N^ x^2 / (^22) N XY N
f x, y 1 e 2
⇒ = −^ −^ + σ^ σ π σ
