Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The solutions to problem 1, 2, 3, 4, 5, and 6 from homework set #5 of the ece 863: analysis of stochastic systems course, fall 2001. The problems involve calculating expectations, integrals, and probabilities related to exponential random variables and their covariance.
Typology: Exercises
1 / 8
This page cannot be seen from the preview
Don't miss anything!
1
Problem 1
4.
fXYZ (^) bx, y, z g =fZY/X (^) bz, y / x g fX (^) b gx
fZY/X (^) bz, y / x g = fZ/XY (^) bz / x, y f g b gY y
⇒ fXYZ (^) ( x, y,z (^) ) =fZ/XY (^) ( z x, y| (^) ) fY X (^) / ( y x| (^) ) fX ( x)
2
Problem 2
4.
Since X and Y are exponential random variables, then X ≥ 0 and Y ≥ 0.
Now we start with FZ (^) b gz = P Z £z
F (^) Z b gz = P X - Y £z
when X Y X Y z X Y z Y x z
when X Y X Y z Y X z Y x z
y y = x +z
y = x - z
z
z x
4
Problem 3
4.
In this case, both X and Y only take on non-negative values: X ≥ 0 and Y ≥ 0. In addition, Y is never bigger than X: Y ≤X
F Z z P Z z P X Y z P Y z x Y x X
The desired area in the 2-dimensional x-y space is shown in the figure below. Note that since X and Y are non-negative, Z=X+Y is also non-negative.
y
z x
( ) (^ )^ (^ ) − + − − + = (^) ∫ ∫ + ∫ ∫
z/2 x (^) x y z z x x y Z 0 0 z/2 0
F z 2e dydx 2e dydx
⇒ FZ ( z ) = (^) ( 1 − 2e −^ z/2^ + e −z^ ) + (^) ( 2e −^ z/2^ − 2e −z^ −ze−z)
⇒ FZ (^) ( z (^) ) = 1 − e −^ z^ −ze−z
fi fZ (^) b gz = ze-z
y = x (^) y = z – x
z/
5
Problem 4
4.59 (a)
E (^) b X + y (^) g^2 = E (^) dX 2 + 2XY +Y^2 i
fi E (^) bX + Y (^) g^2 = E X 2 + 2E XY +E Y^2
4.59 (b)
s^2 X^ + Y = E (^) b X + Y (^) g^2 - (^) cE X +y h^2
fi sX^2 +Y = E X^2 - (^) cE X h^2 + E Y^2 - (^) cE Y h^2 + 2 E XYc - E X E Yh
s (^2) X + Y = s (^2) x+ sY^2 + 2 cov (^) bX, Y g
4.59 (c)
when cov X, Yb g = 0 (i.e., when X and Y are uncorrelated)
fi s (^) X^2 +Y = s (^) X^2 +s^2 Y
7
Problem 6 – 4.69 – Continued:
It can be shown that E N 2 =^22 a
fi sY^2 = E X^2 +E N^2
= 1 + 22 fi (^) Y = 1 +^22 a
s a
s s
X^2
2
X
= - = fi =
c (^) h
r s s
a
XY X Y
2
F + HG^
I KJ
b g
fi =
r
a
XY 2
8
Problem 7
4.
h x, y 1 2 1
e 1
2
x^2 2 1 xy y^2 2 1 12 b g = e^ j
r / r
g x, y 1 1
e 2
2
x^2 2 2 xy y^2 2 1 22 b g = e^ j
/ p r
r r
4.81 (a)
( ) ( )
( ) ( )
∞
−∞ ∞ ∞
− ∞ − ∞ − − −
π^ π^ π
∫
∫ ∫ 2 2 2
X XY
x /2 x /2 x /
f x f x, y dy
(^1) h x, y dy g x, y dy 2
1 e e e (^2 2 2 )
Similarly, f y e Y 2
y^2 / b g =
4.81 (b)
f x, y
1 e 1 e 2 1 1
XY
2
2 x^2 2 1 xy^ y^2 /2 1^12 1
2 x^2 2 2 xy^ y^2 /2 1^22
1
2 2
b g (^2)
r r r r
fi fXY does not have the required form for jointly Gaussian random variables.
