Massachusetts Institute of Technology
Department of Electrical Engineering & Computer Science
6.041/6.431: Probabilistic Systems Analysis
(Spring 2006)
Recitation 08 Answers
March 09, 2006
1. (a) The marginal distributions are obtained by integrating the joint distribution along the X
and Y axes and is shown in the following figure.
y0
y0
2.0
2.0
f x,y(x 0,y 0) = 0.1
0.2
1.0
1.0
0.3
fy (y)-1.0
-1.0
1.0
2.0
x0
-1.0
0.2
-2.0
-2.0
f(x)
x
0.4
0.2
x
-1.0
1.0
0
2.0
Figure 1: Marginal probabilities fX (x) and fY (y) obtained by integration along the y and x axes
respectively
The conditional PDFs are as shown in the figure below.
(b) X and Y are NOT indepenent since fXY (x, y) 6= fX (x)fY (y). Also, from the figures we
have fX|Y (x|y) =
6 fX (x).
(c)
fX,Y |A (x, y) =
=
( f ((x,y)
X,Y
P(A)
0
(
0.1
π0.1
0
(x, y) ∈ A
otherwise
(x, y) ∈ A
otherwise
(d)
E[X|Y = y] =


 0
1
2

 0
−2.0 ≤ y ≤ −1.0
−1.0 ≤ y ≤ 1.0
1.0 ≤ y ≤ 2.0





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Massachusetts Institute of Technology
Department of Electrical Engineering & Computer Science
6.041/6.431: Probabilistic Systems Analysis
(Spring 2006)
f (x|y) {1 < y <= 2}
y
x|y
0
1/2
2.0
-1.0
1.0
1.0
f (x|y) {-1<y<=1}
1/3
f x,y(x 0,y 0) = 0.1
x|y
-1.0
2.0
1.0
-1.0
2.0
x
0
f (x|y) {-2<y<=2}
1/2
-1.0
x|y
-1.0
1.0
-2.0
y
y
0
0
2.0
1.0
1/4
f (y|x)
1/2
f (y|x)
y|x
y|x
-1.0
-2.0
{-1<x<=1}
{1<x<=2}
Figure 2: Conditional Probabilities
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Massachusetts Institute of Technology
Department of Electrical Engineering & Computer Science
6.041/6.431: Probabilistic Systems Analysis
(Spring 2006)
The conditional variance var(X|Y = y) is given by
var(X|Y = y) =

4

 12
9
12

 4
12
2. (a) We have a = 1/800, so that
fXY (x, y) =
(
1/1600
0,
−2.0 ≤ y ≤ −1.0
−1.0 ≤ y ≤ 1.0
1.0 ≤ y ≤ 2.0





if 0 ≤ x ≤ 40 and 0 ≤ y ≤ 2x
otherwise.
)
(b) P(Y > X) = 1/2
(c) Let Z = Y − X. We have
fZ (z) =
E[Z] = 0.

1
1

 1600 z + 40 ,
−
1
1600

 0,
+
1
40 ,
if − 40 ≤ z ≤ 0,
if 0 ≤ z ≤ 40,
otherwise.





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