ECSE-2500 Engineering Probability HW#17 Solutions
Due 11/13/14
1. (11 points) Let X and Y be discrete RV’s with joint PMF pX,Y  x, y  and marginal PMF’s
pX x  and pY  y  as given in the following table:
pX,Y  x, y 
x
1
0
−1
pY  y 
−1
0.06
0.14
0.10
0.30
y
0
0.04
0.05
0.03
0.12
1
0.35
0.21
0.02
0.58
pX x 
0.45
0.40
0.15
1.a. (4 points) Compute the conditional PMF of Y given X, pY y x to 3 decimal places. Then
show that for each x  SX we have
 p  y x  1.
ySY
Y
Clearly show your work.
Solution
Using the formula pY y x 
pY y x
1
x
0
−1
pXY x, y 
, we have
pX x
−1
y
0
1
0.06
 0.133
0.45
0.14
 0.350
0.40
0.10
 0.667
0.15
0.04
 0.088
0.45
0.05
 0.125
0.40
0.03
 0.200
0.15
0.35
 0.777
0.45
0.21
 0.525
0.40
0.02
 0.133
0.15
 p  y x
Y
pX x 
1.000
0.45
1.000
0.40
1.000
0.15
ySY
Page 1 of 3
ECSE-2500 Engineering Probability HW#17 Solutions
Due 11/13/14
1.b. (4 points) Compute the conditional PMF of X given Y, pXx y  to 3 decimal places. Then
show that for each y  SY we have
 p x y  1.
xSX
X
Clearly show your work.
Solution
Using the formula pXx y  
pXx y 
1
x
0
−1
 p x y 
xSX
X
pY  y 
pXY x, y 
, we have
pY  y 
−1
y
0
1
0.06
 0.200
0.30
0.14
 0.467
0.30
0.10
 0.333
0.30
0.04
 0.333
0.12
0.05
 0.417
0.12
0.03
 0.250
0.12
0.35
 0.603
0.58
0.21
 0.362
0.58
0.02
 0.034
0.58
1.000
1.000
0.999
0.30
0.12
0.58
pX x 
0.45
0.40
0.15
The last one adds to 0.999 due to rounding error.
1.c. (3 points) Using your results from part b, show clearly whether X and Y are independent.
Solution
From the table in part b, we see pXx y  pX x for each pair  x, y   SX  SY .
For example, 0.2  pX 1 1  pX 1  0.45.
Page 2 of 3
ECSE-2500 Engineering Probability HW#17 Solutions
Due 11/13/14
2. (8 points) Consider two continuous random variables X and Y with joint PDF
f X,Y x, y  
20
x  0, y  0, x  y  1
elsewhere
2.a. (4 points) Draw a picture of the region where f X,Yx, y  is non-zero. Compute the marginal
PDF of X and the marginal PDF of Y. Show your work. Be very clear about your regions.
Solution
y
1
2
0
f X x  
1 x
1 x
 2dy  2 y
1 x
0
 2  2x for 0  x  1
0
fY y  
1 y
 2dx  2x
1 y
0
 2  2 y for 0  y  1
0
2.b. (4 points) Now compute the conditional PDF of Y given X, fY  y x, and show that for every
x with f X x   0, we have

 f  y xdy  1.
Y
Clearly show your work. Be very clear about your

regions.
Solution
fY y x 
f X,Yx, y 
2
1


for x  0, y  0, x  y  1
f X x 
2  2x 1  x
and zero elsewhere.

1 x
 f  y xdy  
Y

0
1 x
 1  dy   1  1dy   1  1  x  1






1 x 
 1  x  0
1 x 
for 0  x  1.
Page 3 of 3