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Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions : Yates and Goodman,3.1.2 3.3.2 3.4.2 3.5.2 3.6.2 3.6.3 and 3.7.
Problem 3.1. On the X , Y plane, the joint PMF is
y
x
PX , Y ( x , y )
3 c
c
c
1 2
1
(a) To find c , we sum the PMF over all possible values of X and Y. We choose c so the sum equals one.
x
y
PX , Y ( x , y ) = ∑
x =− 2 , 0 , 2
y =− 1 , 0 , 1
c | x + y | = 6 c + 2 c + 6 c = 14 c
Thus c = 1 /14.
(b)
P [ Y < X ] = PX , Y ( 0 , − 1 ) + PX , Y ( 2 , − 1 ) + PX , Y ( 2 , 0 ) + PX , Y ( 2 , 1 ) = c + c + 2 c + 3 c = 7 c = 1 / 2
(c)
P [ Y > X ] = PX , Y (− 2 , − 1 ) + PX , Y (− 2 , 0 ) + PX , Y (− 2 , 1 ) + PX , Y ( 0 , 1 ) = 3 c + 2 c + c + c = 7 c = 1 / 2
(d) From the sketch of PX , Y ( x , y ) given above, P [ X = Y ] = 0.
(e)
P [ X < 1 ] = PX , Y (− 2 , − 1 ) + PX , Y (− 2 , 0 ) + PX , Y (− 2 , 1 ) + PX , Y ( 0 , − 1 ) + PX , Y ( 0 , 1 ) = 8 c = 8 / 14
Problem 3.3. On the X , Y plane, the joint PMF is
y
x
PX , Y ( x , y )
3 c
c
c
1 2
1
(a) To find c , we sum the PMF over all possible values of X and Y. We choose c so the sum equals one.
x
y
PX , Y ( x , y ) = ∑
x =− 2 , 0 , 2
y =− 1 , 0 , 1
c | x + y | = 6 c + 2 c + 6 c = 14 c
Thus c = 1 /14. (b) P [ Y < X ] = PX , Y ( 0 , − 1 ) + PX , Y ( 2 , − 1 ) + PX , Y ( 2 , 0 ) + PX , Y ( 2 , 1 ) = c + c + 2 c + 3 c = 7 c = 1 / 2
(c) P [ Y > X ] = PX , Y (− 2 , − 1 ) + PX , Y (− 2 , 0 ) + PX , Y (− 2 , 1 ) + PX , Y ( 0 , 1 ) = 3 c + 2 c + c + c = 7 c = 1 / 2
(d) From the sketch of PX , Y ( x , y ) given above, P [ X = Y ] = 0. (e) P [ X < 1 ] = PX , Y (− 2 , − 1 ) + PX , Y (− 2 , 0 ) + PX , Y (− 2 , 1 ) + PX , Y ( 0 , − 1 ) + PX , Y ( 0 , 1 ) = 8 c = 8 / 14
Problem 3.4. In Problem 3.2.2, we found that the joint PMF of X and Y was
y
x
PX , Y ( x , y )
3 / 14
2 / 14
• 1 /^14
1 / 14
• 1 /^14
1 / 14
2 / 14
• 3 /^14
1 2
1
Problem 3.6. We can make a table of the possible outcomes and the corresponding values of W and Y outcome P [·] W Y hh p^2 0 ht p ( 1 − p ) 1 1 th p ( 1 − p ) − 1 1 tt ( 1 − p )^2 0 In the following table, we write the joint PMF PW , Y ( w , y ) along with the marginal PMFs PY ( y ) and PW ( w ). PW , Y ( w , y ) w = − 1 w = 0 w = 1 PY ( y ) y = 0 0 ( 1 − p )^2 0 ( 1 − p )^2 y = 1 p ( 1 − p ) 0 p ( 1 − p ) 2 p ( 1 − p ) y = 2 0 p^2 0 p^2 PW ( w ) p ( 1 − p ) 1 − 2 p + 2 p^2 p ( 1 − p ) Using the definition PW | Y ( w | y ) = PW , Y ( w , y ) / PY ( y ), we can find the conditional PMFs of W given Y.
PW | Y ( w | 0 ) =
1 w = 0 0 otherwise PW | Y ( w | 1 ) =
1 / 2 w = − 1 , 1 0 otherwise
PW | Y ( w | 2 ) =
1 w = 0 0 otherwise Similarly, the conditional PMFs of Y given W are
PY | W ( y | − 1 ) =
1 y = 1 0 otherwise PY | W ( y | 0 ) =
( 1 − p )^2 1 − 2 p + 2 p^2 y^ =^0 p^2 1 − 2 p + 2 p^2 y^ =^2 0 otherwise
PY | W ( y | 1 ) =
1 y = 1 0 otherwise
Problem 3.6. (a) First we observe that A takes on the values SA = {− 1 , 1 } while B takes on values from SB = { 0 , 1 }. To construct a table describing PA , B ( a , b ) we build a table for all possible values of pairs ( A , B ). The general form of the entries is PA , B ( a , b ) b = 0 b = 1 a = − 1 PB | A ( 0 | − 1 ) PA (− 1 ) PB | A ( 1 | − 1 ) PA (− 1 ) a = 1 PB | A ( 0 | 1 ) PA ( 1 ) PB | A ( 1 | 1 ) PA ( 1 ) Now we fill in the entries using the conditional PMFs PB | A ( b | a ) and the marginal PMF PA ( a ). This yields PA , B ( a , b ) b = 0 b = 1 a = − 1 ( 1 / 3 )( 1 / 3 ) ( 2 / 3 )( 1 / 3 ) a = 1 ( 1 / 2 )( 2 / 3 ) ( 1 / 2 )( 2 / 3 )
which simplifies to
PA , B ( a , b ) b = 0 b = 1 a = − 1 1 / 9 2 / 9 a = 1 1 / 3 1 / 3
(b) If A = 1, then the conditional expectation of B is
E [ B | A = 1 ] =
1
b = 0
bPB | A ( b | 1 ) = PB | A ( 1 | 1 ) = 1 / 2
(c) Before finding the conditional PMF PA | B ( a | 1 ), we first sum the columns of the joint PMF table to find
PB ( b ) =
4 / 9 b = 0 5 / 9 b = 1
The conditional PMF of A given B = 1 is
PA | B ( a | 1 ) =
PA , B ( a , 1 ) PB ( 1 )
2 / 5 a = − 1 3 / 5 a = 1
(d) Now that we have the conditional PMF PA | B ( a | 1 ), calculating conditional expectations is easy.
E [ A | B = 1 ] = ∑
a =− 1 , 1
aPA | B ( a | 1 ) = − 1 ( 2 / 5 ) + ( 3 / 5 ) = 1 / 5
E
[
A^2 | B = 1
]
a =− 1 , 1
a^2 PA | B ( a | 1 ) = 2 / 5 + 3 / 5 = 1
The conditional variance is then
Var [ A | B = 1 ] = E
[
A^2 | B = 1
]
− ( E [ A | B = 1 ])^2 = 1 − ( 1 / 5 )^2 = 24 / 25
(e) To calculate the covariance, we need
E [ A ] = ∑
a =− 1 , 1
aPA ( a ) = − 1 ( 1 / 3 ) + 1 ( 2 / 3 ) = 1 / 3
E [ B ] =
1
b = 0
bPB ( b ) = 0 ( 4 / 9 ) + 1 ( 5 / 9 ) = 5 / 9
E [ AB ] = ∑
a =− 1 , 1
1
b = 0
abPA , B ( a , b )
= − 1 ( 0 )( 1 / 9 ) + − 1 ( 1 )( 2 / 9 ) + 1 ( 0 )( 1 / 3 ) + 1 ( 1 )( 1 / 3 ) = 1 / 9
The covariance is just
Cov [ A , B ] = E [ AB ] − E [ A ] E [ B ] = 1 / 9 − ( 1 / 3 )( 5 / 9 ) = − 2 / 27
