(a)
x
~ Normal[0,3
2
], and 4 <
x
< 4.
(b)
x
~ chisquared, 8 degrees of freedom, 0 <
x
< 16.
The inequality given in (318) states that Prob[
x

μ
 <
k
σ
] >
1  1/
k
2
.
Note that the result is not
informative if
k
is less than or equal to 1.
(a)
The range is 4/3 standard deviations, so the lower limit is 1  (3/4)
2
or 7/16 = .4375.
From the
standard normal table, the actual probability is 1  2Prob[
z
< 4/3] = .8175.
(b)
The mean of the distribution is 8 and the standard deviation is 4.
The range is, therefore,
μ
±
2
σ
.
The lower limit according to the inequality is 1  (1/2)
2
= .75.
The actual probability is the cumulative
chisquared(8) at 16, which is a bit larger than .95. (The actual value is .9576.)
7.
Given the following joint probability distribution,
X

0
1
2
+
0
.05
.1
.03
Y
1
.21
.11
.19
2
.08
.15
.08
(a)
Compute the following probabilities: Prob[
Y
< 2], Prob[
Y
< 2,
X
> 0], Prob[
Y
= 1,
X
>
1].
(b)
Find the marginal distributions of
X
and
Y
.
(c)
Calculate
E
[
X
],
E
[
Y
], Var[
X
], Var[
Y
], Cov[
X
,
Y
], and
E
[
X
2
Y
3
].
(d)
Calculate Cov[Y,X
2
].
(e)
What are the conditional distributions of
Y
given
X
= 2 and of
X
given
Y
> 0?
(f)
Find
E
[
Y

X
] and Var[
Y

X
].
Obtain the two parts of the variance decomposition
Var[
Y
]
=
E
x
[Var[
Y

X
]]
+
Var
x
[
E
[
Y

X
]].
We first obtain the marginal probabilities.
For the joint distribution, these will be
X:
P(0) = .34, P(1) = .36, P(2) = .30
Y:
P(0) = .18, P(1) = .51, P(2)
= .31
Then,
(a)
Prob[
Y
< 2] = .18 + .51 = .69.
Prob[
Y
< 2,
X
> 0] = .1 + .03 + .11 + .19 = .43.
Prob[
Y
= 1, X $ 1] = .11 + .19 = .30.
(b)
They are shown above.
(c)
E
[
X
]
= 0(.34) + 1(.36) + 2(.30) = .96
E
[
Y
]
= 0(.18) + 1(.51) + 2(.31) = 1.13
E
[
X
2
]
= 0
2
(.34) + 1
2
(.36) + 2
2
(.30)
= 1.56
E
[
Y
2
]
= 0
2
(.18) +
1
2
(.51) + 2
2
(.31)
= 1.75
Var[
X
]
=
1.56  .96
2
= .6384
Var[
Y
]
=
1.75  1.13
2
= .4731
E
[
XY
]
=
1(1)(.11)+1(2)(.15)+2(1)(.19)+2(2)(.08)
=
1.11
Cov[
X
,
Y
]
= 1.11  .96(1.13) = .0252
E
[
X
2
Y
3
]
= .11 + 8(.15) + 4(.19) + 32(.08) = 4.63.
(d)
E[
YX
2
]
= 1(12).11+1(22).19+2(12).15+2(22).08 = 1.81
Cov[
Y
,
X
2
]
= 1.81  1.13(1.56) = .0472.
(e)
Prob[
Y
= 0 *
X
= 2]
= .03/.3 = .1
Prob[
Y
= 1 *
X
= 2]
= .19/.3 = .633
Prob[
Y
= 1 *
X
= 2]
= .08/.3 = .267
Prob[
X
= 0 *
Y
> 0]
= (.21 + .08)/(.51 + .31) = .3537
Prob[
X
= 1 *
Y
> 0]
= (.11 + .15)/(.51 + .31) = .3171
Prob[
X
= 2 *
Y
> 0]
= (.19 + .08)/(.51 + .31) = .3292.
(f)
E
[
Y
*
X
=0]
=
0(.05/.34)+1(.21/.34)+2(.08/.34) = 1.088
E
[
Y
2
*
X
=0]
=
1
2
(.21/.34)+2
2
(.08/.34) = 1.559
Var[
Y
*
X
=0]
=
1.559  1.088
2
= .3751
E
[
Y
*
X
=1]
=
0(.1/.36)+1(.11/.36)+2(.15/.36) =
1.139
E
[
Y
2
*
X
=1]
=
1
2
(.11/.36)+2
2
(.15/.36) = 1.972
Var[
Y
*
X
=1]
=
1.972  1.139
2
= .6749
E
[
Y
*
X
=2]
=
0(.03/.30)+1(.19/.30)+2(.08/.30) =
1.167
163