Download E(3 + 6X)

Statistics for PP&A
Spring 2004 Problem Set #1 Suggested Answers
Stock and Watson Exercises 2.2, 2.3, 2.5.d, 2.6.b, 2.8
#" $%
) *',
+
&(' ) Total
2.2 Using the random variables
and from Table 2.2,
and
. Compute (a)
and
corr
.
Rain (
) No Rain (
Long Commute
0.15
Short Commute
0.15
Total
0.30
!
consider two new random variables
; (b)
and ; and (c)
and
0.07
0.63
0.70
0.22
0.78
1.00
'- !.0/1',
a It might be helpful to give
and interpretations. For example, you could think of as the
if the commute
cost of a cab ride contingent on the length of the commute: is
is short and
if the commute is long.
The answer uses the mean of a linear function of a random variable (end of section 2.2).
%
2.3/-4
5 6 7 .
##
equation 2.12
67 9 8:
* bybecause
#;<98:
=8>
? 0 @0 @ equation 2.12
0@ 9+8:
A* bybecause
B98:
A
+
'DCE8>F
C
b The answer uses the variance of a linear function of a random variable (end of section 2.2) and
the variance of a Bernoulli random variable (section 2.2).
9 G 8>4by 9equation
2.13
H
8
*
=8>F by equation 2.7, variance for a Bernoulli variable
2.13
9I 8>byA* equation
9
>
8
4
A98JC* 4A
C by equation 2.7, variance for a Bernoulli variable
c The answer uses the definition of covariance (equation 2.22) and the probability distribution
given in the example.
1
! L KM X N Y + ON T QP
Y
Y
S
T
S
Y
R [Z N )\ ON E]B^ Z " \ UWV UWV
< ) No Rain ( %_ ) Total
Rain (
)
Long Commute ( 0.15
0.07 0.22
`'D )
Short Commute ( 0.15
0.63 0.78
Total
0.70 1.00
N =8> and N ('aCE8:FbC .0.30
Recall that Y
T
Y
Y
X
T
Y
S
S
! UWR V UWV )Z N )\ N E]B^ Z " \ 0.=8>* 3c'DC98:F
C* =8M',F because 0.15 of the time and 0.=8:
* '-3c'DCE8>F
C1 98d',F4 because 0.15 of the time % and `'-
_0.=8:
* 3c'DCE8>F
C1 98H*4 because 0.07 of the time %_ and _0.=8:
* '-3c'DCE8>F
C1 98H4* because 0.63 of the time %_ and e98>F
With covariance in hand, correlation is easy to compute.
#" B ! e98>F*f hg i8:F g A98JC*4A
C*;(98JC4C
and are negatively correlated? and are positively
Is there any intuition for why
j
increases with ,
correlated: short commutes are associated with no rain. Because
klm decreases with , it’s not surprising that and are negatively correlated.
and is up and is down. Is this consistent
When and are both up, which happens frequently,
and ?
with your interpretation of
corr
2.3 I covered this in some detail in class.
n F="o]B4^
B] ^ qp pmA4 .
+ r V ptsu]Bp ^ r p whenp Z is a standard normal variable. Can we
]B^ v p p(A*
+8Jw CEV ' w type problem into this problem? The
4
'
I
.
g
, find
2.5.d If Y is distributed
We know how to compute
convert
, an example of a
standard deviation of ,
is
, or about
2
r V + w0 V .OF4N yfI xf I
z 9 8>=' g
r +wA0 .OF4N yfI xf I
z = 8d', g
]B^ =8:='qp{s|p{=8d',4 and we can turn to the standard
So we have converted the problem into normal tables.
]B^ 98>='2p<sp<=8d',4B];^ sp=8d',4}@]B^ spm=8:='-B98>4='-A
F="4 will fall between 6 and 8.
Around 22 percent of the time, (~nO
]B^ p=8:
A* .
2.6.b If Y is distributed €  , find )
This problem is a direct look-up from the € table and takes advantage of the fact that 7.78
happens to be a critical value. Ten percent of the time, ‚~u€  will exceed 7.78; so 90 percent
p<=8>A .
%ƒ
2.8 In any year, the weather can inflict storm damage to a home. From year to year, the damage
is random. Let denote the dollar value of damage in any given year. Suppose that in 95 percent
of the years,
, but in 5 percent of the years
.
ƒ49"„44
+ + ? … 9QVH8 _*†1F V ‡ƒ
*… 7† =8>*F ƒ4="o44*
ƒ='4"o 4 ) y P
The variance of the damage is Kd+
KM) + x Pˆ … 9V8H_* †iF V ƒ
3@)ƒ= '4x"o 4‡* … . † 9 8>4 F ƒ4
) 9"ox4 4 6.ƒ='4"o4*
'-_9"o 49"o4
a What is the mean and standard deviation of the damage in any year?
The mean of the damage is
.
The standard deviation of the damage is the square root of the variance.
I Š‰
var
+ ; ‰ '-_9"o49"o44 z ƒ
CE"o*F
A98>_
3
b A key feature of the insurance pool is that the damage to each house is uncorrelated with
damage to any other house. Consider: homeowners on the Florida coast do not want to form an
insurance pool together.
In this case, the average damage can be thought of as adding up the total damage (
) and dividing the cost over the 100 participating households, or
/-/-/- V)‹h‹
i
V ' ST Ž T
(Œ  UWV T
Ž
'
S
T
?  UWV T
Ž
'
S
T
 ) UWV
T
T
'
< T   ) because all of the + are equal.
) T
ƒ='4"„44 (we computed ) Bƒ='4"„44 in part a)
]B^
44* . We need to know the probability distribution
We would like to compute ?
First we can compute the expected value of the average damage:
ii
for the average damage.
Is it possible that this is a normal distribution problem? is a mean of random variables, and
the Central Limit Theorem states that means of random variables are normally distributed. So is
normally distributed with mean
and variance . Before we worry about these values, let’s
convert the normal problem into a standard normal problem.
I
]B^6‘ ’ 44
“ with `~n+ " I is the same as
];^!”)s  443 with s ~nO ="D',
I •
" I ,
Again, don’t worry about and variance I yet. We know that if –~–n+ then a standardization of (by subtracting the mean and dividing by the standard deviation) is
9"D'- .
distributed n
e’ƒi'4"o44 in part
Now, let’s fill in the pieces that we need. We have already computed b.i.
The tricky part is I , but the Central Limit Theorem gives us guidance. Because is an
f
‰ f
average of i.i.d. random variables, its variance is I  . The standard deviation of is I  .
C**F
A98>_*f='-2ƒ
C**F=8HA4_
We have the elements that we need: I
Before we proceed with the analytic answer, this is a good opportunity for some estimation.
The standard deviation of average damage is slightly under $500. How often will average damage
4
 4ƒ F
4 )? This should
]B^ ” s  44C43*F=L8HA4'-_ 4 %]B^ s  =8>_*B`'—]B^ s˜m=8>_*B=8>9'D4_98
•
T
Indeed, this is a very rare event: average damage exceeds $2,000 around 1 percent of the time.
(Consider how important independence of each was for this result.)
exceeds its average ($1,000) by $1,000, or about two standard deviations (
not happen often, definitely less than 2.5 percent of the time.
5