Winter 06
1
Econometrics
ASSIGNMENT 1  SOLUTIONS
100 POINTS TOTAL
DUE: Thursday, January 26, 3:20 p.m.
1.
[5]
Let X be the random variable distributed as Normal (5,4). Find the probabilities of the following events:
(i)
P(
X
≤
6)
(ii)
P(
X
> 4)
(iii)
P(
X
– 5 > 1)
(i) P(
X
≤
6) = P[(
X
– 5)/2
≤
(6 – 5)/2] = P(
Z
≤
.5)
≈
.692, where
Z
denotes a Normal (0,1) random variable.
[We obtain P(
Z
≤
.5) from Table 1.]
(ii) P(
X
> 4) = P[(
X
– 5)/2 > (4 – 5)/2] = P(
Z
>
−
.5) = P(
Z
≤
.5)
≈
.692.
(iii) P(
X
– 5 > 1) = P(
X
– 5 > 1) + P(
X
– 5 < –1) = P(
X
> 6) + P(
X
< 4)
≈
(1 – .692) + (1 – .692) = .616,
where we have used answers from parts (i) and (ii).
2.
[5]
Let
X
denote the prison sentence, in years, for people convicted of auto theft in California. Suppose that the
pdf
of
X
is given by
2
1
9
() () ,0
3
.
=
<<
fx
x
x
Use integration to find the expected prison sentence. [
hint
: requires calculus.]
E(
X
)
=
3
0
()
x
fxd
x
∫
=
3
0
x
x
∫
= (1/9)
3
3
0
x dx
∫
.
But
3
3
0
x dx
∫
= (1/4)
x
4
3
0

= 81/4.
Therefore, E(
X
) =
(1/9)(81/4) = 9/4, or 2.25 years.
3.[5]
Let
X
denote the annual salary on university professors in the United States, measured in
thousands of dollars
.
Suppose that the average salary is 52.3, with a standard deviation of 14.6. Find the mean and standard deviation
when salary is measured in
dollars
.
If
Y
is salary in dollars then
Y
= 1000
⋅
X
, and so the expected value of
Y
is 1,000 times the expected value of
X
, and
the standard deviation of
Y
is 1,000 times the standard deviation of
X
.
Therefore, the expected value and standard
deviation of salary, measured in dollars, are $52,300 and $14,600, respectively.
4.
[20]
Suppose a fair coin is tossed 3 times and let a random variable X denote the number of heads. Define Y, a
Bernoulli random variable, as Y=1 if the first tossed coin turns up head, and Y=0 otherwise.
(i) Determine and sketch the pdf of X and cdf of X.
(ii) Determine the joint probability density function of (X, Y)
(iii) Are random variables X and Y independent? Explain.
(iv) Compute P(X=2Y=1), Cov(X,Y) and Corr(X, Y)