Solutions to AMS312.1 Test 2
1. A certain college gives aptitude tests in the science and the humanities to all entering
freshmen. If X and Y are, respectively, the proportions of correct answers that a student
gets on the tests in the two subjects, the joint probability distribution of these random
variables can be approximated with the joint probability density
2
 (2 x  3y), 0  x  1, 0  y  1
f ( x , y)   5
0
elsewhere
What is the probability that a student will get a total score of less than 0.80 from both
tests? (Or equivalently, what proportion of the students will get a total score of less than
0.80 from both tests?)
Solution:
P(X+Y<0.80) =
2
(2 x  3y)dxdy
5
x  y  0 .8

0 x , y 1
1 0.8 x
2
(2 x  3y)dydx
5
=

=
 (0.384  0.32x  0.2x
0 0
1
0
1
48
8
1
 (125  25 x  5 x
or:
0
2
2
)dx
)dx
= 64/375
or: 0.1707
2. Suppose that the life of a certain light bulb is exponentially distributed with mean 100
hours. If 10 such light bulbs are installed simultaneously, what is the distribution of the life
of the light bulb that fails last? Please show the entire derivation.
Solution:
Let X1 … X10 denote the life of the 10 bulbs respectively. Let Y denote the life of the bulb
that fails last. Then X1 … X10 are i.i.d. Exp(1/100) and Y = max { X1 … X10 }.
FY(y) = P(Y  y) = P(max { X1 … X10 }  y)
= P( X1  y , …, X10  y)
= P( X1  y) …P( X10  y)
= (1 – e

y
100 10
)
And the PDF is
y
fY(y) =
y

1 100
e (1  e 100 ) 9
10
Page 1 of 4
Solutions to AMS312.1 Test 2
Where y0.
3. Let X1~N(1=100, 12=9), X2~N(2=200, 22=16), and the two random variables are
independent to each other.
(a). What is the distribution of (X1-X2)/2? Prove it.
(b). What is the distribution of (X1+X2)/2? Prove it.
(c). What is the joint distribution of X1 and X2? Prove it.
(d). What is the joint distribution of (X1-X2)/2 and (X1+X2)/2? Prove it.
Solution:
(a). (X1-X2)/2 ~N(=-50, 2=25/4).
M(t) = E[exp(t(X1-X2)/2)] = E[exp(tX1/2) exp(-tX2/2)]
= E[exp(X1 t/2)] E{exp[X2 (-t/2)]}
= MX1(t/2) MX2(-t/2)
= e
= e
t 9 t
100  ( ) 2
2 2 2
50 t 
e
t 16 t
200(  )  (  ) 2
2 2 2
25 2
t
8
So (X1-X2)/2~N(-50, 25/4).
(b). (X1+X2)/2~N(150,25/4).
M(t) = E[exp(t(X1+X2)/2)] = E[exp(tX1/2) exp(tX2/2)]
= E[exp(X1 t/2)] E{exp[X2 (t/2)]}
= MX1(t/2) MX2(-t/2)
= e
= e
t 9 t
100  ( ) 2
2 2 2
150t 
e
t 16 t
200( )  ( ) 2
2 2 2
25 2
t
8
So (X1+X2)/2~N(150,25/4).
(c). X1 and X2 are independent normal random variables. So the joint moment generating
function is M( t 1 , t 2 )=MX1(t1) MX2(t2) and thus it can be shown easily that the joint
distribution is BN(1=100, 2=200, 12=9, 22=16, =0).
Equivalently, one can also derive the joint density function f(x1,x2)=fX1(x1) fX2(x2).
(d). The joint moment generating function of (X 1-X2)/2 and (X1+X2)/2 is:
M( t 1 , t 2 )= E[exp( t 1
X1  X 2
X  X2
)]
 t2 1
2
2
Page 2 of 4
Solutions to AMS312.1 Test 2
= E[exp( X1
t1  t 2
t t
 X 2 2 1 )]
2
2
= E[exp( X1
t1  t 2
t  t1
)] E[exp( X 2 2
)]
2
2
= exp( 100
t1  t 2 9 t1  t 2 2
t  t 16 t  t
 (
) )exp( 200 2 1  ( 2 1 ) 2 )
2
2
2
2
2
2
=exp(  50 t 1  150 t 2 
25 2 25 2 7
t1 
t 2  t1t 2 )
8
8
4
Recall the joint mgf of a bivariate normal RV is
M( t1 , t2 )= exp
1 22



t


t

 1 t1   22t22  2  1 2t1t2   .

1
1
2
2

2

So the joint distribution is BN(1=-50, 2=150, 12=22 =25/4, =-7/25).
4. A man and a woman decide to meet at a certain location. If the man arrives at t time
uniformly distributed between 12 noon and 12:30 PM, and the woman arrives,
independently, at a time uniformly distributed between 12 noon and 12:40 PM, find the
probability that the man has to wait longer than 10 minutes.
Solution:
Let X and Y denote the time the man and the woman arrives respectively, counting the
minute elapsing from 12:00 PM. So X~U[0, 30], Y~U[0, 40]. Since they are independent,
their joint density is therefore uniform on the range [0,30][0,40]. The event that the man
has to wait longer than 10 minutes is equivalent to the event that Y – X>10, as shown in
the shaded area below.
Y
40
10
30
X
Since the joint density is uniform, the fraction of area indicates the probability. The
fraction is 302/2/(30*40)=3/8. So the probability that man has to wait longer than 10
minutes is 3/8.
Page 3 of 4
Solutions to AMS312.1 Test 2
5. (Extra Credit) For a random sample X1, X2, …, Xn from a normal population N((,
can be shown that the sample mean
n
S2 
 (X
i 1
i
2), it
X and the sample variance
 X) 2
n 1
are indeed, independent. Please prove this for the simple case of n=2. That is, when we
have a sample of size 2.
Proof:
When n=2, we have X1, X2 only and
X
X1  X 2
2
2
S 
2
 (X
i 1
i
 X) 2
(X 1  X 2 ) 2
X  X2 2

 2( 1
)
2
2
2 1
Let’s consider the joint moment generating function of (X1-X2)/2 and (X1+X2)/2:
M( t1 , t2 )= E[exp( t 1
X1  X 2
X  X2
)]
 t2 1
2
2
= E[exp( X1
t1  t 2
t t
 X 2 2 1 )]
2
2
= E[exp( X1
t1  t 2
t  t1
)] E[exp( X 2 2
)]
2
2
= exp( 
t1  t 2  2 t1  t 2 2
t  t1  2 t 2  t1 2

(
) )exp(  2

(
) )
2
2
2
2
2
2
=exp( t 2

2
4
( t 1  t 2 ) 2 ) (Roy, please delete the last square – my pc keeps on
2
2
dying when I use the math editor –Wei)
Recall the joint mgf of a bivariate normal RV is
M( t1 , t2 )= (exp
1 22


2 2
 1t1  2t2  2  1 t1   2 t2  2  1 2t1t2   .
So (X1-X2)/2 and (X1+X2)/2 follow bivariate normal distribution and the correlation
coefficient is 0, which implied that they are independent. Since S2 is a function of (X1X2)/2, so it is of course independent from
X
Page 4 of 4
X1  X 2
too.
2