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AMS 312 Practice Test 2 – Solutions
1. Let the joint density function of X and Y be given by
x  y,0  x  1,0  y  1
f ( x , y)  
elsewhere
0,
What is P[Y<2X]?
Solution:
y
1
y=2x
1
1 1
P[Y<2X] =

0 y/2
1
=
x
( x  y)dxdy
1
5
 (2  y  8 y
0
2
)dy
= 19/24
2. Suppose that X is a random variable with density function f(x) =3x2/8 for 0<x<2, and zero
otherwise. Let Y=X2. What is the density function of Y, where nonzero?
Solution:
Since 0<x<2, so 0<Y=X2<4. In this interval
FY(y) = P(Yy) = P(X2y) = P(
So fY(y) = dFY(y)/dy =
y X y ) = P(0<X y ) =

0
y
3 2
y3/ 2
x dx =
8
8
3 y
, where 0<y<4.
16
3. 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 first, and what is its expected life?
Solution:
Let X1 … X10 denote the life of the 10 bulbs respectively. Let Y denote the life of the bulb
that fails first. Then X1 … X10 are i.i.d. Exp(1/100) and Y = min { X1 … X10 }.
FY(y) = P(Yy) = 1 – P(Y>y) = 1 – P(min { X1 … X10 }>y)
= 1 – P( X1 >y , …, X10 >y)
= 1 – P( X1 >y) …P( X10 >y)
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AMS 312 Practice Test 2 – Solutions
= 1 – (e
=1– e


y
100 10
)
y
10
So Y~Exp(1/10). The expected life is 10 hours.
4. Let X1, X2 be a random sample from N(0,1).
(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-X2)/
2 and (X1+X2)/ 2 ? Prove it.
Solution:
(a). (X1-X2)/
2 ~N(0,1).
M(t) = E[exp(t(X1-X2)/
= E[exp(X1 t/
=
e(t /
=
et
2
2 )] = E[exp(tX1/ 2 ) exp(-tX2/ 2 )]
2 )] E{exp[X2 (-t/ 2 )]}
2 )2 / 2 (  t / 2 )2 / 2
e
/2
2 ~N(0,1).
So (X1-X2)/
(b). (X1+X2)/
2 ~N(0,1).
M(t) = E[exp(t(X1+X2)/
= E[exp(X1 t/
=
=
e(t /
et
2
2
2 )] = E[exp(tX1/ 2 ) exp(tX2/ 2 )]
2 )] E{exp[X2 (t/ 2 )]}
2 ) / 2 ( t / 2 )2 / 2
e
/2
So (X1+X2)/
2 ~N(0,1).
(c). The joint moment generating function of (X1-X2)/
M( t1 , t2 )= E[exp( t1
X1  X 2
X  X2
)]
 t2 1
2
2
= E[exp(
t1  t2
t t
X 1  2 1 X 2 )]
2
2
= E[exp(
t1  t2
t t
X 1 )] E[exp( 2 1 X 2 )]
2
2
Page 2 of 4
2 and (X1+X2)/ 2 is:
AMS 312 Practice Test 2 – Solutions
 1  t1  t2 2 
 1  t2  t1 2 
1 2 2 
exp

 
 =exp   t1  t2  



2

 2  2  
 2  2  
= exp 
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 the joint distribution is BN

1
 2  0,  12   22  1;   0  . Equivalently, we can
say that these two random variables are iid N(0,1).
5. The marginal density of X is ( x+1/2) and the conditional density of Y given X is
(x+y)/(x+1/2)
for 0<x<1, and 0<y<1. Find the marginal distribution of Y.
Solution:
f X (x)  x 
1
2
f Y| X ( y | x ) 
xy
x  1/ 2
So f XY ( x , y)  f Y|X ( y | x )f X ( x )  x  y , 0<x, y<1.
1
f Y ( y)   ( x  y)dx  y 
0
1
, 0<y<1, and 0 elsewhere.
2
6. A man and a woman decide to meet at a certain location. If each person independently
arrives at a time uniformly distributed between 12 noon and 1 P.M., find the probability
that the first to arrive has to wait longer than 10 minutes.
Solution:
Let X and Y denote the time they arrive respectively, counting the minute elapsing from
12:00 PM. So X and Y are i.i.d., uniform R.V.’s on interval [0, 60], and their density is
1/60 on [0, 60] and 0 elsewhere. Their joint density is therefore uniform with
f XY ( x, y )  1/3600, 0  x, y  60. As shown below the shaded area is |X-Y|>10 in the
square [0, 60] by [0, 60].
Y
60
10
10
60
Page 3 of 4
X
AMS 312 Practice Test 2 – Solutions
Since the joint density is uniform over the square, so the fraction of area indicates the
probability. The fraction is 502/602=25/36. So the probability that the first to arrive has to
wait longer than 10 minutes is 25/36.
Page 4 of 4