1. The diameter of a metal cylinder is a random variable X with a probability density function given by
fX (x) = c[1 − 4(x − 50)2 ],
49.5 ≤ x ≤ 50.5.
(a) What is the value of c that makes fX (x) a probability density function? (5%)
Ans. c = 32 .
Sol)
Z
1 =
·Z
50.5
2
c
Z
50.5
[1 − 4(x − 50) ]dx = c
dx − 4
49.5
"
#
¯.5
·
¸
Z .5
4 ¯
2
c 1−4
x2 dx = c 1 − x3 ¯¯
= c
3
3
−.5
−.5
49.5
=
⇒c =
50.5
¸
(x − 50) dx
2
49.5
3
2
(b) What is the probability that the diameter X is less than or equal to 50, using your answer to part (a)?
(5%)
Ans. P (X ≤ 50) = 0.5.
Sol)
P (X ≤ 50) = 0.5
(since fX is symmetrical to X = 50.)
2. Suppose that X has a exponential distribution with p.d.f. f (x) = λe−λx , x > 0. Find the p.d.f. of Y = X 2 .
(10%)
Ans. fY (y) = λ2 y −1/2 e−λy
1/2
,
y > 0.
Sol) Fact: Y is monotonically increasing with respect to X. Let Y = g(X) = X 2 . Then, I(y) = (0, ∞), and
¯
¯
¡ −1 ¢ ¯ d −1 ¯
¯
fY (y) = fX g (y) · ¯ g (y)¯¯
dy
• g −1 (y) = y 1/2 .
= 12 y −1/2 .
¯
¯
¯ d −1 ¯ 1 −1/2
• ¯ dy
g (y)¯ = 2 y
.
¡
¢
1/2
• fX g −1 (y) = λe−λy .
•
d −1
(y)
dy g
• So,
fY (y) =
λ −1/2 −λy1/2
y
e
,
2
y > 0.
3. Let Y have a continuous probability distribution with p.d.f. fY (y) = ye−y , 0 < y < ∞; then let X have a
conditional distribution (given Y ) that is uniform on the interval from 0 to Y .
1
(a) Give the joint density function for X and Y ; be sure to give your answer correctly for all x and y. (5%)
Ans. f (x, y) = fX|Y (x|y)fY (y) = e−y ,
0 < x < y < ∞..
Sol) From the problem setting, we have
fX|Y (x|y) =
Hence,
1
,
y
0<x<y<∞
f (x, y) = fX|Y (x|y)fY (y) = e−y ,
0 < x < y < ∞.
(b) Find the marginal density functions for X and Y . (5%)
Ans. fX (x) = e−x , 0 < x < ∞ and fY (y) = ye−y , 0 < y < ∞.
Sol)
Z
fX (x)
∞
=
Zxy
fY (y)
=
¯∞
e−y dy = −e−y ¯x = e−x ,
e−y dx = ye−y ,
0 < x < ∞.
0 < y < ∞.
0
(c) Find the conditional distribution of X given Y . (5%)
Ans. fX|Y = ex−y , 0 < x < y < ∞.
Sol)
fX|Y =
f (x, y)
e−y
= −x = ex−y ,
fX (x)
e
0 < x < y < ∞.
4. The joint density function of X and Y is
½
f (x, y) =
x+y
0
0 < x < 1, 0 < y < 1
otherwise
Find P (X + Y < 1). (5%)
Ans. P (X + Y < 1) = 13 .
Sol)
¶y=1−x
µ
1
dx
xy + y 2
2
0
0
0
y=0
µ
¶x=1
Z
1 1
1
1
1
(1 − x2 )dx =
x − x3
=
2 0
2
3
3
x=0
Z
P (X + Y < 1) =
=
1
Z
Z
1−x
1
(x + y)dydx =
5. Suppose that the life distribution of an item has hazard rate function h(t) = t3 , t > 0. Find the probability that
2
(a) the item survives to age 2; (5%)
Ans. e−4 = 0.01832.
Sol) Denote the lifetime of the item as X. Then,
Z t
Z t
1
H(t) =
h(x)dx =
x3 dx = t4
4
0
0
R(t)
⇒ P (X > 2)
1 4
= e−H(t) = e− 4 t , t > 0
= R(2) = e−4 = 0.01832
(b) the item’s lifetime is between .4 and 1.4; (5%)
1
4
4
1
Ans. e− 4 ×0.4 − e− 4 ×1.4 = 0.6109.
Sol)
F (t) =
⇒ P (.4 < X < 1.4) =
1 4
1 − R(t) = 1 − e− 4 t ,
F (1.4) − F (0.4) = e
t>0
− 14 ×0.44
1
4
− e− 4 ×1.4 = 0.9936 − 0.3827 = 0.6109
(c) the 1-year-old item will survive to age 2. (5%)
Ans. e−15/4 = 0.02352.
Sol)
1
P (X > 2|X > 1) =
4
P (X > 2 and X > 1)
P (X > 2)
e− 4 ×2
=
= − 1 ×14 = e−15/4 = 0.02352
P (X > 1)
P (X > 1)
e 4
6. A randomly chosen IQ test taker obtains a score that is approximately a normal random variable with mean 100
and standard deviation 15. What is the probability that the test score of such a person is
(a) above 125; (5%)
Ans. 0.04779.
Sol) Denote the score of the test as X. Since X ∼ (100, 152 ), we have
µ
¶
125 − 100
P (X > 125) = 1 − Φ
= 1 − Φ(1.667) = 1 − 0.95221 = 0.04779
15
(b) between 90 and 110. (5%)
Ans. 0.4952.
Sol)
¶
µ
¶
110 − 100
90 − 100
−Φ
P (90 < X < 110) = Φ
15
15
= Φ(0.667) − Φ(−0.667)
= 2Φ(0.667) − 1
µ
=
2 × 0.7476 − 1 = 0.4952
7. If X1 and X2 are independent exponential random variables each having parameter λ, find the joint density
function of Y1 = X1 + X2 and Y2 = eX1 . (10%)
Ans. fY1 ,Y2 (y1 , y2 ) =
λ2 −λy1
,
y2 e
0 < ln y2 < y1 < ∞.
Sol) The joint density function of X1 and X2 is
fX1 ,X2 (x1 , x2 ) = λ2 e−λ(x1 +x2 ) ,
fY1 ,Y2 (y1 , y2 ) can be found through the following steps:
3
x1 , x2 > 0.
• Expressing X1 and X2 in terms of Y1 and Y2 , then
X1
X2
=
=
• The Jacobian matrix is then
·
J=
0
1
ln Y2
Y1 − ln Y2
1/y2
−1/y2
¸
• From the definition of Y1 and Y2 , we have
| det(J)| = 1/y2
• Then,
fY1 ,Y2 (y1 , y2 )
= | det(J)| · fX1 ,X2 (ln y2 , y1 − ln y2 )
λ2 −λy1
=
e
,
0 < ln y2 < y1 < ∞
y2
8. Suppose that the number of fishes I catch up to time t is a Poisson process Xt with rate λ = 10 fishes/hour.
(a) What is the probability that I catch no fish in the first 15 minutes? (5%)
Ans. e−2.5 = 0.08209.
Sol) Let X0.25 represent the number of fish being catched within the 0.25 hour (i.e., 15 minutes). Then,
X0.25 ∼ P (2.5). Hence,
2.50 e−2.5
P (X0.25 = 0) =
= e−2.5 = 0.08209.
0!
(b) Given that I catch 3 fishes in the time period 6:00-6:15, what is the probability I’ll catch 5 fishes in the
time interval 6:00-6:30? (5%)
Ans. 3.125e−2.5 = 0.2565.
Sol) The desired probability is the same as the probability to catch 2 fishes in time interval 6:15-6:30, i.e.,
in 15 minutes. Hence, it is
P (X0.25 = 2) =
2.52 e−2.5
= 3.125e−2.5 = 0.2565.
2!
(c) Let T10 be the time it takes for me to catch 10 fish; T10 is of course a random variable. How is T10
distributed? In particular, write down its cumulative distribution function F10 (t) expressed either as a sum
or integral. (5%)
Z
t
Ans. T10 ∼ Γ(10, 10), and F10 (t) =
0
9
X (10t)k e−10t
1010 x9 e−10x
dx = 1 −
,
9!
k!
t > 0.
k=0
Sol) It is clear that T10 ∼ Γ(10, 10). So,
f10 (t) =
Hence,
Z
t
F10 (t) =
0
1010 t9 e−10t
,
9!
t>0
1010 x9 e−10x
dx,
9!
t>0
By relating gamma distribution with Poisson distribution, we have
Z
F10 (t) =
0
t
9
X (10t)k e−10t
1010 x9 e−10x
dx = 1 −
,
9!
k!
k=0
9. Evaluate the following expressions:
4
t > 0.
Z
10
(a)
x4 e−x/2 dx =? (5%)
0
Z
10
Ans.
4 −x/2
x e
0
4!
dx =
0.55
Ã
4
X
5k e−5
1−
!
= 429.7012
k!
k=0
Sol)
Fact:
λr
(r − 1)!
Z
t
xr−1 e−λx dx
= 1−
0
Z
k=0
t
⇒
r−1 −λx
x
e
dx
=
x4 e−x/2 dx
=
0
Z
10
⇒
0
Z
∞
(b)
2
e−(x−9)
9
Z
∞
Ans.
/32
r−1
X
(λt)k e−λt
k!
Ã
!
r−1
X
(r − 1)!
(λt)k e−λt
1−
λr
k!
k=0
!
Ã
4
X
5k e−5
4!
= 429.7012
1
−
0.55
k!
k=0
dx =? (5%)
2
e−(x−9)
/32
dx =
√
8π.
9
Sol)
Fact:
1
√
2πσ
Z
∞
µ
∞
e−(x−µ)
2
/(2σ 2 )
dx
e−(x−µ)
2
/(2σ 2 )
dx
1
2
p
π/2σ
=
dx
=
Z
⇒
µ
Z
⇒
∞
=
Z
2
e−(x−9)
9
/32
∞
9
5
2
e−(x−9)
/(2×42 )
dx =
√
8π
P?@ 9
½
¾
¾
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