Download Probability HW#5

Probability HW#5
Due: May 20, 2010
Problems
5.6 (6 points)
Compute E[X] if X has a density function given by
(
1 −x/2
xe
x>0
;
(a) f (x) =
4
0
otherwise
c(1 − x2 ) −1 < x < 1
;
(b) f (x) =
0
otherwise
( 5
x>5
.
(c) f (x) =
x2
0 x≤5
Solution.
(a) Note that limx→∞ xn e−x/2 = 0 for all n ∈ N.
Z ∞
E[X] =
xf (x) dx
−∞
Z ∞
1 2 −x/2
=
xe
dx
4
0
∞ Z ∞
1 2 −x/2 −
−xe−x/2 dx
=− xe
2
0
0∞
Z ∞
1 2 −x/2 −x/2 ∞
=− xe
)0 −
−2e−x/2 dx
+ (−2xe
2
0
0
−x/2 ∞
= −4e
=4
0
(b)
E[X] =
Z
∞
xf (x) dx =
−∞
3
Z
1
−1
c(x − x3 ) dx = 0
The last equality holds since x − x is an odd function of x.
1
(c)
E[X] =
=
∞
Z
xf (x) dx
Z−∞
∞
5
5
dx = 5 ln x|∞
5 = ∞
x
5.7 (4 points)
The density function of X is given by
a + bx2 0 ≤ x ≤ 1
f (x) =
0
otherwise
If E[X] = 53 , find a and b.
Solution. First, f (x) is a probability density function.
Z
∞
f (x) dx =
−∞
Z
1
0
And we can compute E[X],
E[X] =
Z
∞
xf (x) dx =
−∞
Hence
1
b
b 3 a + bx dx = ax + x = a + = 1
3 0
3
2
1
a b
3
a 2 b 4 ax + bx dx = x + x = + =
2
4 0 2 4
5
Z
1
3a + b = 3
10a + 5b = 12
0
3
We have a = 3/5 and b = 6/5.
5.8 (4 points)
The lifetime in hours of an electronic tube is a random variable having a
probability density function given by
f (x) = xe−x ,
x≥0
Compute the expected lifetime of such a tube.
2
Solution.
Z
∞
xf (x) dx
Z−∞
∞
x2 e−x dx
0
Z
2 −x ∞
= −x e 0 −
=
∞
−2xe−x dx
0
Z ∞
∞
2 −x −x ∞
= −x e 0 + (−2xe ) 0 −
−2e−x dx
0
∞
= −2e−x 0 = 2
The expected lifetime of such a tube is 2.
5.21 (6 points)
Suppose that the height, in inches, of a 25-year-old man is a normal random
variable with parameters µ = 71 and σ 2 = 6.25. What percentage of 25year-old men are over 6 feet, 2 inches tall? What percentage of men in the
6-footer club are over 6 feet, 5 inches?
Solution. Let X be the normal random variable with parameters µ = 71 and
σ 2 = 6.25, and Z be the√
standard normal random variable. Note that 1 inch
equals to 12 feets, σ = 6.25 = 2.5. Then the first problem is to compute
P (X > 6 · 12 + 2).
X − 71
74 − 71
P (X > 74) = P
= P (Z > 1.2) = 1 − Φ(1.2) ≈ 0.1151
>
2.5
2.5
For the second problem, we need to compute
P (X > 6 · 12 + 5|X > 6 · 12) = P (X > 77|X > 72)
Hence,
P (X > 77)
P (X > 72)
P (Z > 2.4)
=
P (Z > 0.4)
1 − Φ(2.4)
1 − P (Z < 2.4)
=
≈ 0.0238
=
1 − P (Z < 0.4)
1 − Φ(0.4)
P (X > 77|X > 72) =
3
5.24 (4 points)
The lifetimes of interactive computer chips produced by a certain semiconductor manufacturer are normally distributed with parameters µ = 1.4 × 106
hours and σ = 3 × 105 hours. What is the approximate probability that
a batch of 100 chips will contain at least 20 whose lifetimes are less than
1.8 × 106 .
Solution. Let X be the normal random variable with parameters µ = 1.4×106
hours and σ = 3 × 105 hours.
X − 1.4 × 106
1.8 × 106 − 1.4 × 106
6
P (X < 1.8 × 10 ) = P
<
3 × 105
3 × 105
4
4
=P Z <
=Φ
≈ 0.9088
3
3
Let N be the number of chips that will have a lifetime less than 1.8 × 106 .
Then N is a binomial random variable with parameters n = 100 and p =
0.9088. The normal approximation yields that
P (N ≥ 20) = P (N ≥ 19.5)
19.5 − (100)(0.9088)
N − (100)(0.9088)
≥p
=P p
100(0.9088)(0.0912)
100(0.9088)(0.0912)
N − 90.88
=P √
≥ −24.8
8.288256
≈ 1 − Φ(−24.8) ≈ 1
!
5.34 (6 points)
Jones figures that the total number of thousands of miles that an auto can be
driven before it would need to be junked is an exponential random variable
with parameter 1/20. Smith has a used car that he claims has been driven
only 10,000 miles. If Jones purchases the car, what is the probability that
she would get at least 20,000 additional miles out of it? Repeat under the assumption that the lifetime mileage of the car is not exponentially distributed,
but rather is (in thousands of miles) uniformly distributed over (0, 40).
Solution. Let X denote the exponetial random variable with parameter 1/20,
U denote the uniform random variable with parameters 0 and 40. The probability function of an exponential random variable with parameter 1/20 is
F (x) = 1 − e−x/20 ,
4
x≥0
The desired probability is P (X > 30|X > 10). Since exponential random
variables have memoryless property, we have
P (X > 30|X > 10) = P (X > 20) = 1 − P (X ≤ 20) = e−1 ≈ 0.3679
The probability function of an uniform random variable with parameter 0
and 40 is
1
x
F (x) =
x= ,
0 ≤ x ≤ 40
40 − 0
40
Hence
P (U > 30|U > 10) =
P (U > 30)
1 − P (U ≤ 30)
1 − 30/40
1
=
=
=
P (U > 10)
1 − P (U ≤ 10)
1 − 10/40
3
is the desired probability.
5.35 (6 points)
The lung cancer hazard rate λ(t) of a t-year-old male smoker is such that
λ(t) = .027 + .00025(t − 40)2
t ≥ 40
Assuming that a 40-year-old male smoker survives all other hazards, what
is the probability that he survives to (a) age 50 and (b) age 60 without
contracting lung cancer?
Solution. Let X denote the lifetime distribution. We can compute the probability function by
Z t
(
0 t < 40
Rt
1 − F (t) = exp
λ(u) du =
exp − 40 λ(u) du t ≥ 40
0
where
Hence
Z
λ(u) du = 0.027t +
0.00025
(t − 40)3
3
P (X ≥ 40) = 1 − F (40) = exp(−0) = 1
(a) The probability that he survives to age 50 is
P (X ≥ 50) = 1 − F (50)
50 !
0.00025
= exp − 0.027t −
(t − 40)3 3
40
= exp(−0.3533) ≈ 0.7024
5
(b) The probability that he survives to age 60 is
P (X ≥ 60) = 1 − F (60)
60 !
0.00025
= exp − 0.027t −
(t − 40)3 3
40
= exp(−1.2067) ≈ 0.2992
5.36 (6 points)
Suppose that the life distribution of an item has the hazard rate function
λ(t) = t3 , t > 0. What is the probability that
(a) the item survives to age 2?
(b) the item’s lifetime is between .4 and 1.4?
(c) a 1-year-old item will survive to age 2?
Solution. Let X denote the life distribution. First we compute the probability function of the life distribution.
Z t
F (t) = 1 − exp −
λ(u) du
0
1 4
= 1 − exp − t
4
(a) It is to compute P (X ≥ 2)
P (X ≥ 2) = 1 − P (X < 2) = 1 − F (2) = e−4 ≈ 0.0183
(b) The probability is given by
P (0.4 ≤ X ≤ 1.4) = P (X ≤ 1.4) − P (X ≤ 0.4)
= F (1.4) − F (0.4)
= (1 − e−0.9604 ) − (1 − e−0.0064 )
= e−0.0064 − e−0.9604 ≈ 0.6109
(c) The probability that a 1-year-old item will survive to age 2 is
P (X ≥ 2)
P (X ≥ 1)
1 − F (2)
=
1 − F (1)
e−4
= −1/4 = e−15/4 ≈ 0.0235
e
P (X ≥ 2|X ≥ 1) =
6
Theoretical Exercises
5.11 (9 points)
Let Z be a standard normal random variable Z, and let g be a differnetiable
function with derivative g ′.
(a) Show that E[g ′(Z)] = E[Zg(Z)]
(b) Show that E[Z n+1 ] = nE[Z n−1 ]
(c) Find E[Z 4 ].
Solution. Let f (z) denote the probability density function of Z. Hence
1
2
f (z) = √ e−z /2
2π
(a)
′
E[g (Z)] =
2 /2
∞
g ′ (z)f (z) dz
Z−∞
∞
1
2
√ e−z /2 g ′(z) dz
2π
−∞
Z ∞
1
2
e−z /2 g ′ (z) dz
=√
2π −∞
Z ∞
∞
1
−z 2 /2
−z 2 /2
−
−ze
g(z) dz
e
g(z)
=√
−∞
2π
−∞
=
If g(x) ≺ e−x
Z
, the last equation becomes
Z ∞
1
2
√
ze−z /2 g(z) dz = E[Zg(Z)]
2π −∞
Then E[g ′(Z)] = E[Zg(Z)] when g(x) ≺ e−x
(b) Let g(x) = xn ≺ e−x
2 /2
2 /2
.
, then g ′(x) = nxn−1 . By (a), we have
E[Zg(Z)] = E[Z n+1 ] = E[g ′ (Z)] = E[nZ n−1 ] = nE[Z n−1 ]
(c) By (b), we have
E[Z 4 ] = E[Z 3+1 ] = 3E[Z 3−1 ] = 3E[Z 1+1 ] = 3E[Z 1−1 ] = 3E[1] = 3
Hence E[Z 4 ] = 3.
7
5.17 (4 points)
If X has hazard rate function λX (t), compute the hazard rate function of aX
where a is a positive constant.
Solution. Since
FaX (t) = P (aX ≤ t) = P (X ≤ t/a) = FX (t/a)
We have
d
d
1
FaX (t) = FX (t/a) = fX (t/a) ·
dt
dt
a
Then the hazard rate function of aX is
faX (t) =
λaX (t) =
fX (t/a) · a1
1
faX (t)
=
= λX (t/a)
1 − FaX (t)
1 − FX (t/a)
a
5.22 (4 points)
Compute the hazard rate function of a gamma random variable with parameters (α, λ) and show it is increasing when α ≥ 1 and decreasing when
α ≤ 1.
Solution. Let X be a gamma random variable with parameters (α, λ). Then
( −λx α−1
λe
(λx)
x≥0
Γ(α)
f (x) =
0
x<0
Hence the hazard function is
f (t)
λX (t) =
= R∞
1 − F (t)
t
λe−λt (λt)α−1
Γ(α)
λe−λt (λt)α−1
Γ(α)
dx
We can simplify it, then we have
λX (t) = R ∞
t
0
t
e−λt (λt)α−1
e−λx (λx)α−1 dx
1
e−λ(x−t)
Let y = x − t, then we have
λX (t) = R ∞
= R∞
x α−1
t
dx
y α−1
t
dy
1
e−λy 1 +
8
If α − 1 ≥ 0, for t1 ≤ t2 , y > 0, we have
α−1 α−1
y
y
1+
≥ 1+
t1
t2
Hence
λX (t1 ) = R
∞
0
≤R
∞
0
1
e−λy 1 +
1
e−λy 1 +
y
t1
y
t2
α−1
dy
α−1
dy
= λX (t2 )
That is, t1 ≤ t2 implies λX (t1 ) ≤ λX (t2 ) when α − 1 ≥ 0. λX (t) is increasing
when α ≥ 1. Similarly, λX (t) is decreasing when α ≤ 1.
5.28 (9 points)
Consider the beta distribution with parameters (a, b). Show that
(a) when a > 1 and b > 1, the density is unimodal (that is, it has a unique
mode) with mode equal to (a − 1)/(a + b − 2).
(b) when a ≤ 1, b ≤ 1, and a + b < 2, the density is either unimodal with
mode at 0 or 1 or U-shaped with modes at both 0 and 1;
(c) when a = 1 = b, all points in [0, 1] are modes.
Solution. The probability density function of a gamma distribution with parameters (a, b) is
f (x) =
1
xa−1 (1 − x)b−1
B(a, b)
0≤x≤1
Note that a mode of a probability density function occurs at local maximum.
(a) When a > 1 and b > 1
1
((a − 1)xa−2 (1 − x)b−1 − (b − 1)xa−1 (1 − x)b−2 )
B(a, b)
1
xa−2 (1 − x)b−2 ((a − 1) − (a + b − 2)x) = 0
=
B(a, b)
f ′ (x) =
9
We can attain that f ′ (x) = 0 when x = (a − 1)/(a + b − 2). Since a > 1
and b > 1, we note that
0<
a−1
a−1
a−1
=
<
=1
a+b−2
(a − 1) + (b − 1)
a−1
When 0 < x < (a−1)/(a+ b−2), 1 −x > 0 and (a−1) −(a+ b−2)x > 0,
hence
a−1
f ′ (x) > 0,
0<x<
a+b−2
′
Similarly, f (x) < 0 when (a − 1)/(a + b − 2) < x < 1.
a−1
a+b−2
0
+
ր
x
′
f (x)
f ′ (x)
0
0
1
−
ց
Hence, f (x) has a maximum at x = (a − 1)/(a + b − 2). Therefore, f (x)
is unimodal with mode equal to (a − 1)/(a + b − 2).
(b) If a = 1, then we have
f (x) =
1
(1 − x)b−1
B(1, b)
0≤x≤1
The derivative of f (x) is
f ′ (x) =
1
(1 − b)(1 − x)b−2
B(1, b)
Then f ′ (x) > 0 for 0 < x < 1. That is, f (x) is a strictly increasing
function on (0, 1). Therefore, f (x) has maximum at x = 1. Now if b = 1
,then the density function becomes
f (x) =
1
xa−1
B(a, 1)
0≤x≤1
The dreivative of f (x) is
f ′ (x) =
1
(a − 1)xa−2
B(a, 1)
Then f ′ (x) < 0 for 0 < x < 1. That is, f (x) is a strictly decreasing
function on (0, 1). Therefore, f (x) has maximum at x = 0. For a < 1
and b < 1, we have
0<
a−1
1−a
(1 − a) + (1 − b)
=
<
=1
a+b−2
(1 − a) + (1 − b)
(1 − a) + (1 − b)
10
When 0 < x < (a−1)/(a+ b−2), 1 −x > 0 and (a−1) −(a+ b−2)x < 0,
hence
a−1
f ′ (x) < 0,
0<x<
a+b−2
′
Similarly, f (x) > 0 when (a − 1)/(a + b − 2) < x < 1.
x
f (x)
f ′ (x)
′
a−1
a+b−2
0
−
ց
0
0
1
+
ր
Hence, f (x) is a U-shaped when a < 1 and b < 1. If f (0) 6= f (1), then
f (x) has mode at x = 0 or x = 1. If f (0) = f (1), then f (x) is U-shaped
with mode at both 0 and 1.
(c) When a = 1 = b, since B(1, 1) = 1, the probability density function is
f (x) =
1
= 1,
B(1, 1)
0<x<1
That is, f (x) is a contant function. Then all points in [0, 1] are modes.
11